NEET 2023 Question Paper with Answers and Solutions PDF H4 in English

Step 1: Understanding the Question:

We are given the torque experienced by an electric dipole in a uniform electric field, along with the field strength, the angle, and the dipole length. We need to calculate the magnitude of the charge on the dipole.


Step 2: Key Formula or Approach:

The torque (\(\tau\)) on an electric dipole in an electric field (\(E\)) is given by:
\[ \tau = pE\sin\theta \]
where \(p\) is the magnitude of the electric dipole moment and \(\theta\) is the angle between the dipole moment vector and the electric field vector.

The electric dipole moment \(p\) is defined as the product of the magnitude of one of the charges (\(q\)) and the separation between the charges (\(d\), or dipole length):
\[ p = qd \]

Step 3: Detailed Explanation:

Given values:

Angle, \(\theta = 30^\circ\).

Electric field, \(E = 2 \times 10^5\) N/C.

Torque, \(\tau = 4\) Nm.

Dipole length, \(d = 2\) cm = 0.02 m.


First, combine the two formulas:
\[ \tau = (qd)E\sin\theta \]
Now, rearrange the formula to solve for the charge \(q\):
\[ q = \frac{\tau}{dE\sin\theta} \]
Substitute the given values:
\[ q = \frac{4}{(0.02) \times (2 \times 10^5) \times \sin(30^\circ)} \]
We know that \(\sin(30^\circ) = 0.5\).
\[ q = \frac{4}{(0.02) \times (2 \times 10^5) \times 0.5} \] \[ q = \frac{4}{(0.04 \times 10^5) \times 0.5} \] \[ q = \frac{4}{0.02 \times 10^5} = \frac{4}{2 \times 10^3} \] \[ q = 2 \times 10^{-3} \, C \]
Since \(1 \, mC = 10^{-3} \, C\), the charge is:
\[ q = 2 \, mC \]

Step 4: Final Answer:

The magnitude of the charge on the dipole is 2 mC.
Quick Tip: Always ensure your units are consistent before calculation. In this problem, the dipole length was given in cm and needed to be converted to meters to match the SI units of other quantities (Nm, N/C).

Step 1: Understanding the Question:

We are asked to calculate the maximum vertical height reached by a projectile, given its initial speed and launch angle.


Step 2: Key Formula or Approach:

The motion can be analyzed by separating it into horizontal and vertical components. The maximum height is reached when the vertical component of the velocity becomes zero.

The initial vertical velocity is \(u_y = u \sin\theta\).

Using the kinematic equation \(v_y^2 = u_y^2 + 2a_y s_y\):

At maximum height (\(H\)), the final vertical velocity \(v_y = 0\). The acceleration is \(a_y = -g\). The vertical displacement is \(s_y = H\).
\[ 0^2 = (u\sin\theta)^2 + 2(-g)H \] \[ 2gH = u^2 \sin^2\theta \] \[ H = \frac{u^2 \sin^2\theta}{2g} \]

Step 3: Detailed Explanation:

Given values:

Initial speed, \(u = 280\) m/s.

Launch angle, \(\theta = 30^\circ\).

Acceleration due to gravity, \(g = 9.8\) m/s\(^2\).

We are also given \(\sin 30^\circ = 0.5\).


Substitute these values into the formula for maximum height:
\[ H = \frac{(280)^2 \times (\sin 30^\circ)^2}{2 \times 9.8} \] \[ H = \frac{(280)^2 \times (0.5)^2}{19.6} \] \[ H = \frac{78400 \times 0.25}{19.6} \] \[ H = \frac{19600}{19.6} \] \[ H = 1000 \, m \]

Step 4: Final Answer:

The maximum height attained by the bullet is 1000 m.
Quick Tip: Memorize the standard formulas for projectile motion:
- Maximum Height: \(H = \frac{u^2 \sin^2\theta}{2g}\)
- Time of Flight: \(T = \frac{2u \sin\theta}{g}\)
- Range: \(R = \frac{u^2 \sin(2\theta)}{g}\)
Knowing these formulas saves valuable time during exams.

Step 1: Understanding the Question:

We need to calculate the work done (or energy required) to create a soap bubble of a given radius. The surface tension of the soap solution is provided.


Step 2: Key Formula or Approach:

The energy required to form a liquid surface is equal to the product of the surface tension (S) and the increase in the surface area (\(\Delta A\)).
\[ W = S \times \Delta A \]
A crucial point for a soap bubble is that it has two surfaces: an inner surface and an outer surface. Therefore, the total surface area is twice the area of a single sphere.


Step 3: Detailed Explanation:

Given values:

Radius, \(r = 2 \, cm = 0.02 \, m\).

Surface tension, \(S = 0.03 \, N \, m^{-1}\).


First, calculate the total surface area of the soap bubble. Since it has two surfaces, the area is:
\[ A = 2 \times (4 \pi r^2) = 8 \pi r^2 \]
The bubble is formed from a solution, so the initial area is zero. Thus, the increase in area is \(\Delta A = A\).
\[ \Delta A = 8 \pi (0.02 \, m)^2 = 8 \pi (0.0004 \, m^2) = 0.0032 \pi \, m^2 \]

Now, calculate the energy required (work done):
\[ W = S \times \Delta A \] \[ W = 0.03 \, N \, m^{-1} \times 0.0032 \pi \, m^2 \] \[ W = 0.000096 \pi \, J \]
Using the approximation \(\pi \approx 3.14\):
\[ W \approx 0.000096 \times 3.14 \, J \approx 0.00030144 \, J \]
Expressing this in scientific notation:
\[ W \approx 3.0144 \times 10^{-4} \, J \]
This is approximately \(3.01 \times 10^{-4} \, J\).


Step 4: Final Answer:

The amount of energy required is nearly \(3.01 \times 10^{-4} \, J\).
Quick Tip: A common mistake is forgetting that a soap bubble has two surfaces. A liquid drop in air has only one surface. Always check if the object is a bubble (two surfaces) or a drop (one surface) when dealing with surface tension energy problems.

Step 1: Understanding the Question:

We are given the half-life of a radioactive substance and asked to find the total time it takes for its activity to decrease to 1/16th of its original activity.


Step 2: Key Formula or Approach:

The activity (A) of a radioactive substance after a certain time is related to its initial activity (\(A_0\)) by the formula:
\[ A = A_0 \left(\frac{1}{2}\right)^n \]
where \(n\) is the number of half-lives that have passed. The total time elapsed (\(t\)) is related to the number of half-lives (\(n\)) and the half-life period (\(T_{1/2}\)) by:
\[ t = n \times T_{1/2} \]

Step 3: Detailed Explanation:

Given values:

Half-life, \(T_{1/2} = 20\) minutes.

The final activity is \(\frac{1}{16}\) of the initial activity, so \(\frac{A}{A_0} = \frac{1}{16}\).


First, let's find the number of half-lives, \(n\).
\[ \frac{A}{A_0} = \left(\frac{1}{2}\right)^n \] \[ \frac{1}{16} = \left(\frac{1}{2}\right)^n \]
We need to express 16 as a power of 2. Since \(2^4 = 16\), we have:
\[ \left(\frac{1}{2}\right)^4 = \left(\frac{1}{2}\right)^n \]
Therefore, the number of half-lives is \(n = 4\).


Now, calculate the total time elapsed:
\[ t = n \times T_{1/2} \] \[ t = 4 \times 20 \, minutes = 80 \, minutes \]

Step 4: Final Answer:

It will take 80 minutes for the activity to drop to 1/16th of its initial value.
Quick Tip: For fractions that are integer powers of 1/2 (like 1/2, 1/4, 1/8, 1/16, etc.), you can quickly count the number of half-lives. 1 half-life \(\rightarrow\) 1/2, 2 half-lives \(\rightarrow\) 1/4, 3 half-lives \(\rightarrow\) 1/8, 4 half-lives \(\rightarrow\) 1/16. This avoids formal calculation for simple cases.

Step 1: Understanding the Question:

We have an ideal transformer with given primary voltage, and secondary voltage and power. We need to find the current in the primary coil.


Step 2: Key Formula or Approach:

For an ideal transformer, there is no power loss. This means the power in the primary coil (\(P_p\)) is equal to the power in the secondary coil (\(P_s\)).
\[ P_p = P_s \]
The power in a coil is given by \(P = V \times I\), where V is the voltage and I is the current.

Therefore, for an ideal transformer:
\[ V_p I_p = V_s I_s = P_s \]

Step 3: Detailed Explanation:

Given values:

Secondary Voltage, \(V_s = 12\) V.

Secondary Power (power of the lamp), \(P_s = 60\) W.

Primary Voltage, \(V_p = 220\) V.


Since the transformer is ideal, the power drawn by the primary winding from the mains is equal to the power delivered by the secondary winding to the lamp.
\[ P_p = P_s = 60 \, W \]
Now we can use the power formula for the primary coil to find the primary current (\(I_p\)):
\[ P_p = V_p \times I_p \] \[ 60 \, W = 220 \, V \times I_p \]
Solve for \(I_p\):
\[ I_p = \frac{60}{220} \, A = \frac{6}{22} \, A = \frac{3}{11} \, A \]
Now, convert the fraction to a decimal:
\[ I_p \approx 0.2727... \, A \]
This is approximately 0.27 A.


Step 4: Final Answer:

The current in the primary winding is approximately 0.27 A.
Quick Tip: For ideal transformers, the core principle is \(Power_{in} = Power_{out}\). This directly relates the primary voltage and current to the output power. You don't always need to calculate the secondary current or the turns ratio if the power is known.

Step 1: Understanding the Question:

We need to evaluate two statements about the angular separation of fringes in a Young's double-slit experiment (YDSE) and determine their validity.


Step 2: Key Formula or Approach:

In YDSE, two important quantities describe the fringe pattern:

1. Fringe Width (\(\beta\)): This is the linear distance between two consecutive bright or dark fringes. It is given by \(\beta = \frac{\lambda D}{d}\), where \(\lambda\) is the wavelength of light, \(D\) is the distance between the slits and the screen, and \(d\) is the distance between the two slits.

2. Angular Separation or Angular Fringe Width (\(\theta\)): This is the angle subtended by one fringe width at the slits. For small angles, it is given by \(\theta \approx \frac{\beta}{D} = \frac{(\lambda D/d)}{D}\). This simplifies to:
\[ \theta = \frac{\lambda}{d} \]

Step 3: Detailed Explanation:

Analysis of Statement I:

"If screen is moved away from the plane of slits, angular separation of the fringes remains constant."

The formula for angular separation is \(\theta = \frac{\lambda}{d}\). This formula depends only on the wavelength (\(\lambda\)) and the slit separation (\(d\)). It does not depend on the screen distance (\(D\)). Therefore, moving the screen away (increasing D) does not change the angular separation.

Conclusion: Statement I is true.


Analysis of Statement II:

"If the monochromatic source is replaced by another monochromatic source of larger wavelength, the angular separation of fringes decreases."

Again, the formula is \(\theta = \frac{\lambda}{d}\). This shows that the angular separation \(\theta\) is directly proportional to the wavelength \(\lambda\). If the wavelength is increased (using a source of "larger wavelength"), the angular separation \(\theta\) must also increase. The statement claims it decreases.

Conclusion: Statement II is false.


Step 4: Final Answer:

Based on the analysis, Statement I is true and Statement II is false.
Quick Tip: A common point of confusion is between linear fringe width (\(\beta\)) and angular fringe width (\(\theta\)). Remember: \(\beta\) depends on the screen distance D (\(\beta \propto D\)), but \(\theta\) is independent of D. Both are directly proportional to the wavelength \(\lambda\).

Step 1: Understanding the Question:

The question asks to find the new potential energy of a spring when its stretch is increased, given its initial potential energy at a smaller stretch.


Step 2: Key Formula or Approach:

The potential energy (\(U_{sp}\)) stored in a spring is given by:
\[ U_{sp} = \frac{1}{2} k x^2 \]
where \(k\) is the spring constant and \(x\) is the extension or compression from the equilibrium position. This shows that the potential energy is directly proportional to the square of the extension (\(U_{sp} \propto x^2\)).


Step 3: Detailed Explanation:

Let the initial state be State 1 and the final state be State 2.

In State 1:

Extension, \(x_1 = 2 \, cm\).

Potential energy, \(U_1 = U\).

So, \(U = \frac{1}{2} k (2)^2 = \frac{1}{2} k \times 4 = 2k\).


In State 2:

Extension, \(x_2 = 8 \, cm\).

Potential energy, \(U_2\).

So, \(U_2 = \frac{1}{2} k (8)^2 = \frac{1}{2} k \times 64 = 32k\).


To find \(U_2\) in terms of U, we can use a ratio method.
\[ \frac{U_2}{U_1} = \frac{\frac{1}{2} k x_2^2}{\frac{1}{2} k x_1^2} = \left(\frac{x_2}{x_1}\right)^2 \]
Substituting the values:
\[ \frac{U_2}{U} = \left(\frac{8 \, cm}{2 \, cm}\right)^2 = (4)^2 = 16 \] \[ U_2 = 16U \]

Step 4: Final Answer:

The potential energy stored in the spring when stretched by 8 cm will be 16U.
Quick Tip: For problems involving changes in spring extension, using the proportionality \(U \propto x^2\) is much faster than calculating the spring constant explicitly. If the stretch is increased by a factor of 'n', the potential energy increases by a factor of 'n\(^2\)'. Here, the stretch increases by a factor of 4 (from 2 cm to 8 cm), so the energy increases by a factor of \(4^2 = 16\).

Step 1: Understanding the Question:

We are given the amplitude of the electric field component of an electromagnetic wave and asked to find the amplitude of the magnetic field component.


Step 2: Key Formula or Approach:

In an electromagnetic wave traveling in a vacuum (free space), the magnitudes of the electric field (E) and magnetic field (B) at any instant are related by the speed of light (c). The same relationship holds for their amplitudes (\(E_0\) and \(B_0\)):
\[ \frac{E_0}{B_0} = c \]

Step 3: Detailed Explanation:

Given values:

Amplitude of electric field, \(E_0 = 48\) V/m.

Speed of light, \(c = 3 \times 10^8\) m/s.

The frequency information is not needed to find the magnetic field amplitude.


Rearrange the formula to solve for the amplitude of the magnetic field, \(B_0\):
\[ B_0 = \frac{E_0}{c} \]
Substitute the given values:
\[ B_0 = \frac{48 \, V/m}{3 \times 10^8 \, m/s} \] \[ B_0 = 16 \times 10^{-8} \, T \]
To express this in standard scientific notation, we can write it as:
\[ B_0 = 1.6 \times 10^1 \times 10^{-8} \, T = 1.6 \times 10^{-7} \, T \]

Step 4: Final Answer:

The amplitude of the oscillating magnetic field is \(1.6 \times 10^{-7}\) T.
Quick Tip: A simple way to remember the E, B, and c relationship is \(E = cB\). Note that the electric field value is much larger than the magnetic field value because the speed of light 'c' is a very large number. The frequency is irrelevant for this calculation, so watch out for extraneous information in problems.

Step 1: Understanding the Question:

We are given the efficiency of a Carnot engine and the temperature of its source. We need to find the temperature of the sink.


Step 2: Key Formula or Approach:

The efficiency (\(\eta\)) of a Carnot engine is given by:
\[ \eta = 1 - \frac{T_L}{T_H} \]
where \(T_L\) is the temperature of the sink and \(T_H\) is the temperature of the source. It is crucial that both temperatures are in an absolute scale (Kelvin).

The conversion from Celsius (\(T_C\)) to Kelvin (\(T_K\)) is \(T_K = T_C + 273\).


Step 3: Detailed Explanation:

Given values:

Efficiency, \(\eta = 50% = 0.5\).

Source temperature, \(T_{H(C)} = 327^\circ C\).


First, convert the source temperature to Kelvin:
\[ T_H = 327 + 273 = 600 \, K \]
Now, substitute the values into the efficiency formula:
\[ 0.5 = 1 - \frac{T_L}{600} \]
Rearrange the formula to solve for \(T_L\):
\[ \frac{T_L}{600} = 1 - 0.5 = 0.5 \] \[ T_L = 0.5 \times 600 = 300 \, K \]
The question asks for the temperature of the sink in degrees Celsius. Convert \(T_L\) back to Celsius:
\[ T_{L(C)} = T_L - 273 = 300 - 273 = 27^\circ C \]

Step 4: Final Answer:

The temperature of the sink is 27\(^\circ\) C.
Quick Tip: Always convert temperatures to Kelvin before using them in any thermodynamics formula involving ratios or absolute temperatures, such as the ideal gas law, Carnot efficiency, or Stefan-Boltzmann law. Forgetting this conversion is a very common mistake.

Step 1: Understanding the Question:

We need to determine the direction of the net force acting on a player who changes their direction of motion from south to east, while keeping their speed constant.


Step 2: Key Formula or Approach:

According to Newton's Second Law of Motion, the net force \(\vec{F}\) acting on an object is proportional to its acceleration \(\vec{a}\), and hence to the change in its velocity \(\Delta \vec{v}\).
\[ \vec{F} = m\vec{a} = m \frac{\Delta \vec{v}}{\Delta t} \]
The direction of the force is the same as the direction of the change in velocity, \(\Delta \vec{v}\). The change in velocity is calculated as \(\Delta \vec{v} = \vec{v}_f - \vec{v}_i\), where \(\vec{v}_f\) is the final velocity and \(\vec{v}_i\) is the initial velocity.


Step 3: Detailed Explanation:

Let's set up a coordinate system. Let the \(+\hat{j}\) direction be North and the \(+\hat{i}\) direction be East. Consequently, South is \(-\hat{j}\) and West is \(-\hat{i}\).

Let the speed of the player be \(v\).


Initial Velocity (\(\vec{v}_i\)):

The player is moving southward. So, \(\vec{v}_i = -v\hat{j}\).


Final Velocity (\(\vec{v}_f\)):

The player turns and moves eastward with the same speed. So, \(\vec{v}_f = v\hat{i}\).


Change in Velocity (\(\Delta \vec{v}\)):
\[ \Delta \vec{v} = \vec{v}_f - \vec{v}_i \] \[ \Delta \vec{v} = (v\hat{i}) - (-v\hat{j}) \] \[ \Delta \vec{v} = v\hat{i} + v\hat{j} \]

The direction of the force is the direction of \(\Delta \vec{v}\). The vector \(\Delta \vec{v} = v(\hat{i} + \hat{j})\) has a positive component in the East direction (\(\hat{i}\)) and a positive component in the North direction (\(\hat{j}\)). A vector with equal positive components along East and North points in the north-east direction.


Step 4: Final Answer:

The force that acts on the player is along the north-east direction.
Quick Tip: You can visualize vector subtraction \(\vec{v}_f - \vec{v}_i\) as vector addition \(\vec{v}_f + (-\vec{v}_i)\). Here, \(-\vec{v}_i\) is a vector of magnitude \(v\) pointing North. Adding a North vector to an East vector gives a resultant vector pointing North-East.

Step 1: Understanding the Question:

The question asks to find the magnitude and direction of the current in the given circuit. The circuit appears to be a single loop containing resistors and voltage sources (batteries).


Step 2: Key Formula or Approach:

We will use Kirchhoff's Voltage Law (KVL), which states that the algebraic sum of the potential differences (voltages) around any closed loop is zero. The procedure is:

1. Identify the components in the single loop.

2. Sum the resistances to find the total resistance \(R_{total}\).

3. Sum the electromotive forces (EMFs) to find the net EMF \(E_{net}\), paying attention to their polarities.

4. Calculate the current using Ohm's law for the entire circuit: \( I = \frac{E_{net}}{R_{total}} \).

5. Determine the direction of the current based on the polarity of the net EMF.


Step 3: Detailed Explanation:

1. Total Resistance: The resistors are all in series in the single loop. \[ R_{total} = 2 \, \Omega + 1 \, \Omega + 7 \, \Omega = 10 \, \Omega \]
2. Net EMF: There are two voltage sources, 10 V and 5 V. Tracing the loop, we see that their terminals are connected in opposition (positive to positive or negative to negative). The 10 V source tries to drive the current clockwise (from A to B in the top part), while the 5 V source tries to drive it counter-clockwise. The net EMF is the difference between them, and its direction is determined by the larger source. \[ E_{net} = 10 \, V - 5 \, V = 5 \, V \]
The direction of the net EMF is the same as the 10 V source.

3. Calculate Current: \[ I = \frac{E_{net}}{R_{total}} = \frac{5 \, V}{10 \, \Omega} = 0.5 \, A \]
4. Determine Direction: Since the 10 V source is stronger, it determines the direction of the current. The current will flow out of its positive terminal and into its negative terminal, meaning it flows in a clockwise direction around the loop. In the top part of the circuit (where 'E' is mentioned), this corresponds to a direction from left to right, which is from node A to node B.


Step 4: Final Answer:

The magnitude of the current is 0.5 A, and its direction is from A to B. This corresponds to option (C).
Quick Tip: When applying KVL to a loop with multiple batteries, first determine if they are aiding or opposing each other. If they are connected in series aiding (positive to negative), add their EMFs. If they are in series opposing (positive to positive), subtract the smaller EMF from the larger one. The overall current direction is set by the resulting net EMF.

Step 1: Understanding the Question:

We are given the measured values and absolute errors for the mass, radius, and length of a wire. We need to calculate the maximum percentage error in the calculated density.


Step 2: Key Formula or Approach:

1. The density (\(\rho\)) is mass (\(m\)) per unit volume (\(V\)): \(\rho = \frac{m}{V}\).

2. The wire is a cylinder, so its volume is \(V = \pi r^2 l\), where \(r\) is the radius and \(l\) is the length.

3. The formula for density is \(\rho = \frac{m}{\pi r^2 l}\).

4. The rule for propagation of errors for a quantity \(X = \frac{A^a B^b}{C^c}\) is:
\[ \frac{\Delta X}{X} = a\frac{\Delta A}{A} + b\frac{\Delta B}{B} + c\frac{\Delta C}{C} \]
For density, the maximum relative error is \(\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta r}{r} + \frac{\Delta l}{l}\). The percentage error is this value multiplied by 100.


Step 3: Detailed Explanation:

First, calculate the relative error for each measurement:

- Mass (m): \(\frac{\Delta m}{m} = \frac{0.002 \, g}{0.4 \, g} = \frac{2}{400} = 0.005\).

- Radius (r): \(\frac{\Delta r}{r} = \frac{0.001 \, mm}{0.3 \, mm} = \frac{1}{300} \approx 0.00333\).

- Length (l): \(\frac{\Delta l}{l} = \frac{0.02 \, cm}{5 \, cm} = \frac{2}{500} = 0.004\).


Now, use the error propagation formula for density. Note the power of 2 for the radius term.
\[ \frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\left(\frac{\Delta r}{r}\right) + \frac{\Delta l}{l} \] \[ \frac{\Delta \rho}{\rho} \approx 0.005 + 2(0.00333) + 0.004 \] \[ \frac{\Delta \rho}{\rho} \approx 0.005 + 0.00666 + 0.004 = 0.01566 \]
To find the percentage error, multiply by 100:
\[ Percentage Error = 0.01566 \times 100% \approx 1.566% \]
This value is nearly 1.6%.


Step 4: Final Answer:

The maximum possible percentage error in the measurement of density will be nearly 1.6%.
Quick Tip: When calculating percentage error, remember to multiply the relative error of each variable by the magnitude of its power in the formula. For density of a wire (\(\rho \propto m r^{-2} l^{-1}\)), the errors are added, and the error for radius is multiplied by 2.

Step 1: Understanding the Question:

We need to assess the correctness of two separate statements related to semiconductor devices.


Step 2: Detailed Explanation:

Analysis of Statement I:

"Photovoltaic devices can convert optical radiation into electricity."

This statement describes the fundamental principle of the photovoltaic effect. Devices like solar cells and photodiodes are designed to absorb photons (optical radiation) and generate a potential difference (voltage) or a current. This is a direct conversion of light energy into electrical energy.

Conclusion: Statement I is correct.


Analysis of Statement II:

"Zener diode is designed to operate under reverse bias in breakdown region."

A Zener diode is a special type of diode that is engineered to have a precise and sharp reverse breakdown voltage, known as the Zener voltage. Its primary application is as a voltage regulator. In these circuits, it is intentionally operated in the reverse breakdown region, where it maintains a nearly constant voltage across its terminals despite changes in current.

Conclusion: Statement II is correct.


Step 3: Final Answer:

Since both statements are factually correct descriptions of their respective devices, the correct option is that both Statement I and Statement II are correct.
Quick Tip: Remember the specific operating regions for different diodes: - Rectifier diode: Forward bias (conduction), Reverse bias (blocking). - Zener diode: Reverse bias in the breakdown region (voltage regulation). - LED: Forward bias (light emission). - Photodiode/Solar cell: Reverse bias or unbiased (light detection/power generation).

Step 1: Understanding the Question:

We need to identify the component in a full-wave rectifier circuit that is responsible for smoothing the output voltage, i.e., removing the AC ripple.


Step 2: Detailed Explanation:

Let's analyze the role of each component mentioned:

- Centre-tapped transformer and p-n junction diodes: These components together perform the rectification. The transformer steps down the AC voltage and provides two out-of-phase inputs. The diodes allow current to flow in only one direction, converting the AC input into a pulsating DC output. This output, however, still has significant voltage variations, known as AC ripple.

- Load resistance: This is the component across which the output voltage is delivered. It does not perform any filtering.

- Capacitor: A capacitor placed in parallel with the load resistance acts as a filter. It works by storing charge when the rectified voltage is increasing and releasing that charge to the load when the rectified voltage is decreasing. This process smooths out the peaks and valleys of the pulsating DC, significantly reducing the AC ripple and providing a more stable DC output. This is often called a smoothing capacitor or filter capacitor.


Step 3: Final Answer:

The capacitor is the component used to remove the AC ripple from the rectified output.
Quick Tip: In rectifier circuits, the process of converting AC to DC has two main stages: 1. \textbf{Rectification:} Using diodes to convert AC to pulsating DC. 2. \textbf{Filtering:} Using components like capacitors (or inductors) to smooth the pulsating DC into a more constant DC. The capacitor is the key element for filtering.

Step 1: Understanding the Question:

We are given the values of L, C, and R for a series LCR circuit and asked to find the resonant frequency.


Step 2: Key Formula or Approach:

Resonance in a series LCR circuit occurs when the inductive reactance (\(X_L\)) equals the capacitive reactance (\(X_C\)). The frequency at which this happens is the resonant frequency (\(f_0\)). The formula for resonant frequency is:
\[ f_0 = \frac{1}{2\pi\sqrt{LC}} \]
The angular resonant frequency (\(\omega_0\)) is \(\omega_0 = \frac{1}{\sqrt{LC}}\). Note that the resistance R does not affect the resonant frequency itself.


Step 3: Detailed Explanation:

Given values:

L = 10 mH = \(10 \times 10^{-3}\) H = \(10^{-2}\) H.

C = 1 \(\mu\)F = \(1 \times 10^{-6}\) F.

R = 100 \(\Omega\).


First, calculate the product LC:
\[ LC = (10^{-2} \, H) \times (10^{-6} \, F) = 10^{-8} \, s^2 \]
Next, calculate the square root of LC:
\[ \sqrt{LC} = \sqrt{10^{-8} \, s^2} = 10^{-4} \, s \]
Now, substitute this into the formula for the resonant frequency \(f_0\):
\[ f_0 = \frac{1}{2\pi (10^{-4})} = \frac{10^4}{2\pi} \, Hz \] \[ f_0 = \frac{10000}{2\pi} \approx \frac{10000}{2 \times 3.14159} \approx \frac{10000}{6.283} \approx 1591.5 \, Hz \]
To express this in kHz, we divide by 1000:
\[ f_0 \approx 1.59 \, kHz \]

Step 4: Final Answer:

The frequency at which resonance occurs is approximately 1.59 kHz.
Quick Tip: Remember that the resistance R in a series LCR circuit affects the "sharpness" or quality factor (Q-factor) of the resonance, but not the resonant frequency itself. The resonant frequency depends only on L and C.

Step 1: Understanding the Question:

We need to find the ratio of the radius of gyration of a solid sphere to that of a thin hollow sphere, both having the same mass M and radius R, rotating about an axis passing through their centers.


Step 2: Key Formula or Approach:

1. The moment of inertia (\(I\)) is related to the radius of gyration (\(k\)) by \(I = Mk^2\), which means \(k = \sqrt{\frac{I}{M}}\).

2. The moment of inertia of a solid sphere about its central axis is \(I_{solid} = \frac{2}{5}MR^2\).

3. The moment of inertia of a thin hollow sphere (spherical shell) about its central axis is \(I_{hollow} = \frac{2}{3}MR^2\).


Step 3: Detailed Explanation:

First, find the radius of gyration for the solid sphere (\(k_{solid}\)):
\[ k_{solid} = \sqrt{\frac{I_{solid}}{M}} = \sqrt{\frac{\frac{2}{5}MR^2}{M}} = \sqrt{\frac{2}{5}}R \]
Next, find the radius of gyration for the hollow sphere (\(k_{hollow}\)):
\[ k_{hollow} = \sqrt{\frac{I_{hollow}}{M}} = \sqrt{\frac{\frac{2}{3}MR^2}{M}} = \sqrt{\frac{2}{3}}R \]
Now, find the required ratio:
\[ \frac{k_{solid}}{k_{hollow}} = \frac{\sqrt{\frac{2}{5}}R}{\sqrt{\frac{2}{3}}R} = \sqrt{\frac{2/5}{2/3}} = \sqrt{\frac{2}{5} \times \frac{3}{2}} = \sqrt{\frac{3}{5}} \]
The exact ratio is \(\sqrt{3} : \sqrt{5}\). Quick Tip: Be aware that exam questions can sometimes be flawed. If your correct derivation leads to an answer not in the options, re-read the question carefully. If the issue persists, check if a simpler, related quantity (like the ratio of moments of inertia instead of radii of gyration) matches an option. This often reveals the intended, albeit imprecisely stated, question.

Step 1: Understanding the Question:

We are examining how the capacitive reactance and displacement current in a capacitor change when the frequency of the connected AC source decreases.


Step 2: Key Formula or Approach:

1. Capacitive Reactance (\(X_C\)): The opposition offered by a capacitor to the flow of alternating current. It is given by:
\[ X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C} \]
where \(\omega\) is the angular frequency, f is the operating frequency, and C is the capacitance.

2. Current in a Capacitive Circuit (I): According to Ohm's law for AC circuits, the current is given by \(I = \frac{V}{X_C}\), where V is the RMS voltage of the source.

3. Displacement Current (\(I_D\)): In a capacitor, the displacement current is equal to the conduction current in the connecting wires. Thus, \(I_D = I\).


Step 3: Detailed Explanation:

The problem states that the operating frequency (f) decreases.

Let's analyze the effect on capacitive reactance (\(X_C\)).

From the formula \(X_C = \frac{1}{2\pi f C}\), we can see that \(X_C\) is inversely proportional to f (\(X_C \propto \frac{1}{f}\)).

Therefore, as frequency (f) decreases, the capacitive reactance (\(X_C\)) will increase.

This eliminates options (A) and (B).


Now, let's analyze the effect on the current. The current in the circuit is \(I = \frac{V}{X_C}\).

Since \(X_C\) increases and the source voltage V is assumed to be constant, the current (I) will decrease.

The displacement current (\(I_D\)) inside the capacitor is equal to the conduction current (I) in the circuit.

Since the conduction current (I) decreases, the displacement current (\(I_D\)) also decreases.

This confirms that option (D) is correct and (C) is incorrect.


Step 4: Final Answer:

Due to a decrease in operating frequency, the capacitive reactance increases, which causes the current (and thus the displacement current) to decrease.
Quick Tip: Remember the frequency dependence for capacitors and inductors: - Capacitor: \(X_C \propto 1/f\). At low frequencies (like DC), \(X_C\) is very high (infinite), blocking the current. - Inductor: \(X_L = \omega L = 2\pi f L\), so \(X_L \propto f\). At low frequencies, \(X_L\) is low, offering little opposition.

Step 1: Understanding the Question:

The question asks for the calculation of the magnetic potential energy stored in an inductor with given inductance and current.


Step 2: Key Formula or Approach:

The energy (U) stored in an inductor is given by the formula:
\[ U = \frac{1}{2} L I^2 \]
where L is the inductance and I is the current flowing through it.


Step 3: Detailed Explanation:

We are given the following values:

Inductance, \(L = 4 \, \mu H = 4 \times 10^{-6} \, H\).

Current, \(I = 2 \, A\).


Substitute these values into the formula:
\[ U = \frac{1}{2} \times (4 \times 10^{-6} \, H) \times (2 \, A)^2 \] \[ U = \frac{1}{2} \times 4 \times 10^{-6} \times 4 \] \[ U = 2 \times 10^{-6} \times 4 \] \[ U = 8 \times 10^{-6} \, J \]
Since \(1 \, \mu J = 10^{-6} \, J\), the energy is:
\[ U = 8 \, \mu J \]

Step 4: Final Answer:

The magnetic energy stored in the inductor is 8 \(\mu\)J.
Quick Tip: Pay close attention to the units and prefixes (\(\mu\) for micro = \(10^{-6}\), m for milli = \(10^{-3}\)). This is a common source of errors. Also, remember the analogous formula for energy stored in a capacitor: \(U_C = \frac{1}{2} C V^2 = \frac{Q^2}{2C}\).

Step 1: Understanding the Question:

The question asks for the final temperature required to increase the root-mean-square (rms) speed of a gas. The phrase "increased by 3 times" is crucial to interpret correctly. It means the final speed is the initial speed plus three times the initial speed.


Step 2: Key Formula or Approach:

The rms speed (\(v_{rms}\)) of gas molecules is related to the absolute temperature (T) by the formula:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
where R is the ideal gas constant and M is the molar mass. From this, we can see the proportionality:
\[ v_{rms} \propto \sqrt{T} \]
where T must be in Kelvin.


Step 3: Detailed Explanation:

Let the initial state be 1 and the final state be 2.

Initial State (1):

Initial temperature, \(T_1 = -50^\circ C\). First, convert to Kelvin:
\[ T_1 = -50 + 273 = 223 \, K \]
Let the initial rms speed be \(v_1\).


Final State (2):

The rms speed is "increased by 3 times". This means:
\[ v_2 = v_1 + 3v_1 = 4v_1 \]
So, the final speed is 4 times the initial speed.


Now, use the proportionality \(v_{rms} \propto \sqrt{T}\):
\[ \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} \]
Substitute the known values:
\[ \frac{4v_1}{v_1} = \sqrt{\frac{T_2}{223 \, K}} \] \[ 4 = \sqrt{\frac{T_2}{223}} \]
Square both sides to solve for \(T_2\):
\[ 16 = \frac{T_2}{223} \] \[ T_2 = 16 \times 223 = 3568 \, K \]
The options are given in both K and \(^\circ\)C. Let's convert \(T_2\) to Celsius:
\[ T_{2(C)} = 3568 - 273 = 3295^\circ C \]

Step 4: Final Answer:

The gas should be heated to a temperature of 3295\(^\circ\) C.
Quick Tip: Be very careful with phrasing like "increased by X times" versus "increased to X times". "Increased by X times" means \(v_{final} = v_{initial} + X \cdot v_{initial} = (1+X)v_{initial}\). "Increased to X times" means \(v_{final} = X \cdot v_{initial}\). Here, X=3, so the factor is (1+3)=4.

Step 1: Understanding the Question:

We are given the shortest wavelength (\(\lambda\)) for the Balmer series of the hydrogen spectrum and asked to find the shortest wavelength for the Brackett series in terms of \(\lambda\).


Step 2: Key Formula or Approach:

The Rydberg formula for the wavelength of spectral lines in the hydrogen spectrum is:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]
where \(R_H\) is the Rydberg constant, \(n_f\) is the principal quantum number of the final state, and \(n_i\) is the principal quantum number of the initial state.

The shortest wavelength in any series corresponds to the maximum energy transition, which occurs when the electron transitions from \(n_i = \infty\) to the final state \(n_f\). This is also known as the series limit.


Step 3: Detailed Explanation:

For the Balmer series, the final state is \(n_f = 2\). The shortest wavelength (\(\lambda\)) occurs for the transition from \(n_i = \infty\) to \(n_f = 2\).

Using the Rydberg formula:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = R_H \left( \frac{1}{4} - 0 \right) \] \[ \frac{1}{\lambda} = \frac{R_H}{4} \quad \Rightarrow \quad R_H = \frac{4}{\lambda} \quad (Equation 1) \]

For the Brackett series, the final state is \(n_f = 4\). Let the shortest wavelength be \(\lambda_{Br}\). This occurs for the transition from \(n_i = \infty\) to \(n_f = 4\).

Using the Rydberg formula:
\[ \frac{1}{\lambda_{Br}} = R_H \left( \frac{1}{4^2} - \frac{1}{\infty^2} \right) = R_H \left( \frac{1}{16} - 0 \right) \] \[ \frac{1}{\lambda_{Br}} = \frac{R_H}{16} \quad (Equation 2) \]

Now, substitute the value of \(R_H\) from Equation 1 into Equation 2:
\[ \frac{1}{\lambda_{Br}} = \frac{1}{16} \left( \frac{4}{\lambda} \right) \] \[ \frac{1}{\lambda_{Br}} = \frac{1}{4\lambda} \] \[ \lambda_{Br} = 4\lambda \]

Step 4: Final Answer:

The shortest wavelength in the Brackett series is \(4\lambda\).
Quick Tip: Remember the final states for the first few series in the hydrogen spectrum: Lyman (\(n_f=1\)), Balmer (\(n_f=2\)), Paschen (\(n_f=3\)), Brackett (\(n_f=4\)), Pfund (\(n_f=5\)). "Shortest wavelength" means "highest energy," which corresponds to the transition from \(n_i = \infty\) (the series limit).

Step 1: Understanding the Question:

The question asks for the physical implication of the net electric flux through a closed surface being zero.


Step 2: Key Formula or Approach:

The given equation, \( \oint_S \vec{E} \cdot d\vec{S} = 0 \), is a mathematical statement about the net electric flux through a closed surface S. According to Gauss's Law for electrostatics, the net electric flux through any closed surface is proportional to the net electric charge enclosed by that surface:
\[ \oint_S \vec{E} \cdot d\vec{S} = \frac{Q_{enc}}{\epsilon_0} \]
where \(Q_{enc}\) is the net charge inside the surface and \(\epsilon_0\) is the permittivity of free space.


Step 3: Detailed Explanation:

If \( \oint_S \vec{E} \cdot d\vec{S} = 0 \), then from Gauss's Law, it implies that \( \frac{Q_{enc}}{\epsilon_0} = 0 \), which means the net charge enclosed within the surface is zero (\(Q_{enc} = 0\)).

Let's analyze this in terms of electric field lines (flux lines):
- Electric flux is a measure of the number of electric field lines passing through a surface.
- By convention, flux lines leaving a closed surface contribute positively to the total flux, while lines entering contribute negatively.
- A net flux of zero means that the total positive contribution (flux out) is exactly cancelled by the total negative contribution (flux in).
- Therefore, the number of electric field lines entering the surface must be equal to the number of electric field lines leaving it.


Let's evaluate the given options:

- (A) is incorrect. The field inside could be zero or non-uniform (e.g., from an electric dipole placed inside the surface, where \(Q_{enc} = +q -q = 0\)).

- (B) is correct. This is the direct physical interpretation of zero net flux

- (C) is incorrect. The field strength can vary across the surface.

- (D) is incorrect and poorly phrased. The statement implies that there cannot be any charge outside, which is false. Also, it's about the net charge inside being zero, not that all charges are inside.


Step 4: Final Answer:

The condition of zero net electric flux means that the number of flux lines entering the surface must be equal to the number of flux lines leaving it.
Quick Tip: Gauss's Law is a powerful tool. Remember its core implications: - Net flux depends ONLY on the enclosed charge. - Charges outside the surface contribute to the electric field at the surface but do not contribute to the net flux through the surface. - Zero net flux implies zero net enclosed charge.

Step 1: Understanding the Question:

We are given a circuit and told that the galvanometer shows no deflection. This is a key condition that simplifies the circuit, and we need to find the value of the unknown resistance R.


Step 2: Key Formula or Approach:

The condition of "no deflection in the galvanometer" means that no current is flowing through it. This implies that the potential difference across the galvanometer is zero. In this circuit, this means the potential difference across the resistor R must be equal and opposite to the EMF of the 2V battery. This is the principle of a potentiometer.


Step 3: Detailed Explanation:

Since no current flows through the galvanometer, we can analyze the main circuit loop (the one with the 10V battery and resistors 400 \(\Omega\) and R) as if the galvanometer and the 2V battery branch were not there.


The resistors 400 \(\Omega\) and R are in series with the 10V battery.

The total resistance in this primary loop is \(R_{total} = 400\Omega + R\).

The current flowing through the primary loop is given by Ohm's law:
\[ I = \frac{V_{total}}{R_{total}} = \frac{10}{400 + R} \]

This current 'I' flows through the resistor R. The potential difference (voltage drop) across R is:
\[ V_R = I \times R = \left(\frac{10}{400 + R}\right)R \]

For the galvanometer to show no deflection, the potential at the points across it must be equal. This means the potential drop across resistor R must exactly balance the EMF of the 2V battery in the secondary loop.
\[ V_R = 2V \]
Now we can set up the equation:
\[ \left(\frac{10}{400 + R}\right)R = 2 \]
Now, solve for R:
\[ 10R = 2(400 + R) \] \[ 10R = 800 + 2R \] \[ 10R - 2R = 800 \] \[ 8R = 800 \] \[ R = \frac{800}{8} = 100 \, \Omega \]

Step 4: Final Answer:

The value of R for which the galvanometer shows no deflection is 100 \(\Omega\).
Quick Tip: This circuit is an application of the potentiometer principle. The main loop with the 10V battery acts as the potentiometer wire, creating a potential gradient. The galvanometer shows a null point when the potential drop across a portion of the main circuit (resistor R here) equals the EMF of the cell in the secondary circuit (the 2V battery).

Step 1: Understanding the Question:

The question asks for the value of the net magnetic flux passing through any arbitrary closed surface.


Step 2: Key Formula or Approach:

This question relates to one of Maxwell's equations, specifically Gauss's Law for Magnetism. The law is stated mathematically as:
\[ \Phi_B = \oint \vec{B} \cdot d\vec{A} = 0 \]
where \(\Phi_B\) is the magnetic flux, \(\vec{B}\) is the magnetic field, and the integral is taken over a closed surface A.


Step 3: Detailed Explanation:

Gauss's Law for Magnetism states that the net magnetic flux out of any closed surface is zero.

This is a fundamental law of physics with a deep physical meaning:

1. Magnetic monopoles do not exist: Unlike electric charges (which can be positive or negative), there are no isolated "magnetic charges" (north or south poles).

2. Magnetic field lines are continuous loops: Every magnetic field line that enters a closed surface must also exit that surface. Therefore, the total incoming flux (which can be considered negative) is always perfectly balanced by the total outgoing flux (positive), making the net flux zero.

This holds true for any closed surface, regardless of its shape or size, or the magnetic fields present.


Step 4: Final Answer:

The net magnetic flux through any closed surface is always zero.
Quick Tip: Remember the contrast with Gauss's Law for electricity: \(\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0}\). The net electric flux is proportional to the enclosed charge because isolated electric charges (monopoles) exist. For magnetism, the equivalent of enclosed charge is zero.

Step 1: Understanding the Question:

We need to calculate the average speed of a vehicle that covers two equal halves of its total journey at two different constant speeds.


Step 2: Key Formula or Approach:

The definition of average speed is:
\[ Average Speed = \frac{Total Distance Travelled}{Total Time Taken} \]
It is important not to simply average the speeds, as the time spent at each speed is different.


Step 3: Detailed Explanation:

Let the total distance of the journey be \(2D\).

The first half of the distance is \(D\), and the second half is also \(D\).


For the first half of the journey:

Distance = \(D\).

Speed = \(v\).

Time taken, \(t_1 = \frac{Distance}{Speed} = \frac{D}{v}\).


For the second half of the journey:

Distance = \(D\).

Speed = \(2v\).

Time taken, \(t_2 = \frac{Distance}{Speed} = \frac{D}{2v}\).


Now, we can calculate the average speed for the entire journey.

Total Distance = \(D + D = 2D\).

Total Time = \(t_1 + t_2 = \frac{D}{v} + \frac{D}{2v}\).

To add the fractions, find a common denominator:

Total Time = \(\frac{2D}{2v} + \frac{D}{2v} = \frac{3D}{2v}\).


Average Speed = \(\frac{Total Distance}{Total Time} = \frac{2D}{3D / 2v}\).

Average Speed = \(2D \times \frac{2v}{3D}\).

The 'D' terms cancel out.

Average Speed = \(\frac{4v}{3}\).


Step 4: Final Answer:

The average speed of the vehicle is \(\frac{4v}{3}\).
Quick Tip: When an object travels two equal distances at speeds \(v_1\) and \(v_2\), the average speed is the harmonic mean of the two speeds: \(v_{avg} = \frac{2v_1v_2}{v_1+v_2}\). In this case, \(v_1=v\) and \(v_2=2v\), so \(v_{avg} = \frac{2(v)(2v)}{v+2v} = \frac{4v^2}{3v} = \frac{4v}{3}\). This formula is a useful shortcut.

Step 1: Understanding the Question:

We need to find the relationship between the minimum wavelength of produced X-rays and the accelerating potential difference applied to the electrons.


Step 2: Key Formula or Approach:

1. Energy of an accelerated electron: When an electron (charge e) is accelerated through a potential difference V, it gains kinetic energy \(E_k = eV\).

2. Energy of a photon: The energy of a photon (like an X-ray) is given by \(E_{photon} = hf = \frac{hc}{\lambda}\), where h is Planck's constant, c is the speed of light, f is the frequency, and \(\lambda\) is the wavelength.

The minimum wavelength (\(\lambda_{min}\)) of the X-ray is produced when the electron loses all of its kinetic energy in a single interaction to create one photon. This photon will have the maximum possible energy.


Step 3: Detailed Explanation:

Equating the maximum kinetic energy of the electron to the maximum energy of the X-ray photon:
\[ E_{k, max} = E_{photon, max} \] \[ eV = \frac{hc}{\lambda_{min}} \]
We want to find the proportionality of \(\lambda_{min}\) with respect to V. Rearranging the equation for \(\lambda_{min}\):
\[ \lambda_{min} = \frac{hc}{e} \frac{1}{V} \]
Since h (Planck's constant), c (speed of light), and e (electron charge) are all constants, we can see the relationship between \(\lambda_{min}\) and V:
\[ \lambda_{min} \propto \frac{1}{V} \]
The minimum wavelength is inversely proportional to the accelerating voltage.


Step 4: Final Answer:

The minimum wavelength of the produced X-rays is proportional to \(\frac{1}{V}\).
Quick Tip: This minimum wavelength is also known as the "cutoff wavelength." A higher accelerating voltage means electrons have more energy, which allows them to produce higher-energy (and thus shorter-wavelength) X-ray photons.

Step 1: Understanding the Question:

We are given the value of a carbon resistor and need to determine the color of the third band based on the standard resistor color code.


Step 2: Key Formula or Approach:

The value of a four-band carbon resistor is given by the formula: \(R = (AB \times 10^C) \pm D%\), where:

- A is the digit corresponding to the first color band.

- B is the digit corresponding to the second color band.

- C is the multiplier corresponding to the third color band.

- D is the tolerance corresponding to the fourth color band.

The color code mnemonic is "BB ROY of Great Britain has a Very Good Wife" for Black(0), Brown(1), Red(2), Orange(3), Yellow(4), Green(5), Blue(6), Violet(7), Grey(8), White(9).


Step 3: Detailed Explanation:

The given resistance is \(22000 \, \Omega\). We can write this in scientific notation to match the formula:
\[ R = 22 \times 1000 \, \Omega = 22 \times 10^3 \, \Omega \]
Comparing this with \(R = AB \times 10^C\):

- The first significant digit, A, is 2. The color for digit 2 is Red.
- The second significant digit, B, is 2. The color for digit 2 is Red.
- The multiplier is \(10^C = 10^3\). The value of C is 3. The color for multiplier \(10^3\) is Orange.
- The tolerance is \(\pm 5%\). The color for this tolerance is Gold.


The question asks for the color of the third band, which corresponds to the multiplier C. Since C=3, the color is Orange.


Step 4: Final Answer:

The colour of the third band must be Orange.
Quick Tip: To quickly find the multiplier band, write the resistance value in standard engineering notation (\(XY \times 10^Z\)). The power 'Z' directly gives you the color number for the third band (e.g., \(10^3 \rightarrow 3 \rightarrow\) Orange).

Step 1: Understanding the Question:

The question asks to classify the type of measurement error that results from unpredictable changes in environmental or experimental conditions.


Step 2: Detailed Explanation:

Let's define the different types of errors:

- Random errors: These are errors that occur irregularly and are unpredictable. They are random in both magnitude and direction. They are caused by uncontrolled variables, such as fluctuations in temperature, voltage, pressure, or mechanical vibrations. Because they are random, their effects can be minimized by taking multiple measurements and calculating the average.

- Systematic errors: These errors have a consistent direction or magnitude. They are typically caused by a flaw in the experimental setup or the instrument itself.
- Instrumental errors: Arise from imperfections in the measuring instrument, such as incorrect calibration (a zero error) or faulty design.
- Personal errors: Occur due to the observer's bias, carelessness in taking readings, or incorrect experimental procedure (e.g., parallax error).
- Least count errors: This error is associated with the resolution of the instrument. The least count is the smallest value that can be measured by the instrument, and any measurement is only accurate up to this value.


The question specifically mentions "unpredictable fluctuations," which is the defining characteristic of random errors.


Step 3: Final Answer:

Errors arising from unpredictable fluctuations in temperature and voltage supply are classified as random errors.
Quick Tip: A key distinction: Systematic errors can, in principle, be identified and corrected for, as they are consistent. Random errors cannot be eliminated but can be reduced by repeated measurements and statistical analysis.

Step 1: Understanding the Question:

The question asks for the underlying physical principle behind the operation of a venturi-meter.


Step 2: Detailed Explanation:

A venturi-meter is a device used for measuring the rate of flow of a fluid through a pipe. It consists of a converging section, a throat (the narrowest part), and a diverging section.

1. As the fluid flows from the wider section into the narrow throat, the cross-sectional area decreases. According to the principle of continuity (\(A_1 v_1 = A_2 v_2\)), the speed of the fluid (\(v\)) must increase.

2. Bernoulli's principle states that for a horizontal flow, an increase in the speed of a fluid occurs simultaneously with a decrease in pressure. The principle is expressed as \(P + \frac{1}{2}\rho v^2 = constant\) for a horizontal pipe.

3. In the venturi-meter, since the speed is highest in the throat, the pressure is lowest there. By measuring the pressure difference between the main pipe and the throat using a manometer, one can calculate the fluid's velocity and thus the flow rate.

Therefore, the operation of the venturi-meter is a direct application of Bernoulli's principle, combined with the principle of continuity.


The other options are irrelevant:
- Principles of perpendicular and parallel axes are theorems related to the moment of inertia in rotational mechanics.
- Huygen's principle is a concept in wave optics that describes how wavefronts propagate.


Step 3: Final Answer:

The venturi-meter works on Bernoulli's principle.
Quick Tip: Many practical applications in fluid dynamics, such as the lift of an airplane wing, the curve of a spinning ball (Magnus effect), and atomizers/sprayers, are explained by Bernoulli's principle: where speed is high, pressure is low.

Step 1: Understanding the Question:

We need to identify the direction of the angular acceleration vector (\(\vec{\alpha}\)) for an object undergoing circular motion.


Step 2: Detailed Explanation:

Let's distinguish between linear and angular quantities in circular motion.

- Linear quantities describe the motion of the particle itself and lie in the plane of rotation:

- Tangential velocity (\(\vec{v}_t\)): Always tangent to the circular path.

- Tangential acceleration (\(\vec{a}_t\)): Also tangent to the path, responsible for changing the speed of the particle.

- Centripetal (or radial) acceleration (\(\vec{a}_c\)): Always points along the radius towards the center of the circle, responsible for changing the direction of the velocity.

- Angular quantities describe the rotation of the body as a whole. These are axial vectors, meaning their direction is along the axis of rotation, perpendicular to the plane of motion. The direction is determined by the right-hand thumb rule.

- Angular velocity (\(\vec{\omega}\)): Points along the axis of rotation.
- Angular acceleration (\(\vec{\alpha}\)): Defined as the rate of change of angular velocity, \(\vec{\alpha} = \frac{d\vec{\omega}}{dt}\). Since \(\vec{\omega}\) is an axial vector, its change (\(d\vec{\omega}\)) and hence \(\vec{\alpha}\) must also be directed along the axis of rotation. If the body speeds up, \(\vec{\alpha}\) is in the same direction as \(\vec{\omega}\). If it slows down, \(\vec{\alpha}\) is in the opposite direction to \(\vec{\omega}\). In both cases, it is along the axis of rotation.


Therefore, options (B), (C), and (D) describe linear acceleration components, not angular acceleration.


Step 3: Final Answer:

The angular acceleration of a body in circular motion is directed along the axis of rotation.
Quick Tip: Remember the rule of thumb: If the quantity's name starts with "angular" (angular velocity, angular acceleration, angular momentum) or is a "moment" (torque, moment of inertia), it's related to rotation, and its vector representation (if it's a vector) is typically along the axis of rotation.

Step 1: Understanding the Question:

We are given the work functions for three different metals and the energy of incident photons. We need to determine which of these metals will exhibit the photoelectric effect.


Step 2: Key Formula or Approach:

The condition for the photoelectric effect to occur is that the energy of the incident photon (E) must be greater than or equal to the work function (\(\phi\)) of the metal.
\[ E \geq \phi \]
The work function is the minimum energy required to remove an electron from the surface of the material.


Step 3: Detailed Explanation:

Given values:

Incident photon energy, \(E = 2.20\) eV.

Work function of Caesium, \(\phi_{Cs} = 2.14\) eV.

Work function of Potassium, \(\phi_{K} = 2.30\) eV.

Work function of Sodium, \(\phi_{Na} = 2.75\) eV.


Let's check the condition \(E \geq \phi\) for each metal:

1. Caesium (Cs): Is \(2.20 \, eV \geq 2.14 \, eV\)? Yes. Therefore, photoemission will occur from Caesium.

2. Potassium (K): Is \(2.20 \, eV \geq 2.30 \, eV\)? No. Therefore, photoemission will not occur from Potassium.

3. Sodium (Na): Is \(2.20 \, eV \geq 2.75 \, eV\)? No. Therefore, photoemission will not occur from Sodium.


Only Caesium satisfies the condition for photoelectron emission.


Step 4: Final Answer:

Of the given surfaces, only Cs may emit photoelectrons.
Quick Tip: Remember that the work function is a "threshold" energy. If the incoming photon's energy is below this threshold, no electrons will be emitted, no matter how intense the light is. If the energy is above the threshold, the excess energy (\(E - \phi\)) becomes the maximum kinetic energy of the emitted photoelectron.

Step 1: Understanding the Question:

We need to find the equivalent capacitance between points A and B for the given arrangement of four capacitors, each with a capacitance of 3 \(\mu\)F.


Step 2: Key Formula or Approach:


1. Capacitors in Series: The equivalent capacitance \(C_{eq}\) for two capacitors \(C_1\) and \(C_2\) in series is given by \(\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}\) or \(C_{eq} = \frac{C_1C_2}{C_1+C_2}\).


2. Capacitors in Parallel: The equivalent capacitance for capacitors in parallel is the sum of their individual capacitances: \(C_{eq} = C_1 + C_2 + \dots\).


3. Circuit Analysis: We need to identify which parts of the circuit are in series and which are in parallel. Let's label the junctions to analyze the circuit. Let the junction after the first capacitor be P, the top junction be Q, and the bottom junction be R. From the diagram, we can see that point B is connected to both Q and R.


Step 3: Detailed Explanation:

Let's analyze the circuit structure.

- There is a 3 \(\mu\)F capacitor between A and a central node P.

- From node P, the circuit splits. There is a 3 \(\mu\)F capacitor between P and Q (top path), and another 3 \(\mu\)F capacitor between P and R (bottom path).

- There is a 3 \(\mu\)F capacitor connected between nodes Q and R.

- Both nodes Q and R are connected to the terminal B.

Since both Q and R are connected to the same point B, they are at the same potential. This means the potential difference across the capacitor between Q and R is zero. A capacitor with zero potential difference across it carries no charge and can be removed from the circuit for analysis. It is effectively short-circuited.


The simplified circuit becomes:

- A capacitor \(C_1 = 3 \, \mu F\) from A to P.

- Two capacitors, \(C_{top} = 3 \, \mu F\) (from P to Q/B) and \(C_{bottom} = 3 \, \mu F\) (from P to R/B), are now in parallel with each other between node P and node B.


The equivalent capacitance of the parallel combination (\(C_p\)) is: \[ C_p = C_{top} + C_{bottom} = 3 \, \mu F + 3 \, \mu F = 6 \, \mu F \]
Now, this equivalent capacitance \(C_p\) is in series with the first capacitor \(C_1\).


The total equivalent capacitance of the system (\(C_{eq}\)) is: \[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_p} = \frac{1}{3} + \frac{1}{6} \] \[ \frac{1}{C_{eq}} = \frac{2+1}{6} = \frac{3}{6} = \frac{1}{2} \] \[ C_{eq} = 2 \, \mu F \]

Step 4: Final Answer:

The equivalent capacitance of the system is 2 \(\mu\)F.
Quick Tip: When analyzing complex capacitor networks, always look for points that are at the same potential. Capacitors connected between such points can be removed from the circuit as they are shorted out, which often simplifies the problem significantly.

Step 1: Understanding the Question:

We are asked to find the gravitational potential at a specific point on the line connecting two masses, m and 9m. This point is where the net gravitational field due to the two masses is zero.


Step 2: Key Formula or Approach:

1. Gravitational Field (E): The magnitude of the gravitational field due to a mass M at a distance r is \(E = \frac{GM}{r^2}\). It is a vector quantity.

2. Gravitational Potential (V): The gravitational potential due to a mass M at a distance r is \(V = -\frac{GM}{r}\). It is a scalar quantity.

The approach is to first find the point where the net gravitational field is zero and then calculate the net potential at that point.


Step 3: Detailed Explanation:

Let the two masses, m and 9m, be placed along the x-axis at x=0 and x=R, respectively. Let P be the point on the line joining them where the gravitational field is zero. Let the distance of P from mass m be 'x'. Then, the distance of P from mass 9m will be (R-x).


For the net gravitational field at P to be zero, the magnitudes of the fields due to both masses must be equal.
\[ E_m = E_{9m} \] \[ \frac{Gm}{x^2} = \frac{G(9m)}{(R-x)^2} \]
Taking the square root on both sides:
\[ \frac{1}{x} = \frac{3}{R-x} \] \[ R-x = 3x \] \[ R = 4x \] \[ x = \frac{R}{4} \]
So, the point where the field is zero is at a distance of R/4 from mass m. The distance from mass 9m is \(R-x = R - \frac{R}{4} = \frac{3R}{4}\).


Now, we calculate the net gravitational potential at this point P. Since potential is a scalar, we simply add the potentials due to each mass.
\[ V_{net} = V_m + V_{9m} \] \[ V_{net} = \left(-\frac{Gm}{x}\right) + \left(-\frac{G(9m)}{R-x}\right) \]
Substitute the values of x and (R-x):
\[ V_{net} = \left(-\frac{Gm}{R/4}\right) - \left(\frac{G(9m)}{3R/4}\right) \] \[ V_{net} = -\frac{4Gm}{R} - \frac{36Gm}{3R} \] \[ V_{net} = -\frac{4Gm}{R} - \frac{12Gm}{R} \] \[ V_{net} = -\frac{16Gm}{R} \]

Step 4: Final Answer:

The gravitational potential at the point where the gravitational field is zero is \(-\frac{16Gm}{R}\).
Quick Tip: The point where the gravitational field is zero (the null point) between two masses always lies closer to the smaller mass. Remember that gravitational potential is a scalar quantity, so you add the potentials algebraically (including the negative sign), whereas the gravitational field is a vector, requiring vector addition.

Step 1: Understanding the Question:

We need to find the critical angle for a denser medium, given the time it takes for light to travel certain distances in air and in that medium.


Step 2: Key Formula or Approach:

1. Velocity of light: \(v = \frac{distance}{time}\).

2. Refractive index (n): \(n = \frac{speed of light in vacuum (c)}{speed of light in medium (v)}\). For air, we can approximate \(n_{air} \approx 1\).

3. Critical angle (\(i_c\)): When light travels from a denser medium (\(n_d\)) to a rarer medium (\(n_r\)), the critical angle is given by Snell's law: \(\sin(i_c) = \frac{n_r}{n_d}\).


Step 3: Detailed Explanation:

First, let's find the speed of light in air (\(v_{air}\)) and in the denser medium (\(v_{medium}\)).

Speed in air: \(v_{air} = \frac{x}{t_1}\). We'll consider air as the rarer medium, so \(n_r = n_{air} = 1\).

Speed in the denser medium: \(v_{medium} = \frac{10x}{t_2}\).


Next, let's find the refractive index of the denser medium (\(n_d\)). The refractive index is the ratio of the speed of light in air to the speed of light in the medium.
\[ n_d = \frac{v_{air}}{v_{medium}} = \frac{x/t_1}{10x/t_2} \] \[ n_d = \frac{x}{t_1} \times \frac{t_2}{10x} = \frac{t_2}{10t_1} \]

Now, we can find the critical angle (\(i_c\)) for light going from the denser medium to air.
\[ \sin(i_c) = \frac{n_{rarer}}{n_{denser}} = \frac{n_{air}}{n_d} \] \[ \sin(i_c) = \frac{1}{t_2 / (10t_1)} \] \[ \sin(i_c) = \frac{10t_1}{t_2} \]
Therefore, the critical angle is:
\[ i_c = \sin^{-1}\left(\frac{10t_1}{t_2}\right) \]

Step 4: Final Answer:

The critical angle for the medium is \(\sin^{-1}\left(\frac{10t_1}{t_2}\right)\).
Quick Tip: The critical angle only exists when light travels from a denser medium to a rarer medium. The value of \(\sin(i_c)\) must be less than or equal to 1. In this problem, \(\frac{n_r}{n_d} = \frac{v_d}{v_r}\), which can be a useful shortcut if you calculate the velocities first.

Step 1: Understanding the Question:

We need to find the formula for longitudinal stress in a wire that is supporting a weight W.


Step 2: Key Formula or Approach:

Stress is defined as the internal restoring force (\(F_{restoring}\)) acting per unit of cross-sectional area (A).
\[ Stress = \frac{F_{restoring}}{A} \]
In equilibrium, the internal restoring force is equal in magnitude to the external deforming force.

Longitudinal stress (or tensile stress) occurs when the force is applied perpendicular to the cross-section, causing a change in length.


Step 3: Detailed Explanation:

- A wire of cross-sectional area A is suspended vertically.
- A weight W is attached to its free end. This weight W is the external deforming force that stretches the wire.
- The wire is in static equilibrium. This means that at any cross-section of the wire, the upward internal restoring force (tension, T) must balance the downward external force.
- If we neglect the weight of the wire itself, the only downward force is the attached weight W.
- Therefore, the tension at any point in the wire is \(T = W\).
- This tension T is the internal restoring force.
- Now, we can apply the formula for stress: \[ Longitudinal Stress = \frac{Internal Restoring Force}{Cross-sectional Area} = \frac{T}{A} \]
Substituting \(T=W\): \[ Longitudinal Stress = \frac{W}{A} \]

Step 4: Final Answer:

The longitudinal stress at any point in the wire is W/A.
Quick Tip: In problems like this, unless the mass or density of the wire is mentioned and you are asked to account for it, you should assume the wire is massless. The stress is then uniform throughout the wire and depends only on the attached weight.

Step 1: Understanding the Question:

The question asks for the ratio of the fundamental frequency of an open organ pipe to that of a closed organ pipe, given that they both have the same length.


Step 2: Key Formula or Approach:

The fundamental frequency (\(f_1\)) is the lowest frequency at which a system can resonate.

1. Open Organ Pipe: An open pipe has antinodes at both ends. The fundamental mode of vibration has a wavelength \(\lambda_{open} = 2L\). The fundamental frequency is given by:
\[ f_{open} = \frac{v}{\lambda_{open}} = \frac{v}{2L} \]
where \(v\) is the speed of sound and L is the length of the pipe.

2. Closed Organ Pipe: A closed pipe has a node at the closed end and an antinode at the open end. The fundamental mode of vibration has a wavelength \(\lambda_{closed} = 4L\). The fundamental frequency is given by:
\[ f_{closed} = \frac{v}{\lambda_{closed}} = \frac{v}{4L} \]

Step 3: Detailed Explanation:

We are asked to find the ratio \(\frac{f_{open}}{f_{closed}}\).

Using the formulas from Step 2:
\[ \frac{f_{open}}{f_{closed}} = \frac{v/2L}{v/4L} \]
The terms \(v\) and \(L\) are the same for both pipes, so they cancel out.
\[ \frac{f_{open}}{f_{closed}} = \frac{1/2}{1/4} = \frac{1}{2} \times \frac{4}{1} = \frac{4}{2} = 2 \]
So, the ratio of the frequencies is 2:1.


Step 4: Final Answer:

The ratio of the fundamental frequency of the open pipe to the closed pipe is 2:1.
Quick Tip: A simple way to remember the harmonics: - Open pipe: All harmonics are present (\(f_n = nf_1\), where n=1, 2, 3,...). Frequencies are in the ratio 1:2:3... - Closed pipe: Only odd harmonics are present (\(f_n = (2n-1)f_1\), where n=1, 2, 3,...). Frequencies are in the ratio 1:3:5...

Step 1: Understanding the Question:

We are given the initial upward velocity of a ball thrown from a bridge and the total time until it hits the water below. We need to find the height of the bridge.


Step 2: Key Formula or Approach:

We will use the second equation of motion for displacement under constant acceleration (gravity). We must be careful with the sign convention. Let's choose the point of projection (on the bridge) as the origin (s=0) and the upward direction as positive.

The formula is:
\[ s = ut + \frac{1}{2}at^2 \]
where \(s\) is the displacement, \(u\) is the initial velocity, \(t\) is the time, and \(a\) is the acceleration.


Step 3: Detailed Explanation:

According to our sign convention:

- Initial velocity, \(u = +4\) m/s (since it's thrown upwards).

- Time, \(t = 4\) s.

- Acceleration, \(a = -g = -10\) m/s\(^2\) (since gravity acts downwards).


Now, we calculate the displacement (\(s\)) of the ball from the bridge to the water surface:
\[ s = (4)(4) + \frac{1}{2}(-10)(4)^2 \] \[ s = 16 - 5(16) \] \[ s = 16 - 80 \] \[ s = -64 \, m \]
The negative sign indicates that the final position (the water surface) is 64 meters below the initial position (the bridge). Therefore, the height of the bridge above the water is the magnitude of this displacement.

Height \(H = |s| = 64\) m.


Step 4: Final Answer:

The height of the bridge above the water surface is 64 m.
Quick Tip: In projectile motion problems, carefully establishing a coordinate system and consistently applying a sign convention (e.g., up is positive, down is negative) is crucial to avoid errors. The displacement 's' is a vector quantity, while height is a scalar distance.

Step 1: Understanding the Question:

The given setup consists of a biconvex lens made of material \(n_1=1.5\) enclosed within a container filled with a medium of refractive index \(n_2=1.6\). This combination forms a system of three thin lenses in contact. We need to find the equivalent focal length of this system, assuming it is placed in air (\(n_{air}=1\)).


Step 2: Key Formula or Approach:

We can treat the system as three thin lenses placed in contact:
1. A plano-concave lens of material \(n_2\). (\(L_1\))
2. A biconvex lens of material \(n_1\). (\(L_2\))
3. Another plano-concave lens of material \(n_2\). (\(L_3\))

The equivalent focal length \(F\) is given by: \( \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3} \).
The focal length of each lens is calculated using the Lens Maker's Formula:
\[ \frac{1}{f} = (n_{lens} - n_{medium}) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Here, the surrounding medium is air, so \(n_{medium}=1\).


Step 3: Detailed Explanation:

Given: \(n_1 = 1.5\), \(n_2 = 1.6\), and \(|R_1| = |R_2| = 20\) cm.
Using the standard sign convention (light travels from left to right):


For lens \(L_1\) (plano-concave, \(n_2 = 1.6\)):

The first surface is plane (\(R_{1,1} = \infty\)). The second surface is concave with radius 20 cm (\(R_{1,2} = +20\) cm).
\[ \frac{1}{f_1} = (1.6 - 1) \left(\frac{1}{\infty} - \frac{1}{20}\right) = 0.6 \times \left(-\frac{1}{20}\right) = -\frac{0.6}{20} = -\frac{3}{100} \]

For lens \(L_2\) (biconvex, \(n_1 = 1.5\)):

The first surface is convex (\(R_{2,1} = +20\) cm). The second surface is also convex from the right side, so its radius of curvature is (\(R_{2,2} = -20\) cm).
\[ \frac{1}{f_2} = (1.5 - 1) \left(\frac{1}{20} - \frac{1}{-20}\right) = 0.5 \times \left(\frac{2}{20}\right) = 0.5 \times \frac{1}{10} = \frac{5}{100} \]

For lens \(L_3\) (plano-concave, \(n_2 = 1.6\)):

The first surface is concave (\(R_{3,1} = -20\) cm). The second surface is plane (\(R_{3,2} = \infty\)).
\[ \frac{1}{f_3} = (1.6 - 1) \left(\frac{1}{-20} - \frac{1}{\infty}\right) = 0.6 \times \left(-\frac{1}{20}\right) = -\frac{0.6}{20} = -\frac{3}{100} \]

For the combination:
\[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3} = \left(-\frac{3}{100}\right) + \left(\frac{5}{100}\right) + \left(-\frac{3}{100}\right) \] \[ \frac{1}{F} = \frac{-3 + 5 - 3}{100} = \frac{-1}{100} \] \[ F = -100 \, cm \]

Step 4: Final Answer:

The equivalent focal length of the combination is \(-100\) cm.
Quick Tip: Complex lens arrangements can often be simplified by treating them as multiple thin lenses in contact. Remember to apply the sign convention for radii of curvature carefully for each lens component. A convex surface has a positive radius if light hits it from the left, while a concave surface has a negative radius.

Step 1: Understanding the Question:

We are given the radius of the first orbit (n=1) of a hydrogen atom and asked to find the radius of the third orbit (n=3).


Step 2: Key Formula or Approach:

According to the Bohr model for the hydrogen atom, the radius of the n-th allowed orbit is given by:
\[ r_n = r_1 \times n^2 \]
where \(r_1\) is the radius of the first orbit (also known as the Bohr radius, \(a_0\)) and \(n\) is the principal quantum number.


Step 3: Detailed Explanation:

Given values:

- Radius of the innermost orbit, \(r_1 = 5.3 \times 10^{-11}\) m.

- We need to find the radius of the third orbit, so \(n=3\).


Substitute the values into the formula:
\[ r_3 = r_1 \times (3)^2 = r_1 \times 9 \] \[ r_3 = (5.3 \times 10^{-11} \, m) \times 9 \] \[ r_3 = 47.7 \times 10^{-11} \, m \]
The options are in Angstroms (Å). We know that \(1 \, Å = 10^{-10}\) m. Let's convert our answer to Angstroms.
\[ r_3 = 4.77 \times 10^{-10} \, m = 4.77 \, Å \]

Step 4: Final Answer:

The radius of the third allowed orbit of the hydrogen atom is 4.77 Å.
Quick Tip: For hydrogen-like atoms, the key proportionalities from the Bohr model are very useful: - Radius: \(r_n \propto \frac{n^2}{Z}\) - Velocity: \(v_n \propto \frac{Z}{n}\) - Energy: \(E_n \propto -\frac{Z^2}{n^2}\) For hydrogen, Z=1, so the relationships simplify.

Step 1: Understanding the Question:

We have a combination of two thin lenses in contact: a convex lens and a concave lens, both having the same magnitude of focal length, \(f\). We need to find the equivalent focal length of this combination.


Step 2: Key Formula or Approach:

The equivalent focal length, \(F_{eq}\), of two thin lenses with focal lengths \(f_1\) and \(f_2\) placed in contact is given by:
\[ \frac{1}{F_{eq}} = \frac{1}{f_1} + \frac{1}{f_2} \]
By convention, the focal length of a convex lens is positive, and that of a concave lens is negative.


Step 3: Detailed Explanation:

Let \(f_1\) be the focal length of the convex lens and \(f_2\) be the focal length of the concave lens.

According to the sign convention:

- \(f_1 = +f\) (for the convex lens)

- \(f_2 = -f\) (for the concave lens)


Now, substitute these into the combination formula:
\[ \frac{1}{F_{eq}} = \frac{1}{+f} + \frac{1}{-f} \] \[ \frac{1}{F_{eq}} = \frac{1}{f} - \frac{1}{f} = 0 \]
If the reciprocal of the equivalent focal length is zero, the focal length itself must be infinitely large.
\[ F_{eq} = \frac{1}{0} \rightarrow \infty \]

Step 4: Final Answer:

The equivalent focal length of the combination will be infinite. Such a combination acts like a plane glass plate and has zero power.
Quick Tip: The power of a lens is the reciprocal of its focal length (\(P = 1/f\)). For lenses in contact, the powers add up: \(P_{eq} = P_1 + P_2\). In this case, \(P_1 = +1/f\) and \(P_2 = -1/f\), so \(P_{eq} = 0\). A system with zero power has an infinite focal length.

Step 1: Understanding the Question:

We need to find what the expression \(\frac{3\pi}{Gd}\) represents in terms of the orbital period T of a satellite orbiting just above the Earth's surface.


Step 2: Key Formula or Approach:

1. The orbital period \(T\) of a satellite is given by Kepler's third law applied to circular orbits: \( T^2 = \frac{4\pi^2 r^3}{GM} \), where \(r\) is the orbital radius and \(M\) is the mass of the central body (Earth).

2. For a satellite orbiting "just above the surface," the orbital radius \(r\) is approximately equal to the Earth's radius \(R\).

3. The mass of the Earth \(M\) can be expressed in terms of its density \(d\) and radius \(R\): \(M = Volume \times Density = \frac{4}{3}\pi R^3 d\).


Step 3: Detailed Explanation:

Start with the formula for the orbital period squared:
\[ T^2 = \frac{4\pi^2 R^3}{GM} \]
Now, substitute the expression for the mass of the Earth \(M\):
\[ T^2 = \frac{4\pi^2 R^3}{G \left(\frac{4}{3}\pi R^3 d\right)} \]
We can cancel out several terms: \(4\), \(\pi\), and \(R^3\).
\[ T^2 = \frac{\pi}{G \left(\frac{1}{3} d\right)} \] \[ T^2 = \frac{3\pi}{Gd} \]
This shows that the given quantity \(\frac{3\pi}{Gd}\) is exactly equal to the square of the orbital period, \(T^2\).


Step 4: Final Answer:

The quantity \( \frac{3\pi}{Gd} \) represents \(T^2\).
Quick Tip: This is a classic derivation. It shows that for a satellite orbiting at the surface of any spherical body, the square of the orbital period is inversely proportional to the body's density (\(T^2 \propto 1/d\)), regardless of its size.

Step 1: Understanding the Question:

We have two scenarios with 10 identical resistors and the same battery. First, they are in series, and second, they are in parallel. We need to find the factor 'n' by which the current increases in the parallel case compared to the series case.


Step 2: Key Formula or Approach:

1. Series Combination: The equivalent resistance \(R_s\) of N resistors each of resistance R in series is \(R_s = NR\).

2. Parallel Combination: The equivalent resistance \(R_p\) of N resistors each of resistance R in parallel is \(R_p = R/N\).

3. Ohm's Law: The current \(I\) from a battery of emf E is \(I = E/R_{eq}\).


Step 3: Detailed Explanation:

Let N = 10 and each resistor has resistance R. The battery emf is E.


Case 1: Series Connection

The equivalent resistance is \(R_{series} = 10R\).

The current flowing from the battery is:
\[ I_{series} = \frac{E}{R_{series}} = \frac{E}{10R} \]

Case 2: Parallel Connection

The equivalent resistance is \(R_{parallel} = \frac{R}{10}\).

The current flowing from the battery is:
\[ I_{parallel} = \frac{E}{R_{parallel}} = \frac{E}{R/10} = \frac{10E}{R} \]

Finding the value of n

We are given that the current is increased n times, which means \(I_{parallel} = n \times I_{series}\).
\[ \frac{10E}{R} = n \times \left(\frac{E}{10R}\right) \]
We can cancel E and R from both sides of the equation:
\[ 10 = n \times \frac{1}{10} \] \[ n = 10 \times 10 = 100 \]

Step 4: Final Answer:

The value of n is 100.
Quick Tip: For N identical resistors, the ratio of series to parallel equivalent resistance is \(R_s/R_p = (NR)/(R/N) = N^2\). Since current is inversely proportional to resistance, the ratio of currents will be \(I_p/I_s = R_s/R_p = N^2\). In this case, \(n = 10^2 = 100\).

Step 1: Understanding the Question:

We need to find the magnitude of the magnetic force on a straight current-carrying wire placed in a uniform magnetic field.


Step 2: Key Formula or Approach:

The magnetic force \(\vec{F}\) on a straight wire of length vector \(\vec{L}\) carrying a current \(I\) in a uniform magnetic field \(\vec{B}\) is given by the Lorentz force law:
\[ \vec{F} = I (\vec{L} \times \vec{B}) \]
The magnitude of the force is \(|\vec{F}| = I |\vec{L} \times \vec{B}|\).


Step 3: Detailed Explanation:

First, we need to define the length vector \(\vec{L}\).

The wire has length L and carries current along the positive x-axis. Therefore, the length vector is:
\[ \vec{L} = L\hat{i} \]
The magnetic field vector is given as:
\[ \vec{B} = (2\hat{i} + 3\hat{j} - 4\hat{k}) \, T \]
Next, we calculate the cross product \(\vec{L} \times \vec{B}\):
\[ \vec{L} \times \vec{B} = (L\hat{i}) \times (2\hat{i} + 3\hat{j} - 4\hat{k}) \]
Using the distributive property of the cross product:
\[ \vec{L} \times \vec{B} = L [(\hat{i} \times 2\hat{i}) + (\hat{i} \times 3\hat{j}) + (\hat{i} \times -4\hat{k})] \]
Recall the properties of unit vector cross products: \(\hat{i} \times \hat{i} = 0\), \(\hat{i} \times \hat{j} = \hat{k}\), and \(\hat{i} \times \hat{k} = -\hat{j}\).
\[ \vec{L} \times \vec{B} = L [0 + 3(\hat{i} \times \hat{j}) - 4(\hat{i} \times \hat{k})] \] \[ \vec{L} \times \vec{B} = L [3\hat{k} - 4(-\hat{j})] = L(4\hat{j} + 3\hat{k}) \]
Now, calculate the force vector \(\vec{F}\):
\[ \vec{F} = I (\vec{L} \times \vec{B}) = I [L(4\hat{j} + 3\hat{k})] = IL(4\hat{j} + 3\hat{k}) \]
Finally, find the magnitude of the force vector:
\[ |\vec{F}| = |IL(4\hat{j} + 3\hat{k})| = IL |4\hat{j} + 3\hat{k}| \] \[ |\vec{F}| = IL \sqrt{(4)^2 + (3)^2} = IL \sqrt{16 + 9} = IL \sqrt{25} \] \[ |\vec{F}| = 5IL \]

Step 4: Final Answer:

The magnitude of the magnetic force acting on the wire is 5 IL.
Quick Tip: Remember that the component of the magnetic field that is parallel to the current (\(2\hat{i}\) in this case) does not contribute to the magnetic force, as the cross product of parallel vectors is zero. Only the components of \(\vec{B}\) perpendicular to \(\vec{L}\) contribute to the force.

Step 1: Understanding the Question:

We need to find the maximum acceleration a car can have without an object on its floor slipping. This is a problem involving static friction.


Step 2: Key Formula or Approach:

The object accelerates along with the car because of the force of static friction (\(f_s\)) between the object and the car floor. According to Newton's second law:
\[ F_{net} = ma \]
Here, the net horizontal force on the body is the static friction force, so \(f_s = ma\).

The static friction force has a maximum possible value, \(f_{s,max} = \mu_s N\), where \(\mu_s\) is the coefficient of static friction and \(N\) is the normal force.

For the object not to slip, \(f_s \leq f_{s,max}\). The maximum acceleration (\(a_{max}\)) occurs when the required force is equal to the maximum available friction force, i.e., \(ma_{max} = f_{s,max}\).


Step 3: Detailed Explanation:

On a horizontal floor, the normal force \(N\) balances the weight of the body \(mg\). So, \(N = mg\).

The maximum static friction force is:
\[ f_{s,max} = \mu_s N = \mu_s mg \]
The condition for maximum acceleration is:
\[ ma_{max} = f_{s,max} \] \[ ma_{max} = \mu_s mg \]
The mass \(m\) cancels out from both sides:
\[ a_{max} = \mu_s g \]
Given values:

- \(\mu_s = 0.15\)

- \(g = 10\) m/s\(^2\)

Substitute the values:
\[ a_{max} = 0.15 \times 10 = 1.5 \, m/s^2 \]

Step 4: Final Answer:

The maximum acceleration of the car is 1.5 m/s\(^{-2}\).
Quick Tip: This result, \(a_{max} = \mu_s g\), is a standard and useful one for problems involving an object on an accelerating horizontal surface. Notice that the maximum acceleration is independent of the mass of the object.

Step 1: Understanding the Question:

We are asked to find the net impedance (Z) of a series LCR circuit with given values for the inductor (L), capacitor (C), resistor (R), and the AC source frequency (f).


Step 2: Key Formula or Approach:

The impedance Z of a series LCR circuit is given by:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
where \(R\) is the resistance, \(X_L\) is the inductive reactance, and \(X_C\) is the capacitive reactance.

The reactances are calculated as:
\[ X_L = \omega L = 2\pi f L \] \[ X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C} \]

Step 3: Detailed Explanation:

Given values:

- \(R = 10 \, \Omega\)

- \(L = \frac{50}{\pi} \, mH = \frac{50}{\pi} \times 10^{-3} \, H\)

- \(C = \frac{10^3}{\pi} \, \muF = \frac{10^3}{\pi} \times 10^{-6} \, F = \frac{10^{-3}}{\pi} \, F\)

- \(f = 50 \, Hz\)


First, calculate the inductive reactance \(X_L\):
\[ X_L = 2\pi f L = 2\pi (50) \left(\frac{50}{\pi} \times 10^{-3}\right) = 100\pi \left(\frac{50}{\pi} \times 10^{-3}\right) = 5000 \times 10^{-3} = 5 \, \Omega \]
Next, calculate the capacitive reactance \(X_C\):
\[ X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi (50) \left(\frac{10^{-3}}{\pi}\right)} = \frac{1}{100\pi \left(\frac{10^{-3}}{\pi}\right)} = \frac{1}{100 \times 10^{-3}} = \frac{1}{10^{-1}} = 10 \, \Omega \]
Now, calculate the impedance Z:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{(10)^2 + (5 - 10)^2} \] \[ Z = \sqrt{100 + (-5)^2} = \sqrt{100 + 25} = \sqrt{125} \] \[ Z = \sqrt{25 \times 5} = 5\sqrt{5} \, \Omega \]

Step 4: Final Answer:

The net impedance of the circuit is \(5\sqrt{5} \, \Omega\).
Quick Tip: In LCR circuit calculations, the values of L, C, and f are often chosen to give simple integer values for \(X_L\) and \(X_C\). Always calculate the reactances first before finding the impedance. The voltage of the source (220 V) is extra information not needed for calculating impedance.

Step 1: Understanding the Question:

We need to calculate the net electric potential at point P due to an electric dipole. Point P lies on the axial line of the dipole.


Step 2: Key Formula or Approach:

Electric potential is a scalar quantity. The total potential at a point due to multiple charges is the algebraic sum of the potentials due to individual charges. The potential V at a distance r from a point charge Q is given by:
\[ V = \frac{KQ}{r} \]
where \(K = \frac{1}{4\pi\epsilon_0}\).


Step 3: Detailed Explanation:

From the figure, we have:

- A negative charge \(-q\) at \(x = -3\) cm.
- A positive charge \(+q\) at \(x = +3\) cm.
- Point P is on the x-axis at \(x = +5\) cm.

1. Calculate the distance of P from each charge:

- Distance of P from the positive charge \(+q\):
\(r_+ = (5 \, cm) - (3 \, cm) = 2 \, cm\).

- Distance of P from the negative charge \(-q\):
\(r_- = (5 \, cm) - (-3 \, cm) = 5 \, cm + 3 \, cm = 8 \, cm\).


2. Calculate the potential at P:

The total potential \(V_P\) is the sum of the potential from \(+q\) (\(V_+\)) and the potential from \(-q\) (\(V_-\)).
\[ V_P = V_+ + V_- = \frac{K(+q)}{r_+} + \frac{K(-q)}{r_-} \] \[ V_P = Kq \left( \frac{1}{r_+} - \frac{1}{r_-} \right) \]
Substitute the distances (we can keep them in cm as we are looking for a ratio which makes the units cancel):
\[ V_P = Kq \left( \frac{1}{2} - \frac{1}{8} \right) \]
Find a common denominator:
\[ V_P = Kq \left( \frac{4}{8} - \frac{1}{8} \right) = Kq \left( \frac{3}{8} \right) \] \[ V_P = \left(\frac{3}{8}\right)qK \]
The phrase "in 10\(^2\) V" seems to be extraneous information, as the options are symbolic expressions.


Step 4: Final Answer:

The electric potential at point P due to the dipole is \( \left(\frac{3}{8}\right)qK \).
Quick Tip: Remember that electric potential is a scalar, so you perform algebraic addition (including signs of the charges). The distances used are the absolute distances from the charge to the point of interest. For points on the axis of a dipole, be careful to add or subtract distances correctly from the center.

Step 1: Understanding the Question:

The diagram shows a current configuration consisting of a very long straight wire and a semi-circular arc. We need to find the net magnetic field at point P, which is the center of the semi-circle. The wording "A very long conducting wire is bent..." is ambiguous. A more plausible interpretation, matching the options, is that the total field is the superposition of the field from a long straight wire and the field from the semi-circular arc.


Step 2: Key Formula or Approach:

We use the principle of superposition. The total magnetic field \(\vec{B}_{net}\) is the vector sum of the magnetic field from the straight wire (\(\vec{B}_{straight}\)) and the semi-circular arc (\(\vec{B}_{arc}\)).

1. Field from a long straight wire: At a perpendicular distance \(R\), the magnetic field is \(B_{straight} = \frac{\mu_0 i}{2\pi R}\).

2. Field from a semi-circular arc: At its center, the magnetic field is \(B_{arc} = \frac{1}{2} \left( \frac{\mu_0 i}{2R} \right) = \frac{\mu_0 i}{4R}\).

The direction of each field is found using the right-hand rule.


Step 3: Detailed Explanation:

Let's analyze the contribution from each part based on the diagram:

1. The very long straight wire (top part):

- The current \(i\) flows to the left.
- Point P is at a perpendicular distance \(R\) from this wire.
- Using the right-hand grip rule (point your thumb in the direction of the current, i.e., to the left), your fingers curl such that the magnetic field at point P (which is below the wire in the plane of the page) points into the page.
- Magnitude: \(B_{straight} = \frac{\mu_0 i}{2\pi R}\).


2. The semi-circular arc (A to B):

- The current flows from A to B, which is a counter-clockwise direction.
- Using the right-hand curl rule (curl the fingers of your right hand in the direction of the current flow around the arc), your thumb points away from the page (outward).
- Magnitude: \(B_{arc} = \frac{\mu_0 i}{4R}\).


3. Net Field:

The two fields are in opposite directions. Let's take the direction "away from the page" (outward) as positive.
\[ B_{net} = B_{arc} - B_{straight} \] \[ B_{net} = \frac{\mu_0 i}{4R} - \frac{\mu_0 i}{2\pi R} \]
Factor out the common term \(\frac{\mu_0 i}{4R}\):
\[ B_{net} = \frac{\mu_0 i}{4R} \left( 1 - \frac{4R}{2\pi R} \right) = \frac{\mu_0 i}{4R} \left( 1 - \frac{2}{\pi} \right) \]
To determine the final direction, we compare the magnitudes. \(1 > 2/\pi\) (since \(\pi \approx 3.14\), \(2/\pi \approx 0.637\)). So the net result is positive, which means the direction is away from the page.


Step 4: Final Answer:

The net magnetic field is \( \frac{\mu_0 i}{4R} \left[ 1 - \frac{2}{\pi} \right] \) pointed away from the page.
Quick Tip: When a problem presents a complex shape, always break it down into simpler, standard shapes (like straight lines, arcs, loops) for which you know the magnetic field formulas. Then, apply the principle of superposition, paying close attention to the direction of the field from each part.

Step 1: Understanding the Question:

We are given the position-time (x-t) graph for a particle in Simple Harmonic Motion (SHM) and asked to find its acceleration at a specific time, t = 2 s.


Step 2: Key Formula or Approach:

The acceleration \(a\) of a particle in SHM is related to its displacement \(x\) by the formula:
\[ a = -\omega^2 x \]
where \(\omega\) is the angular frequency. The angular frequency can be found from the time period \(T\) using \(\omega = \frac{2\pi}{T}\). We can determine \(A\) and \(T\) from the graph.


Step 3: Detailed Explanation:

1. Read parameters from the graph:

- Amplitude (A): The maximum displacement from the mean position. From the graph, the maximum value of x is 1 m. So, \(A = 1\) m.

- Time Period (T): The time taken for one complete oscillation. The graph shows one full cycle is completed at t = 8 s. So, \(T = 8\) s.


2. Calculate angular frequency (\(\omega\)):
\[ \omega = \frac{2\pi}{T} = \frac{2\pi}{8} = \frac{\pi}{4} \, rad/s \]

3. Find the displacement at t = 2 s:

From the graph, at \(t = 2\) s, the particle is at its maximum positive displacement.
\[ x(t=2s) = +1 \, m \]

4. Calculate the acceleration at t = 2 s:

Using the formula \(a = -\omega^2 x\):
\[ a(t=2s) = - \left(\frac{\pi}{4}\right)^2 \times (1) \] \[ a(t=2s) = -\frac{\pi^2}{16} \, m/s^2 \]

Step 4: Final Answer:

The acceleration of the particle at t=2 s is \(-\frac{\pi^2}{16}\) m s\(^{-2}\).
Quick Tip: In SHM, acceleration is maximum in magnitude at the extreme positions (where displacement is maximum) and is directed towards the mean position. At the positive extreme (\(x = +A\)), acceleration is maximum negative (\(a = -\omega^2 A\)). At the negative extreme (\(x = -A\)), acceleration is maximum positive (\(a = +\omega^2 A\)).

Step 1: Understanding the Question:

We are given the resistance of a wire at two different temperatures and asked to find its temperature coefficient of resistance, \(\alpha\).


Step 2: Key Formula or Approach:

The relationship between resistance and temperature is given by the formula:
\[ R_T = R_0 (1 + \alpha \Delta T) \]
where \(R_T\) is the resistance at temperature T, \(R_0\) is the resistance at a reference temperature (here, 0°C), \(\alpha\) is the temperature coefficient of resistance, and \(\Delta T\) is the change in temperature.


Step 3: Detailed Explanation:

Given values:

- Resistance at 0°C, \(R_0 = 2 \, \Omega\).

- Resistance at 80°C, \(R_{80} = 6.8 \, \Omega\).

- The change in temperature is \(\Delta T = 80°C - 0°C = 80°C\).


Substitute these values into the formula:
\[ R_{80} = R_0 (1 + \alpha \times 80) \] \[ 6.8 = 2 (1 + 80\alpha) \]
Divide both sides by 2:
\[ 3.4 = 1 + 80\alpha \]
Subtract 1 from both sides:
\[ 2.4 = 80\alpha \]
Solve for \(\alpha\):
\[ \alpha = \frac{2.4}{80} = \frac{24}{800} = \frac{3}{100} = 0.03 \, °C^{-1} \]
Expressing this in scientific notation:
\[ \alpha = 3 \times 10^{-2} \, °C^{-1} \]

Step 4: Final Answer:

The temperature coefficient of resistance of the wire is \(3 \times 10^{-2}\) °C\(^{-1}\).
Quick Tip: The formula \(R_T = R_0(1+\alpha\Delta T)\) is an approximation that works well for metals over a limited temperature range. \(\alpha\) itself can vary slightly with temperature. For exam purposes, assume \(\alpha\) is constant unless stated otherwise.

Step 1: Understanding the Question:

A bullet enters a wooden block and slows down. We are given its initial velocity, its velocity after traveling a certain distance, and that it stops at the end of the block. We need to find the total length of the block, assuming constant resistive force (and thus constant deceleration).


Step 2: Key Formula or Approach:

We can use the third equation of motion, which relates initial velocity (u), final velocity (v), acceleration (a), and displacement (s):
\[ v^2 = u^2 + 2as \]
Alternatively, we can use the work-energy theorem, which states that the work done by the net force is equal to the change in kinetic energy: \(W = \Delta K\). Here, the work is done by the resistive force of the wood.


Step 3: Detailed Explanation (Using Work-Energy Theorem):

Let \(F\) be the constant resistive force exerted by the block on the bullet. Let \(m\) be the mass of the bullet.

Part 1: Bullet travels the first 24 cm.

Initial velocity = \(u\).

Final velocity = \(u/3\).

Distance, \(s_1 = 24\) cm.

Work done by resistive force, \(W_1 = -F \times s_1 = -F \times 24\).

Change in kinetic energy, \(\Delta K_1 = \frac{1}{2}m(\frac{u}{3})^2 - \frac{1}{2}mu^2 = \frac{1}{2}m(\frac{u^2}{9} - u^2) = -\frac{1}{2}m(\frac{8u^2}{9})\).

According to the work-energy theorem, \(W_1 = \Delta K_1\):
\[ -F \times 24 = -\frac{4mu^2}{9} \quad \Rightarrow \quad F \times 24 = \frac{4mu^2}{9} \quad (Equation 1) \]

Part 2: Bullet travels the remaining distance \(s_2\) and stops.

Initial velocity = \(u/3\).

Final velocity = 0.

Distance = \(s_2\).

Work done by resistive force, \(W_2 = -F \times s_2\).

Change in kinetic energy, \(\Delta K_2 = 0 - \frac{1}{2}m(\frac{u}{3})^2 = -\frac{mu^2}{18}\).

According to the work-energy theorem, \(W_2 = \Delta K_2\):
\[ -F \times s_2 = -\frac{mu^2}{18} \quad \Rightarrow \quad F \times s_2 = \frac{mu^2}{18} \quad (Equation 2) \]

Now, divide Equation 2 by Equation 1:
\[ \frac{F \times s_2}{F \times 24} = \frac{mu^2/18}{4mu^2/9} \] \[ \frac{s_2}{24} = \frac{1}{18} \times \frac{9}{4} = \frac{1}{8} \] \[ s_2 = \frac{24}{8} = 3 \, cm \]

The total length of the block is the sum of the two distances:

Total Length = \(s_1 + s_2 = 24 \, cm + 3 \, cm = 27 \, cm\).


Step 4: Final Answer:

The total length of the block is 27 cm.
Quick Tip: The work-energy theorem is often simpler than kinematics when forces and distances are involved, as it bypasses the need to calculate acceleration and time. For constant force, Work \(\propto\) distance and \(\Delta K \propto (v^2 - u^2)\).

Step 1: Understanding the Question:

We are asked to determine the output Y for all possible combinations of inputs A and B for the given logic circuit.


Step 2: Key Formula or Approach:

1. Identify the logic gates in the circuit.
2. Write the Boolean expression for the output Y in terms of the inputs A and B.
3. Construct the truth table based on the Boolean expression.


Step 3: Detailed Explanation:

The circuit consists of two gates:
- A NOT gate with input A. Its output is \(\bar{A}\).
- A NOR gate with two inputs: one from the output of the NOT gate (\(\bar{A}\)) and the other from input B.


The Boolean expression for the output Y of the NOR gate is:
\[ Y = \overline{\bar{A} + B} \]
Using De Morgan's theorem, which states \(\overline{X+Y} = \bar{X} \cdot \bar{Y}\), we can simplify the expression:
\[ Y = \overline{(\bar{A})} \cdot \bar{B} = A \cdot \bar{B} \]
This expression represents the operation "A AND NOT B".


Let's construct the truth table for \(Y = A \cdot \bar{B}\):

When A=0, B=0: \(Y = 0 \cdot \overline{0} = 0 \cdot 1 = 0\)
When A=0, B=1: \(Y = 0 \cdot \overline{1} = 0 \cdot 0 = 0\)
When A=1, B=0: \(Y = 1 \cdot \overline{0} = 1 \cdot 1 = 1\)
When A=1, B=1: \(Y = 1 \cdot \overline{1} = 1 \cdot 0 = 0\)

The resulting truth table is:

\begin{tabular{|c|c|c|
\hline
A & B & Y

\hline
0 & 0 & 0

0 & 1 & 0

1 & 0 & 1

1 & 1 & 0

\hline
\end{tabular



Step 4: Final Answer:

The logical operation of the circuit is \(Y = A \cdot \bar{B}\). The correct truth table for this circuit is option (C) .
Quick Tip: Always double-check the symbols for logic gates. A D-shape is for AND/NAND, while a curved input line is for OR/NOR. In case of discrepancies between your result and the given options in an exam, re-read the question and re-check your work. If the discrepancy persists, it might indicate an error in the question paper.

Step 1: Understanding the Question:

This Assertion-Reason question tests the fundamental concepts of activation energy in chemical kinetics. We need to evaluate the validity of both statements and the causal link between them.


Step 2: Detailed Explanation:

Analysis of Reason R:

Reason R provides the standard definition of activation energy (\(E_a\)). It is the minimum amount of energy that must be provided to reactant molecules to overcome the energy barrier and form the transition state, which then proceeds to products. The total energy required is the threshold energy. This definition is perfectly correct.

Therefore, Reason R is true.


Analysis of Assertion A:

Assertion A claims that a reaction can have zero activation energy. According to the Arrhenius equation, \( k = Ae^{-E_a/RT} \), if \(E_a = 0\), then \( k = A \), meaning the reaction rate is independent of temperature and every collision is effective. While some reactions, particularly the combination of free radicals in the gas phase (e.g., \( CH_3\cdot + CH_3\cdot \rightarrow C_2H_6 \)), have very low or negligible activation energies, the concept of a truly zero activation energy is an idealization. For the purposes of general chemistry curriculum and examinations, it is generally considered that a reaction involves some form of bond rearrangement or formation, which necessitates overcoming an energy barrier, however small. Therefore, the statement that a reaction can have zero activation energy is considered to be false in this context.


Step 3: Final Answer:

Since Assertion A is considered false and Reason R is true, the correct option is (1).
Quick Tip: In chemical kinetics, activation energy is a fundamental concept representing an energy barrier. While some barrierless reactions exist, for exam purposes, assume that reactions generally have a non-zero activation energy unless dealing with specific exceptions like radical recombination. The definition of activation energy (Reason R) is a core concept you must know.

Step 1: Understanding the Question:

The question shows the reaction of a secondary alcohol, 3-methylbutan-2-ol, with hydrogen bromide (HBr). We need to predict the major product (P) of this reaction. This is a nucleophilic substitution reaction.


Step 2: Key Formula or Approach:

The reaction of an alcohol with HBr proceeds via an S\(_N\)1 mechanism, especially for secondary and tertiary alcohols, which involves the formation of a carbocation intermediate. Carbocation intermediates can undergo rearrangement to form a more stable carbocation. The order of carbocation stability is: tertiary (\(3^{\circ}\)) \(>\) secondary (\(2^{\circ}\)) \(>\) primary (\(1^{\circ}\)).


Step 3: Detailed Explanation:

Mechanism:

1. Protonation of the alcohol: The lone pair of electrons on the oxygen atom of the hydroxyl group attacks the proton (H\(^+\)) from HBr, forming a protonated alcohol (an oxonium ion). This makes the hydroxyl group a good leaving group (water).
\[ CH_3-\underset{CH_3}{\underset{|}{CH}}-\underset{OH}{\underset{|}{CH}}-CH_3 + H^+ \rightleftharpoons CH_3-\underset{CH_3}{\underset{|}{CH}}-\underset{OH_2^+}{\underset{|}{CH}}-CH_3 \]
2. Formation of carbocation: The C-O bond breaks, and the water molecule leaves, resulting in the formation of a secondary (\(2^{\circ}\)) carbocation.
\[ CH_3-\underset{CH_3}{\underset{|}{CH}}-\underset{OH_2^+}{\underset{|}{CH}}-CH_3 \rightarrow CH_3-\underset{CH_3}{\underset{|}{CH}}-\underset{+}{\underset{|}{CH}}-CH_3 + H_2O \]
(This is 3-methylbutan-2-yl cation, a \(2^{\circ}\) carbocation)

3. Carbocation rearrangement: The secondary carbocation can rearrange to a more stable tertiary (\(3^{\circ}\)) carbocation. A hydrogen atom from the adjacent carbon (C-3) shifts with its pair of electrons to the positively charged carbon (C-2). This is called a 1,2-hydride shift.
\[ \underset{\(2^{\circ\) Carbocation (less stable)}}{CH_3-\underset{H}{\underset{|}{\overset{CH_3}{\overset{|}{C}}}}-\overset{+}{C}H-CH_3} \xrightarrow{1,2-Hydride shift} \underset{\(3^{\circ\) Carbocation (more stable)}}{CH_3-\overset{+}{\underset{CH_3}{\underset{|}{C}}}-CH_2-CH_3} \]
4. Nucleophilic attack: The bromide ion (Br\(^-\)), which is a good nucleophile, attacks the more stable tertiary carbocation to form the final product.
\[ CH_3-\overset{+}{\underset{CH_3}{\underset{|}{C}}}-CH_2-CH_3 + Br^- \rightarrow CH_3-\underset{Br}{\underset{|}{\overset{CH_3}{\overset{|}{C}}}}-CH_2-CH_3 \]

Step 4: Final Answer:

The major product formed is 2-bromo-2-methylbutane. This corresponds to option (2). Option (4) would be the product formed without rearrangement, which is the minor product.
Quick Tip: Whenever a reaction involves a carbocation intermediate (like S\(_N\)1, E1, or acid-catalyzed hydration/dehydration), always check for the possibility of rearrangement (1,2-hydride or 1,2-methyl shift) to form a more stable carbocation. The major product will always arise from the most stable carbocation.

Step 1: Understanding the Question:

We need to compare the sizes of the stable ions formed by Na, O, F, and N and identify which one is the largest.


Step 2: Key Formula or Approach:

1. First, determine the stable ion each element forms to achieve a noble gas electron configuration.
2. The species formed are Na\(^+\), O\(^{2-}\), F\(^-\), and N\(^{3-}\). Notice that all these ions have 10 electrons, making them an isoelectronic series.
3. For isoelectronic species (ions with the same number of electrons), the ionic radius decreases as the nuclear charge (atomic number, Z) increases. This is because a stronger pull from the nucleus on the same number of electrons contracts the electron cloud.


Step 3: Detailed Explanation:

- Na (Z=11) loses one electron to form Na\(^+\) (10 e\(^-\)).
- O (Z=8) gains two electrons to form O\(^{2-}\) (10 e\(^-\)).
- F (Z=9) gains one electron to form F\(^-\) (10 e\(^-\)).
- N (Z=7) gains three electrons to form N\(^{3-}\) (10 e\(^-\)).


All ions have 10 electrons. We now compare their nuclear charges (Z):
- N: Z=7
- O: Z=8
- F: Z=9
- Na: Z=11

The ion with the lowest nuclear charge will have the weakest attraction for the 10 electrons, resulting in the largest electron cloud and thus the largest ionic radius.

The order of nuclear charge is N < O < F < Na.

Therefore, the order of ionic size will be the reverse: N\(^{3-}\) > O\(^{2-}\) > F\(^-\) > Na\(^+\).

The largest ion is N\(^{3-}\), which is formed from the element Nitrogen (N).


Step 4: Final Answer:

Nitrogen (N) is the element that forms the largest ion among the given options.
Quick Tip: For isoelectronic species, the rule is simple: higher nuclear charge (more protons) = smaller radius. Anions are always larger than their parent atoms, and cations are always smaller.

Step 1: Understanding the Question:

The question asks us to identify the correct statements among the five given options related to atomic structure and theory.


Step 2: Detailed Explanation:

Let's evaluate each statement for its correctness.


Statement A: "Atoms of all elements are composed of two fundamental particles." This is incorrect. Atoms are composed of three fundamental particles: protons, neutrons, and electrons. An exception is the protium isotope of hydrogen (\(^{1}H\)), which has one proton and one electron but no neutron, but the statement refers to "all elements".

Statement B: "The mass of the electron is \(9.10939 \times 10^{-31}\) kg." This is a factual statement and is correct. The accepted value for the rest mass of an electron is approximately \(9.1093837 \times 10^{-31}\) kg, so the given value is accurate for exam purposes.

Statement C: "All the isotopes of a given element show same chemical properties." This is correct. Isotopes of an element have the same number of protons and, in a neutral atom, the same number of electrons. Since chemical properties are primarily determined by the electron configuration, isotopes exhibit identical chemical behavior. They differ only in the number of neutrons, which affects their mass and nuclear properties.

Statement D: "Protons and electrons are collectively known as nucleons." This is incorrect. Nucleons are the particles that reside in the atomic nucleus. These are protons and neutrons. Electrons orbit the nucleus.

Statement E: "Dalton's atomic theory, regarded the atom as an ultimate particle of matter." This is correct. A fundamental postulate of John Dalton's atomic theory was that atoms are indivisible and indestructible fundamental particles of matter. Although we now know atoms are divisible, this was a cornerstone of his original theory.



Step 3: Final Answer:

The correct statements are B, C, and E. Therefore, the correct option is (1).
Quick Tip: For questions involving multiple statements, evaluate each one individually as true or false. This systematic approach helps eliminate incorrect options and pinpoint the correct combination. Pay close attention to absolute words like "all" or "always".

Step 1: Understanding the Question:

The question asks to identify the single correct statement among the four options concerning the biological roles and requirements of Magnesium (Mg) and Calcium (Ca).


Step 2: Detailed Explanation:

Let's evaluate each statement:


(1) Mg plays roles in neuromuscular function and interneuronal transmission. This statement is correct. Mg\(^{2+}\) ions act as a physiological antagonist to Ca\(^{2+}\) ions at the neuromuscular junction and are crucial for nerve impulse transmission and muscle relaxation.

(2) The daily requirement of Mg and Ca in the human body is estimated to be 0.2 - 0.3 g. This statement is presented as the correct answer. The value 0.2 - 0.3 g (or 200 - 300 mg) corresponds to the estimated daily requirement for Magnesium. However, the daily requirement for Calcium is much higher, around 1.0 - 1.2 g (1000 - 1200 mg). The phrasing "Mg and Ca" makes the statement scientifically imprecise. However, in the context of multiple-choice questions based on specific textbook lines (like NCERT), which states "The daily requirement in the human body has been estimated to be 200 – 300 mg" in a section discussing both elements, this option may be considered correct by the exam setters.

(3) All enzymes that utilise ATP in phosphate transfer require Ca as the cofactor. This statement is incorrect. The vast majority of enzymes that use ATP for phosphate transfer, such as kinases, require Magnesium (Mg\(^{2+}\)) as a cofactor. Mg\(^{2+}\) forms a complex with ATP (MgATP\(^{2-}\)) which is the true substrate for these enzymes.

(4) The bone in human body is an inert and unchanging substance. This statement is incorrect. Bone is a dynamic, living tissue that is constantly undergoing remodeling (resorption and formation). It also serves as the body's primary reservoir for calcium and phosphate.



Step 3: Final Answer:

While statement (1) is biologically accurate, the provided answer key points to (2). The justification for (2) relies on the fact that the given range (200-300 mg) is the approximate daily requirement for Mg, and the question might be flawed by including Ca. Given the clear incorrectness of options (3) and (4), and the potential ambiguity in interpreting the question's intent for (2), it is selected as the intended answer.
Quick Tip: When answering biology-related chemistry questions, be aware that exam questions can sometimes be based on specific, and occasionally ambiguous, statements from prescribed textbooks. While Ca\(^{2+}\) is famous for muscle contraction and nerve signals, Mg\(^{2+}\) is the key cofactor for almost all ATP-related enzymes.

Step 1: Understanding the Question:

The question asks to identify the product (A) of a reaction involving a diketone with zinc amalgam (Zn-Hg) and concentrated hydrochloric acid (conc. HCl).


Step 2: Key Formula or Approach:

The reagent system, Zn-Hg / conc. HCl, is used for the Clemmensen reduction. This reaction specifically reduces a carbonyl group (C=O) of an aldehyde or a ketone to a methylene group (-CH\(_2\)-), effectively converting the carbonyl compound into an alkane. It does not reduce carboxylic acids or their derivatives and does not affect carbon-carbon double or triple bonds.
\[ R-CO-R' \xrightarrow{Zn-Hg, conc. HCl} R-CH_2-R' \]

Step 3: Detailed Explanation:

The starting material has two ketone groups:

1. An acetyl group (\(-COCH_3\)) attached to the benzene ring.

2. A carbonyl group within the cyclohexanone ring.


The Clemmensen reduction will reduce both of these carbonyl groups.


The acetyl group (\(-C(=O)CH_3\)) will be reduced to an ethyl group (\(-CH_2CH_3\)).

The carbonyl group in the cyclohexanone ring will be reduced to a methylene group (\(-CH_2-\)), converting the cyclohexanone ring into a cyclohexane ring.


The overall transformation is from 1-(4-acetylphenyl)cyclohexan-1-one to 1-ethyl-4-cyclohexylbenzene.


Step 4: Final Answer:

Let's examine the options, which represent the structures shown in the image:


Option (1) corresponds to 1-Ethyl-4-(1-hydroxycyclohexyl)benzene, which is incorrect.

Option (2) corresponds to 1-Cyclohexyl-4-ethylbenzene, the correct product of a double Clemmensen reduction.

Option (3) corresponds to 1-(4-(1-hydroxyethyl)phenyl)cyclohexan-1-ol, which is a diol, a typical result from reagents like NaBH\(_4\) or LiAlH\(_4\), not Clemmensen reduction.

Option (4) shows incorrect reduction products.


Thus, the correct product corresponds to structure (2).
Quick Tip: Recognizing named reactions and their specific reagents is key in organic chemistry. Clemmensen (Zn-Hg/HCl) and Wolff-Kishner (N\(_2\)H\(_4\)/KOH) both reduce C=O to CH\(_2\). Remember that Clemmensen reduction uses acidic conditions, while Wolff-Kishner uses basic conditions. Choose the reagent based on the presence of other acid- or base-sensitive groups in the molecule.

Step 1: Understanding the Question:

We are given the conductivity (\(\kappa\)) and resistance (R) of an electrolyte solution in a conductivity cell and asked to calculate the cell constant (\(G^*\)).


Step 2: Key Formula or Approach:

The relationship between conductivity (\(\kappa\)), resistance (R), and the cell constant (\(G^*\)) is given by the formula:
\[ \kappa = \frac{1}{R} \times G^* \]
The cell constant is a property of the conductivity cell, defined as the ratio of the distance between the electrodes (l) to their area of cross-section (A), i.e., \(G^* = l/A\).

Rearranging the formula to find the cell constant:
\[ G^* = \kappa \times R \]

Step 3: Detailed Explanation:

Given values:

- Conductivity, \(\kappa = 0.0210 \, \Omega^{-1} \, cm^{-1}\).

- Resistance, \(R = 60 \, \Omega\).


Substitute these values into the rearranged formula:
\[ G^* = (0.0210 \, \Omega^{-1} \, cm^{-1}) \times (60 \, \Omega) \] \[ G^* = 1.26 \, cm^{-1} \]
The ohm (\(\Omega\)) and inverse ohm (\(\Omega^{-1}\)) units cancel out, leaving the unit for the cell constant as cm\(^{-1}\).


Step 4: Final Answer:

The value of the cell constant is 1.26 cm\(^{-1}\).
Quick Tip: Remember the fundamental relationships in conductivity: - Resistance \(R = \rho \frac{l}{A}\) - Conductance \(G = \frac{1}{R} = \kappa \frac{A}{l}\) - Conductivity \(\kappa = \frac{1}{\rho}\) From these, you can derive the key formula used here: \(\kappa = G \times \frac{l}{A} = \frac{1}{R} \times G^*\).

Step 1: Understanding the Question:

The question asks to identify the chemical species responsible for the blood-red coloration observed in Lassaigne's test when both nitrogen and sulphur are present in the organic compound.


Step 2: Key Formula or Approach:

Lassaigne's test involves fusing the organic compound with sodium metal to convert covalently bonded elements like N, S, and halogens into ionic sodium salts.

1. If both N and S are present: \(Na + C + N + S \xrightarrow{\Delta} NaSCN\) (Sodium thiocyanate).

2. The sodium extract is then treated with a neutral or slightly acidic solution of Ferric chloride (FeCl\(_3\)), which provides Fe\(^{3+}\) ions.

3. The Fe\(^{3+}\) ions react with the thiocyanate ions (SCN\(^-\)) to form a complex ion which has a characteristic blood-red color.


Step 3: Detailed Explanation:

The reaction sequence is as follows:

- Fusion: Organic Compound (containing C, N, S) + Na \(\rightarrow\) NaSCN

- Test: The aqueous extract containing SCN\(^-\) ions is treated with Fe\(^{3+}\).
\[ Fe^{3+} (aq) + SCN^{-} (aq) \rightarrow [Fe(SCN)(H_2O)_5]^{2+} (aq) \]
This complex ion, ferric thiocyanate (or more accurately, pentaaquathiocyanatoiron(III)), is responsible for the intense blood-red color. For simplicity, it is often written as \([Fe(SCN)]^{2+}\).

Let's analyze the other options:
- (B) \(Fe_4[Fe(CN)_6]_3\): This is Ferric ferrocyanide (Prussian blue), which is formed when only nitrogen is present, not when both N and S are present.
- (C) NaSCN: This is the salt formed in the sodium extract, but it is colorless. The color appears only after reacting with Fe\(^{3+}\).
- (D) \([Fe(CN)_5 NOS]^{4-}\): This is the sodium nitroprusside complex, which is used to test for sulphur (as sulphide ions), not for the combined presence of N and S.


Step 4: Final Answer:

The blood-red color is due to the formation of the \([Fe(SCN)]^{2+}\) complex.
Quick Tip: Remember the characteristic colors in Lassaigne's test: - N only: Prussian blue with Fe\(^{2+}\)/Fe\(^{3+}\). - S only: Violet color with sodium nitroprusside. - N and S together: Blood-red color with Fe\(^{3+}\).

Step 1: Understanding the Question:

We need to evaluate an assertion and a reason related to the use of helium in deep-sea diving equipment.


Step 2: Detailed Explanation:

Analysis of Assertion A:

"Helium is used to dilute oxygen in diving apparatus."

This is a factually correct statement. For deep-sea diving, a mixture of helium and oxygen (called heliox) is used instead of compressed air (nitrogen and oxygen). This is done to prevent a condition called nitrogen narcosis, which occurs at high pressures when divers breathe nitrogen.

Conclusion: Assertion A is true.


Analysis of Reason R:

"Helium has high solubility in O\(_2\)."

This statement is irrelevant to the application. The important property is the solubility of the diluent gas (helium or nitrogen) in the diver's blood under pressure. The primary reason for using helium is its very low solubility in blood compared to nitrogen. When a diver ascends, the pressure decreases, and dissolved gases can come out of solution to form bubbles in the bloodstream, leading to a painful and dangerous condition called decompression sickness or "the bends". Because helium is much less soluble in blood, the risk of the bends is significantly reduced. Therefore, the statement that helium is used because it has *high solubility* is incorrect in context and factually wrong regarding its key property.

Conclusion: Reason R is false.


Step 3: Final Answer:

Since Assertion A is true and Reason R is false, the correct option is (4).
Quick Tip: The use of helium in deep-sea diving is a classic example demonstrating the application of Henry's Law, which states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. The key to avoiding "the bends" is using a breathing gas that has low solubility in blood.

Step 1: Understanding the Question:

The question asks to identify the statements that are incorrect regarding hydrogen and its compounds.


Step 2: Detailed Explanation:

Let's analyze each statement:


A. Hydrogen is used to reduce heavy metal oxides to metals. This statement is correct. Hydrogen is a good reducing agent and is used in metallurgy to reduce oxides of less reactive metals like copper, lead, and zinc to their respective metals (e.g., \( CuO + H_2 \rightarrow Cu + H_2O \)).

B. Heavy water is used to study reaction mechanism. This statement is correct. Heavy water (D\(_2\)O) is used as a tracer to study the mechanisms of chemical and biological reactions. The different mass of deuterium compared to hydrogen can lead to a kinetic isotope effect, which provides insights into reaction pathways.

C. Hydrogen is used to make saturated fats from oils. This statement is correct. The process, known as hydrogenation, involves adding hydrogen across the double bonds of unsaturated fats (in oils) using a catalyst (like Ni, Pd, or Pt) to produce saturated fats (like vanaspati ghee or margarine).

D. The H-H bond dissociation enthalpy is lowest as compared to a single bond between two atoms of any element. This statement is incorrect. The H-H bond dissociation enthalpy (\( \approx 436 kJ/mol \)) is actually the highest for a single bond between two atoms of any element. For example, the C-C bond enthalpy is \( \approx 348 kJ/mol \), and the F-F bond enthalpy is \( \approx 159 kJ/mol \).

E. Hydrogen reduces oxides of metals that are more active than iron. This statement is incorrect. Hydrogen can only reduce the oxides of metals that are less reactive (less electropositive) than itself. The reactivity series is K \(>\) Na \(>\) Ca \(>\) Mg \(>\) Al \(>\) Zn \(>\) Fe \(>\) H \(>\) Cu \(>\) Ag \(>\) Au. Since metals like Al, Zn, and Fe are more reactive than hydrogen, hydrogen cannot reduce their oxides under standard conditions.



Step 3: Final Answer:

Statements D and E are incorrect. Therefore, the correct option is (4).
Quick Tip: Be careful with keywords like "NOT correct," "incorrect," or "false" in the question stem. After analyzing all statements, re-read the question to ensure you are selecting the incorrect ones. Remember the position of Hydrogen in the electrochemical/reactivity series to answer questions about its reducing properties.

Step 1: Understanding the Question:

We are given a multi-step reaction starting from benzenediazonium chloride and need to identify the final product.


Step 2: Key Formula or Approach:

We need to analyze each step of the reaction sequence:
- Step (i): Sandmeyer Reaction: Reaction of a diazonium salt with cuprous halide (Cu\(_2\)X\(_2\)) in the presence of the corresponding halogen acid (HX). This reaction replaces the diazonium group (\(-N_2^+\)) with a halogen atom (\(-X\)).
- Step (ii): Grignard Reagent Formation: Reaction of an aryl halide with magnesium metal in dry ether.
- Step (iii): Reaction of Grignard Reagent with Water: Grignard reagents are strong bases and react readily with any source of protons (like water) to form a hydrocarbon.


Step 3: Detailed Explanation:

Step (i): The starting material is benzenediazonium chloride. It is treated with Cu\(_2\)Br\(_2\)/HBr. This is a Sandmeyer reaction which will replace the \(-N_2^+Cl^-\) group with a \(-Br\) atom. \[ C_6H_5N_2^+Cl^- \xrightarrow{Cu_2Br_2/HBr} C_6H_5Br + N_2 + CuCl \]
The product of the first step is bromobenzene.


Step (ii): Bromobenzene is treated with Mg in dry ether. This reaction forms a Grignard reagent. \[ C_6H_5Br + Mg \xrightarrow{dry ether} C_6H_5MgBr \]
The product is phenylmagnesium bromide.


Step (iii): Phenylmagnesium bromide is treated with water (H\(_2\)O). The Grignard reagent is a very strong base (the C\(_6\)H\(_5\)\(^-\) part is a carbanion). It will abstract a proton from water. \[ C_6H_5MgBr + H_2O \rightarrow C_6H_6 + Mg(OH)Br \]
The organic product is benzene (C\(_6\)H\(_6\)).


Looking at the options, option (3) represents the structure of benzene.


Step 4: Final Answer:

The final product of the reaction sequence is benzene.
Quick Tip: Grignard reagents are extremely useful but must be handled under anhydrous (dry) conditions. This is because they react with any protic solvent (water, alcohols, etc.) to form the corresponding alkane or arene, which is often an unwanted side reaction but is the main reaction in this specific problem.

Step 1: Understanding the Question:

The question asks to identify which of the given tranquilizers is a member of the barbiturate class of drugs.


Step 2: Detailed Explanation:

Tranquilizers are neurological drugs used to treat anxiety, fear, tension, and disturbances of the mind. They are classified into different chemical groups. Let's analyze the given options:


(1) Veronal: Veronal is the trade name for barbital. It is a derivative of barbituric acid and was one of the first commercially available barbiturate hypnotics. Therefore, Veronal is a barbiturate.

(2) Chlordiazepoxide: This is sold under the trade name Librium. It belongs to the benzodiazepine class of drugs, not barbiturates.

(3) Meprobamate: This is sold under trade names like Miltown and Equanil. It is a carbamate derivative and is classified as a non-barbiturate anxiolytic.

(4) Valium: Valium is the well-known trade name for Diazepam. It is a classic example of a benzodiazepine, a different class of tranquilizers from barbiturates.



Step 3: Final Answer:

Among the given options, only Veronal is a barbiturate.
Quick Tip: In the chapter "Chemistry in Everyday Life", it's important to memorize the classification of common drugs and at least one or two examples from each class. For tranquilizers, remember the main classes: barbiturates (e.g., Veronal, Luminal) and benzodiazepines (e.g., Valium, Librium).

Step 1: Understanding the Question:

The question presents two statements related to the fundamental components of nucleic acids, nucleosides and nucleotides. We need to evaluate the correctness of each statement.


Step 2: Detailed Explanation:

Analysis of Statement I:

A nucleoside is a glycosylamine that consists of a nitrogenous base (a purine like Adenine or Guanine, or a pyrimidine like Cytosine, Thymine, or Uracil) linked to a sugar (ribose or deoxyribose). The bond forms between the anomeric carbon of the sugar (C1') and a nitrogen atom of the base. This definition matches Statement I exactly.

Therefore, Statement I is true.


Analysis of Statement II:

A nucleotide is formed when a phosphate group is attached to a nucleoside. Specifically, a phosphate group derived from phosphoric acid (H\(_3\)PO\(_4\)) is linked to the 5'-hydroxyl group of the sugar moiety of a nucleoside via a phosphoester bond. Statement II incorrectly states that the linkage is to phosphorous acid (H\(_3\)PO\(_3\)). Phosphorous acid and phosphoric acid are different chemical compounds.

Therefore, Statement II is false.


Step 3: Final Answer:

Since Statement I is true and Statement II is false, the correct option is (4).
Quick Tip: To remember the difference: a nucleo\textbf{s}ide is just \textbf{s}ugar + base. A nucleo\textbf{t}ide has an extra component, the phospha\textbf{t}e group. Pay close attention to chemical names; "phosphoric acid" and "phosphorous acid" are not interchangeable.

Step 1: Understanding the Question:

We need to determine the count of sigma bonds, pi bonds, and lone pairs of electrons in a molecule of pyridine.


Step 2: Key Formula or Approach:

First, we must know the structure of pyridine. Pyridine (C\(_5\)H\(_5\)N) is a six-membered heterocyclic aromatic compound, similar to benzene but with one CH group replaced by a nitrogen atom. We then count the bonds and lone pairs directly from the structure.
- Every single bond is one \(\sigma\) bond.
- Every double bond consists of one \(\sigma\) bond and one \(\pi\) bond.
- Every triple bond consists of one \(\sigma\) bond and two \(\pi\) bonds.
- We count the non-bonding valence electron pairs as lone pairs.


Step 3: Detailed Explanation:

The structure of pyridine is a hexagonal ring with alternating double bonds. The ring contains 5 carbon atoms and 1 nitrogen atom. Each of the 5 carbon atoms is bonded to one hydrogen atom.
Let's count the bonds:
1. \(\sigma\) bonds:
- There are 5 C-H single bonds. (5 \(\sigma\) bonds)
- Within the ring, there are bonds between the 6 atoms (4 C-C bonds and 2 C-N bonds). These form the framework of the ring. So there are 6 \(\sigma\) bonds within the ring.
- Total \(\sigma\) bonds = 5 (C-H) + 6 (in-ring) = 11 \(\sigma\) bonds.

2. \(\pi\) bonds:
- Pyridine is an aromatic system, analogous to benzene. There are 3 delocalized \(\pi\) bonds within the ring. ( 3 \(\pi\) bonds).

3. Lone pairs:
- Each carbon atom uses all 4 of its valence electrons in bonding (2 in the ring, 1 with H, 1 in the pi system). So, no lone pairs on carbon.
- The nitrogen atom (Group 15) has 5 valence electrons. It uses one electron for a \(\sigma\) bond with one carbon, one electron for a \(\sigma\) bond with another carbon, and one electron for the \(\pi\) system. This leaves two electrons as a non-bonding pair.
- Total lone pairs = 1 lone pair on the nitrogen atom.


The final count is 11 \(\sigma\) bonds, 3 \(\pi\) bonds, and 1 lone pair.


Step 4: Final Answer:

The number of \(\sigma\) bonds, \(\pi\) bonds, and lone pairs are 11, 3, and 1, respectively.
Quick Tip: For cyclic compounds, a quick way to count \(\sigma\) bonds is to count all the atoms in the molecule and add the number of rings, then subtract 1. Pyridine has 11 atoms (5C, 5H, 1N) and 1 ring. So, \(\sigma\) bonds = (11+1)-1 = 11. This shortcut works for many structures but always double-check by drawing.

Step 1: Understanding the Question:

We need to identify which of the given reactions does not yield a primary amine as the main product.


Step 2: Key Formula or Approach:

We need to know the outcome of several named organic reactions for the synthesis of amines.
- Reduction of amides: LiAlH\(_4\) reduces the carbonyl group (C=O) of an amide to a methylene group (CH\(_2\)).
- Hoffmann bromamide degradation: An amide is treated with Br\(_2\) and KOH to produce a primary amine with one less carbon atom.
- Reduction of nitriles (cyanides): LiAlH\(_4\) reduces the C\(\equiv\)N group to a CH\(_2\)NH\(_2\) group.
- Reduction of isonitriles (isocyanides): LiAlH\(_4\) reduces the N\(\equiv\)C group to an NHCH\(_3\) group.


Step 3: Detailed Explanation:

Let's analyze each reaction:

1. CH\(_3\)CONH\(_2\) + LiAlH\(_4\): This is the reduction of acetamide. The C=O group is reduced to CH\(_2\).
\[ CH_3CONH_2 \xrightarrow{LiAlH_4} CH_3CH_2NH_2 \]
The product is ethylamine, which is a primary amine.

2. CH\(_3\)CONH\(_2\) + Br\(_2\)/KOH: This is the Hoffmann bromamide degradation of acetamide. The amide is converted to a primary amine with one carbon atom less.
\[ CH_3CONH_2 \xrightarrow{Br_2/KOH} CH_3NH_2 \]
The product is methylamine, which is a primary amine.

3. CH\(_3\)CN + LiAlH\(_4\): This is the reduction of acetonitrile (a nitrile). The triple bond is fully reduced.
\[ CH_3CN \xrightarrow{LiAlH_4} CH_3CH_2NH_2 \]
The product is ethylamine, which is a primary amine.

4. CH\(_3\)NC + LiAlH\(_4\): This is the reduction of methyl isocyanide (an isonitrile).
\[ CH_3NC \xrightarrow{LiAlH_4} CH_3NHCH_3 \]
The product is dimethylamine, which is a secondary amine.


Therefore, the reduction of an isonitrile (isocyanide) does not produce a primary amine.


Step 4: Final Answer:

The reaction of CH\(_3\)NC with LiAlH\(_4\) will NOT give a primary amine as the product.
Quick Tip: A key distinction for amine synthesis: - Reduction of \textbf{nitriles} (-C\(\equiv\)N) gives \textbf{primary} amines (-CH\(_2\)NH\(_2\)). - Reduction of \textbf{isonitriles} (-N\(\equiv\)C) gives \textbf{secondary} amines (-NHCH\(_3\)).

Step 1: Understanding the Question:

The question asks for the correct increasing order of energy for the molecular orbitals (MOs) of the nitrogen molecule (N\(_2\)).


Step 2: Key Formula or Approach:

According to Molecular Orbital Theory (MOT), the order of energy levels of MOs for diatomic molecules of the second period elements depends on s-p mixing.


For diatomic molecules up to N\(_2\) (i.e., Li\(_2\), Be\(_2\), B\(_2\), C\(_2\), N\(_2\)), significant s-p mixing occurs. This mixing raises the energy of the \(\sigma2p_z\) orbital above that of the \(\pi2p_x\) and \(\pi2p_y\) orbitals.

For diatomic molecules after N\(_2\) (i.e., O\(_2\), F\(_2\), Ne\(_2\)), the energy gap between 2s and 2p atomic orbitals is larger, so s-p mixing is less effective. The \(\sigma2p_z\) orbital remains below the \(\pi2p_x\) and \(\pi2p_y\) orbitals.



Step 3: Detailed Explanation:

Since the question is about the N\(_2\) molecule (total 14 electrons), we must use the energy order that accounts for s-p mixing.

The correct sequence of molecular orbitals in increasing order of energy is:
\[ \sigma1s < \sigma^*1s < \sigma2s < \sigma^*2s < (\pi2p_x = \pi2p_y) < \sigma2p_z < (\pi^*2p_x = \pi^*2p_y) < \sigma^*2p_z \]
The \(\pi2p_x\) and \(\pi2p_y\) orbitals are degenerate (have the same energy), as are the \(\pi^*2p_x\) and \(\pi^*2p_y\) orbitals.


Step 4: Final Answer:

Comparing this correct order with the given options:


Option (A) incorrectly places \( (\pi^*2p_x = \pi^*2p_y) \) before \( \sigma2p_z \).

Option (B) matches the correct energy order for N\(_2\).

Option (C) shows the order for O\(_2\) and F\(_2\), where \( \sigma2p_z \) is lower in energy than \( \pi2p_x \) and \( \pi2p_y \).

Option (D) shows a completely incorrect sequence.


Therefore, the correct order is given in option (2).
Quick Tip: A simple mnemonic to remember the MO energy order: For N\(_2\) and lighter diatomic molecules (14 or fewer electrons), the order is "pi-sigma" for the 2p bonding orbitals (\(\pi_{2p} < \sigma_{2p}\)). For O\(_2\) and heavier ones (more than 14 electrons), the order is "sigma-pi" (\(\sigma_{2p} < \pi_{2p}\)). This is a crucial distinction and a frequently tested concept.

Step 1: Understanding the Question:

We need to identify the graph that correctly represents Boyle's Law, including the effect of temperature on the graph.


Step 2: Key Formula or Approach:

Boyle's Law states that for a fixed mass of an ideal gas at constant temperature, the pressure (P) is inversely proportional to the volume (V). \[ P \propto \frac{1}{V} \quad or \quad PV = constant \, (k) \]
From the ideal gas equation, \(PV = nRT\), the constant \(k = nRT\). This shows that the value of the constant is directly proportional to the absolute temperature (T).


Step 3: Detailed Explanation:

Let's analyze the types of plots for Boyle's Law:
- P vs. V plot: A plot of P versus V should be a rectangular hyperbola. The curves are called isotherms. Since \(k = nRT\), a higher temperature corresponds to an isotherm that is further away from the origin. Graph (2) shows hyperbolas, but the isotherm T\(_3\) (highest temperature) is closest to the origin, which is incorrect.
- P vs. 1/V plot: The relationship is \(P = k \times (\frac{1}{V})\). This is of the form \(y = mx\), which represents a straight line passing through the origin. The slope of the line is \(m = k = nRT\).
- This means the slope of the P vs. 1/V graph is directly proportional to the temperature T.
- A higher temperature will result in a line with a greater slope.


Now let's examine the relevant graphs:
- Graph (3): This shows plots of P vs 1/V. They are straight lines passing through the origin, which is correct. The lines are for temperatures T\(_1\), T\(_2\), and T\(_3\). The slope of the line for T\(_3\) is the steepest, followed by T\(_2\), and then T\(_1\). This corresponds to the relationship \(slope_{T3} > slope_{T2} > slope_{T1}\), which correctly implies \(T_3 > T_2 > T_1\). This graph is a correct representation.
- Graph (4): This is a P vs. 1/V plot, but the curves are hyperbolic, which is incorrect. It should be a straight line.

- Graph (1): This is a P vs T plot, which represents Gay-Lussac's Law (at constant volume).


Step 4: Final Answer:

Graph (3) is the correct graphical representation of Boyle's Law.
Quick Tip: To analyze gas law graphs, always relate the plotted variables to the ideal gas equation \(PV=nRT\). Rearrange the equation to match the form of a straight line (\(y=mx+c\)) if possible. This helps in correctly interpreting the slope and intercept and their dependence on other variables like temperature.

Step 1: Understanding the Question:

The question asks for the final product [C] of a two-step reaction sequence starting from cyclohexanone [A].


Step 2: Key Formula or Approach:

The reaction sequence involves two key transformations:
1. Cyanohydrin formation: A ketone reacts with HCN to form a cyanohydrin.
2. Acid hydrolysis and dehydration: The cyanohydrin is treated with concentrated acid (H\(_2\)SO\(_4\)) and heat. The nitrile group (-CN) hydrolyzes to a carboxylic acid (-COOH), and the tertiary alcohol group (-OH) undergoes dehydration to form an alkene.


Step 3: Detailed Explanation:

Step I: Formation of Cyanohydrin [B]

The carbonyl group of cyclohexanone [A] is attacked by the nucleophilic cyanide ion (from HCN) to form cyclohexanone cyanohydrin [B].
\[ [A: Cyclohexanone] \xrightarrow{HCN} [B: 1-hydroxycyclohexanecarbonitrile] \]
The structure of [B] has both a hydroxyl (-OH) group and a nitrile (-CN) group attached to the same carbon atom (C1) of the ring.


Step II: Formation of Product [C]

Product [B] is heated with concentrated sulfuric acid. Two reactions occur simultaneously:

Hydrolysis of Nitrile: The nitrile group (-C\(\equiv\)N) is completely hydrolyzed by the strong acid to a carboxylic acid group (-COOH).

Dehydration of Alcohol: The hydroxyl group (-OH) is on a tertiary carbon, making it susceptible to dehydration (elimination of a water molecule) in the presence of a strong acid like conc. H\(_2\)SO\(_4\) and heat. A double bond is formed between C1 and an adjacent carbon (C2 or C6) of the ring.

The combined result is the formation of cyclohex-1-ene-1-carboxylic acid.
\[ [B] \xrightarrow{conc. H_2SO_4, \Delta} [C: Cyclohex-1-ene-1-carboxylic acid] + NH_4HSO_4 + H_2O \]

Step 4: Final Answer:

The final product [C] is cyclohex-1-ene-1-carboxylic acid, which corresponds to the structure shown in option (1).
Quick Tip: Recognize that concentrated H\(_2\)SO\(_4\) is a powerful dehydrating agent. When you see it used with heat on a molecule containing an alcohol group (especially secondary or tertiary), always anticipate an elimination reaction to form an alkene.

Step 1: Understanding the Question:

We need to identify the homoleptic complex from the given list of coordination compounds.


Step 2: Key Formula or Approach:

- Homoleptic complexes are those in which the central metal atom or ion is coordinated to only one type of ligand.

- Heteroleptic complexes are those in which the central metal atom or ion is coordinated to more than one type of ligand.

We will analyze the ligands attached to the central metal ion in each complex.


Step 3: Detailed Explanation:

1. Triamminetriaquachromium (III) chloride: The central metal is Chromium (Cr\(^{3+}\)). The ligands are 'triammine' (\(3 \times NH_3\)) and 'triaqua' (\(3 \times H_2O\)). Since there are two different types of ligands (ammine and aqua), this is a heteroleptic complex.

2. Potassium trioxalatoaluminate (III): The central metal is Aluminate (Al\(^{3+}\)). The ligand is 'trioxalato' (\(3 \times C_2O_4^{2-}\)). Since only one type of ligand (oxalato) is attached to the central metal, this is a homoleptic complex. The formula is \(K_3[Al(C_2O_4)_3]\).

3. Diamminechloridonitrito-N-platinum (II): The central metal is Platinum (Pt\(^{2+}\)). The ligands are 'diammine' (\(2 \times NH_3\)), 'chlorido' (\(Cl^-\)), and 'nitrito-N' (\(NO_2^-\)). Since there are three different types of ligands, this is a heteroleptic complex.

4. Pentaamminecarbonatocobalt (III) chloride: The central metal is Cobalt (Co\(^{3+}\)). The ligands are 'pentaammine' (\(5 \times NH_3\)) and 'carbonato' (\(CO_3^{2-}\)). Since there are two different types of ligands, this is a heteroleptic complex.


Step 4: Final Answer:

The only homoleptic complex among the options is Potassium trioxalatoaluminate (III).
Quick Tip: To identify a homoleptic complex, look at the name. If the name mentions only one type of ligand (e.g., 'hexaammine', 'tetracarbonyl', 'trioxalato'), it's homoleptic. If it lists multiple ligand names (e.g., 'diamminedichlorido'), it's heteroleptic.

Step 1: Understanding the Question:

The question requires us to classify the given organic halide based on the position of the halogen atom (X) in the molecule.


Step 2: Detailed Explanation:

Let's first define the different types of halides given in the options:


Aryl halide: The halogen atom is directly bonded to an sp\(^2\)-hybridized carbon atom of an aromatic ring.
Vinylic halide: The halogen atom is directly bonded to an sp\(^2\)-hybridized carbon atom of a carbon-carbon double bond.
Benzylic halide: The halogen atom is bonded to an sp\(^3\)-hybridized carbon atom which is directly attached to an aromatic ring.
Allylic halide: The halogen atom is bonded to an sp\(^3\)-hybridized carbon atom which is adjacent to a carbon-carbon double bond.


Now, let's analyze the given structure: \( C_6H_5-CH=CH-\overset{X}{\overset{|}{C}}H-CH_2CH_3 \)

1. The halogen atom (X) is bonded to a carbon atom.

2. This carbon atom is singly bonded to its neighbours (another C and an H), so it is sp\(^3\)-hybridized.

3. This sp\(^3\)-hybridized carbon atom is directly attached to a carbon atom which is part of a carbon-carbon double bond (\(-CH=\)).


This fits the definition of an allylic halide. The presence of the phenyl group (\(C_6H_5\)) does not change this classification, as the immediate environment of the C-X bond defines the type.


Step 3: Final Answer:

The compound is an example of an allylic halide. Therefore, option (4) is correct.
Quick Tip: To classify organic halides, always focus on the carbon atom directly bonded to the halogen. Check its hybridization (sp\(^3\), sp\(^2\)) and what it is attached to (aromatic ring, C=C double bond, etc.). This systematic check will lead you to the correct classification.

Step 1: Understanding the Question:

The question asks to identify the correct stability relationship between compounds of Group 13 elements in different oxidation states. This is related to the inert pair effect.


Step 2: Key Formula or Approach:

The inert pair effect is the tendency of the two electrons in the outermost atomic s-orbital to remain unshared or un-ionised in compounds of post-transition metals. This effect becomes more prominent as we move down a group in the p-block.

For Group 13 (B, Al, Ga, In, Tl), the general outer electronic configuration is ns\(^2\)np\(^1\). They can exhibit +3 and +1 oxidation states.
- For lighter elements (Al, Ga, In), the +3 oxidation state is more stable.
- For the heaviest element, Thallium (Tl), the inert pair effect is very strong, making the +1 oxidation state more stable than the +3 oxidation state.


Step 3: Detailed Explanation:

Let's analyze the stability based on the oxidation state of the metal:
1. TlI vs TlI\(_3\): In TlI, Thallium is in the +1 oxidation state. In TlI\(_3\), Thallium is in the +3 oxidation state. Due to the strong inert pair effect, Tl\(^+\) is significantly more stable than Tl\(^{3+}\). Therefore, TlI is more stable than TlI\(_3\). This statement is correct.
2. TlCl\(_3\) vs TlCl: This is the opposite of the first statement. TlCl (+1 state) is more stable than TlCl\(_3\) (+3 state). This statement is incorrect.
3. InI\(_3\) vs InI: For Indium (In), the +3 oxidation state is more stable than the +1 state, although the inert pair effect starts to become noticeable. So, InI\(_3\) should be more stable than InI. The statement is given as InI\(_3\) > InI. This is correct in terms of thermodynamic stability. However, the question asks for the best representation of the trend. The most dramatic and classic example is Thallium. Let's re-evaluate the options. The provided answer key states (1) is correct. Let's assume the question asks for the most pronounced effect.
4. AlCl vs AlCl\(_3\): For Aluminium (Al), the +3 oxidation state is overwhelmingly more stable. AlCl\(_3\) is much more stable than AlCl. This statement is incorrect.


Comparing the options, the most definitive and correct relationship representing the inert pair effect is the superior stability of Tl(+1) compounds over Tl(+3) compounds.


Step 4: Final Answer:

The correct stability relationship is TlI \(>\) TlI\(_3\).
Quick Tip: Remember the stability trend for oxidation states in Group 13, 14, and 15 due to the inert pair effect: - Group 13: Al\(^{3+}\) > Ga\(^{3+}\) > In\(^{3+}\) > Tl\(^{3+}\) (Stability of +3 decreases) - Group 13: Tl\(^+\) > In\(^+\) > Ga\(^+\) > Al\(^+\) (Stability of +1 increases) The same logic applies to Group 14 (+4 vs +2) and Group 15 (+5 vs +3).

Step 1: Understanding the Question:

We need to identify the Lewis acid among the given chemical species.


Step 2: Key Formula or Approach:

According to the Lewis theory of acids and bases:
- A Lewis acid is a species that can accept a pair of electrons. These are typically electron-deficient species (e.g., have an incomplete octet or a positive charge).
- A Lewis base is a species that can donate a pair of electrons. These are typically electron-rich species (e.g., have lone pairs of electrons or a negative charge).


Step 3: Detailed Explanation:

Let's analyze each option:

1. OH\(^-\): The hydroxide ion has a negative charge and lone pairs of electrons on the oxygen atom. It is electron-rich and readily donates an electron pair, making it a strong Lewis base.

2. NH\(_3\): The ammonia molecule has a lone pair of electrons on the nitrogen atom. It can donate this pair to form a coordinate bond, so it acts as a Lewis base.

3. H\(_2\)O: The water molecule has two lone pairs of electrons on the oxygen atom. It can donate one of these pairs, acting as a Lewis base.

4. BF\(_3\): In boron trifluoride, the central boron atom is bonded to three fluorine atoms. Boron has only 3 valence electrons, so after forming 3 bonds, it has only 6 electrons in its valence shell (an incomplete octet). To complete its octet, it has a strong tendency to accept a pair of electrons, making it a classic example of a Lewis acid.


Step 4: Final Answer:

BF\(_3\) is the species that acts as a Lewis acid.
Quick Tip: To quickly identify Lewis acids, look for molecules with central atoms having an incomplete octet (like compounds of B, Al, Be) or simple cations (like H\(^+\), Ag\(^+\)). For Lewis bases, look for anions or molecules with atoms having lone pairs (like compounds of N, O, F, P, S).

Step 1: Understanding the Question:

We need to identify the Lewis acid among the given chemical species.


Step 2: Key Formula or Approach:

According to the Lewis theory of acids and bases:
- A Lewis acid is a species that can accept a pair of electrons. These are typically electron-deficient species (e.g., have an incomplete octet or a positive charge).
- A Lewis base is a species that can donate a pair of electrons. These are typically electron-rich species (e.g., have lone pairs of electrons or a negative charge).


Step 3: Detailed Explanation:

Let's analyze each option:

1. OH\(^-\): The hydroxide ion has a negative charge and lone pairs of electrons on the oxygen atom. It is electron-rich and readily donates an electron pair, making it a strong Lewis base.

2. NH\(_3\): The ammonia molecule has a lone pair of electrons on the nitrogen atom. It can donate this pair to form a coordinate bond, so it acts as a Lewis base.

3. H\(_2\)O: The water molecule has two lone pairs of electrons on the oxygen atom. It can donate one of these pairs, acting as a Lewis base.

4. BF\(_3\): In boron trifluoride, the central boron atom is bonded to three fluorine atoms. Boron has only 3 valence electrons, so after forming 3 bonds, it has only 6 electrons in its valence shell (an incomplete octet). To complete its octet, it has a strong tendency to accept a pair of electrons, making it a classic example of a Lewis acid.


Step 4: Final Answer:

BF\(_3\) is the species that acts as a Lewis acid.
Quick Tip: To quickly identify Lewis acids, look for molecules with central atoms having an incomplete octet (like compounds of B, Al, Be) or simple cations (like H\(^+\), Ag\(^+\)). For Lewis bases, look for anions or molecules with atoms having lone pairs (like compounds of N, O, F, P, S).

Step 1: Understanding the Question:

We are given information about the crystal lattice of an ionic compound A\(_x\)B\(_y\) and need to determine its empirical formula and then calculate x+y.


Step 2: Key Formula or Approach:

1. In a cubic close-packed (ccp) structure, which is equivalent to a face-centered cubic (fcc) lattice, the effective number of atoms per unit cell is 4.
2. For N atoms forming a close-packed structure, there are N octahedral voids and 2N tetrahedral voids.
3. We will determine the effective number of atoms of A and B in one unit cell and then find their simplest whole-number ratio to get the formula.


Step 3: Detailed Explanation:

- Element B forms the ccp structure. So, the effective number of atoms of B per unit cell is \(N_B = 4\).

- In a ccp lattice with \(N_B = 4\) atoms, the number of tetrahedral voids (TV) is \(2 \times N_B = 2 \times 4 = 8\).

- Atoms of element A occupy 1/3 of these tetrahedral voids.

- So, the effective number of atoms of A per unit cell is \(N_A = \frac{1}{3} \times (Number of TVs) = \frac{1}{3} \times 8 = \frac{8}{3}\).


Now we have the ratio of atoms A : B in the unit cell as:
\[ N_A : N_B = \frac{8}{3} : 4 \]
To get the simplest whole-number ratio, we can multiply both sides by 3:
\[ (\frac{8}{3} \times 3) : (4 \times 3) = 8 : 12 \]
Now, divide by the greatest common divisor, which is 4:
\[ \frac{8}{4} : \frac{12}{4} = 2 : 3 \]
So, the empirical formula of the compound is A\(_2\)B\(_3\).

By comparing this to A\(_x\)B\(_y\), we have \(x = 2\) and \(y = 3\).

The value of \(x + y\) is \(2 + 3 = 5\).


Step 4: Final Answer:

The value of x + y is 5.
Quick Tip: Remember the key numbers for close-packed structures (ccp/fcc and hcp): - Atoms per unit cell (Z) = 4 for fcc/ccp. - Octahedral voids = Z. - Tetrahedral voids = 2Z. Knowing these relationships is essential for solving solid-state stoichiometry problems.

Step 1: Understanding the Question:

The question asks to identify which of the listed forces are classified as intermolecular forces. Intermolecular forces are forces that exist between molecules.


Step 2: Detailed Explanation:

Let's analyze each force type:


A. dipole - dipole forces: These are attractive forces between the positive end of one polar molecule and the negative end of another polar molecule. They are a type of intermolecular force.

B. dipole - induced dipole forces: These forces arise when a polar molecule induces a temporary dipole in a nonpolar molecule, leading to a weak attraction. They are a type of intermolecular force (a category of van der Waals forces).

C. hydrogen bonding: This is a special, strong type of dipole-dipole interaction that occurs between a hydrogen atom bonded to a highly electronegative atom (N, O, or F) and another nearby electronegative atom. It is a very important intermolecular force.

D. covalent bonding: This is the force that holds atoms together \textit{within a molecule by the sharing of electrons. It is an intramolecular force, not an intermolecular force. Intramolecular forces are much stronger than intermolecular forces.

E. dispersion forces (London forces): These are weak intermolecular forces caused by temporary fluctuations in electron distribution within atoms or molecules, creating temporary dipoles. They exist between all atoms and molecules.



Step 3: Final Answer:

The forces A, B, C, and E are all types of intermolecular forces. Force D, covalent bonding, is an intramolecular force. Therefore, the correct combination is A, B, C, and E. This corresponds to option (4).
Quick Tip: Remember the distinction: \textbf{Intermolecular forces are "between" molecules (like an \textbf{inter}national flight is between nations), while \textbf{intra}molecular forces are "within" a molecule (like an \textbf{intra}mural sport is within a school). Covalent, ionic, and metallic bonds are intramolecular.

Step 1: Understanding the Question:

The question requires matching the substances in List-I (allotropes/forms of carbon) with their corresponding properties or uses in List-II.


Step 2: Detailed Explanation:

Let's analyze each item in List-I and find its correct match in List-II.


A. Coke: Coke is a high-carbon content fuel derived from coal. In metallurgy, particularly in blast furnaces for iron extraction, it acts as a crucial reducing agent, reducing iron oxides to molten iron. Therefore, Coke matches with III. Used as a reducing agent.

B. Diamond: Diamond is an allotrope of carbon where each carbon atom is covalently bonded to four other carbon atoms, forming a tetrahedral lattice. This type of bonding corresponds to sp\(^3\) hybridization. Therefore, Diamond matches with I. Carbon atoms are sp\(^3\) hybridised.

C. Fullerene: Fullerenes are allotropes of carbon that consist of molecules composed entirely of carbon, forming hollow spheres, ellipsoids, or tubes. The most famous example, Buckminsterfullerene (C\(_{60}\)), has a structure resembling a soccer ball, which is a cage-like molecule. Therefore, Fullerene matches with IV. Cage like molecules.

D. Graphite: Graphite is another allotrope of carbon with a layered structure. The layers are held by weak van der Waals forces, allowing them to slide easily over one another. This property makes graphite an excellent solid or dry lubricant. Therefore, Graphite matches with II. Used as a dry lubricant.



Step 3: Final Answer:

Based on the analysis, the correct matching is:

A \(\rightarrow\) III

B \(\rightarrow\) I

C \(\rightarrow\) IV

D \(\rightarrow\) II

This combination corresponds to option (4).
Quick Tip: Remembering the structure and key applications of carbon allotropes is crucial. Diamond's hardness comes from its sp\(^3\) tetrahedral network, Graphite's slipperiness from its sp\(^2\) layered structure, and Fullerenes are known for their unique cage-like shapes.

Step 1: Understanding the Question:

This is an Assertion-Reason question about the properties of solutions of alkali metals in liquid ammonia. We must evaluate the truthfulness of both statements and the validity of the reason.


Step 2: Detailed Explanation:

Analysis of Assertion A:

When alkali metals, like sodium (Na), are dissolved in liquid ammonia (NH\(_3\)), they ionize to form metal cations and release electrons. The reaction is:
\[ Na(s) + (x+y)NH_3(l) \rightarrow [Na(NH_3)_x]^+ + [e(NH_3)_y]^- \]
The species \([e(NH_3)_y]^-\) is known as the ammoniated electron or solvated electron. This unpaired electron absorbs energy in the visible region of the spectrum, imparting a deep blue color to the solution. Since it is an unpaired electron, its presence also makes the solution paramagnetic. Therefore, Assertion A is true.


Analysis of Reason R:

The reason states that the deep blue color is due to the formation of amide. Sodium amide (NaNH\(_2\)) can be formed in these solutions, but it happens slowly, and this reaction causes the blue color to fade, not to appear. The formation of amide is represented by:
\[ 2Na(s) + 2NH_3(l) \xrightarrow{catalyst or time} 2NaNH_2(s) + H_2(g) \]
The primary cause of the blue color is the ammoniated electron, not the sodium amide. Therefore, Reason R is false.


Step 3: Final Answer:

Since Assertion A is true and Reason R is false, the correct option is (4).
Quick Tip: Remember the three key properties of alkali metal-liquid ammonia solutions and their common cause, the ammoniated electron: 1. \textbf{Deep blue color}: Due to electronic transitions of the ammoniated electron. 2. \textbf{Paramagnetism}: Due to the unpaired spin of the ammoniated electron. 3. \textbf{High electrical conductivity}: Due to both ammoniated cations and ammoniated electrons being mobile charge carriers.

Step 1: Understanding the Question:

We need to identify the example of heterogeneous catalysis from the given options.


Step 2: Key Formula or Approach:

Catalysis is classified based on the physical phases of the reactants and the catalyst:
- Homogeneous Catalysis: The reactants and the catalyst are in the same phase (e.g., all are gases, or all are in the same liquid solution).
- Heterogeneous Catalysis: The reactants and the catalyst are in different phases (e.g., gaseous reactants and a solid catalyst).


Step 3: Detailed Explanation:

Let's analyze the phases in each reaction:

1. Haber's process for ammonia:
\[ N_2(g) + 3H_2(g) \xrightarrow{Fe(s)} 2NH_3(g) \]
The reactants (N\(_2\), H\(_2\)) are in the gas phase, while the catalyst (iron) is in the solid phase. Since the phases are different, this is an example of heterogeneous catalysis.

2. Lead chamber process for H\(_2\)SO\(_4\):
\[ 2SO_2(g) + O_2(g) \xrightarrow{NO(g)} 2SO_3(g) \]
The reactants (SO\(_2\), O\(_2\)) and the catalyst (NO) are all in the gas phase. This is homogeneous catalysis.

3. Hydrolysis of sugar (sucrose):
\[ C_{12}H_{22}O_{11}(aq) + H_2O(l) \xrightarrow{H^+(aq)} C_6H_{12}O_6(aq) + C_6H_{12}O_6(aq) \]
The reactant (sucrose) and the catalyst (H\(^+\) ions) are both in the aqueous solution phase. This is homogeneous catalysis.

4. Decomposition of ozone:
\[ 2O_3(g) \xrightarrow{NO(g)} 3O_2(g) \]
The reactant (O\(_3\)) and the catalyst (NO) are both in the gas phase. This is homogeneous catalysis.


Step 4: Final Answer:

The formation of ammonia in the Haber's process is the correct example of heterogeneous catalysis.
Quick Tip: Most industrial catalytic processes use heterogeneous catalysts because they are easier to separate from the products, making the process more economical and efficient. Surface catalysis, like in the Haber's process, is a hallmark of heterogeneous catalysis.

Step 1: Understanding the Question:

We need to examine a list of molecules and identify how many of them do not follow the octet rule, meaning the central atom does not have exactly eight valence electrons.


Step 2: Key Formula or Approach:

We will draw the Lewis structure for each molecule and count the total number of electrons (from bonds and lone pairs) around the central atom. A single covalent bond contributes 2 electrons to the valence shell of both atoms involved.


Step 3: Detailed Explanation:

Let's analyze each species:

1. NH\(_3\): The central atom is Nitrogen (N). It forms 3 single bonds with H atoms and has 1 lone pair. Total electrons = (3 bonds \(\times\) 2 e\(^-\)/bond) + 2 e\(^-\) (lone pair) = 6 + 2 = 8 electrons. (Obeys octet rule)

2. AlCl\(_3\): The central atom is Aluminum (Al). It forms 3 single bonds with Cl atoms. Al is in group 13 and has 3 valence electrons. Total electrons = 3 bonds \(\times\) 2 e\(^-\)/bond = 6 electrons. (Incomplete octet, does NOT obey)

3. BeCl\(_2\): The central atom is Beryllium (Be). It forms 2 single bonds with Cl atoms. Be is in group 2 and has 2 valence electrons. Total electrons = 2 bonds \(\times\) 2 e\(^-\)/bond = 4 electrons. (Incomplete octet, does NOT obey)

4. CCl\(_4\): The central atom is Carbon (C). It forms 4 single bonds with Cl atoms. Total electrons = 4 bonds \(\times\) 2 e\(^-\)/bond = 8 electrons. (Obeys octet rule)

5. PCl\(_5\): The central atom is Phosphorus (P). It forms 5 single bonds with Cl atoms. Total electrons = 5 bonds \(\times\) 2 e\(^-\)/bond = 10 electrons. (Expanded octet, does NOT obey)


The species that do not have eight electrons around the central atom are AlCl\(_3\), BeCl\(_2\), and PCl\(_5\).

The total number of such species is 3.


Step 4: Final Answer:

There are 3 species that do not follow the octet rule.
Quick Tip: Exceptions to the octet rule are common. - Incomplete Octet: Often seen with elements like Be (4 e\(^-\)), B, and Al (6 e\(^-\)). - Expanded Octet: Possible for elements in the 3rd period and below (like P, S, Cl, Xe) as they can use their empty d-orbitals for bonding. - Odd-Electron Molecules: Molecules with an odd total number of valence electrons (e.g., NO).

Step 1: Understanding the Question:

The question asks for the thermodynamic reason behind the greater stability of the copper(II) ion (Cu\(^{2+}\)) compared to the copper(I) ion (Cu\(^+\)) in an aqueous environment.


Step 2: Key Formula or Approach:

The stability of an ion in an aqueous solution is determined by the overall Gibbs free energy change for its formation from the elemental state. This involves several energy terms, primarily the enthalpy of atomization, ionization enthalpy (IE), and hydration enthalpy (\(\Delta_{hyd}H\)).
\[ M(s) \xrightarrow{\Delta_{at}H} M(g) \xrightarrow{IE} M^{n+}(g) \xrightarrow{\Delta_{hyd}H} M^{n+}(aq) \]
A more stable ion will have a more negative overall enthalpy change.


Step 3: Detailed Explanation:

Let's compare the formation of Cu\(^+\)(aq) and Cu\(^{2+}\)(aq).


Ionization Enthalpy: The second ionization enthalpy (IE\(_2\)) of copper (the energy required to remove an electron from Cu\(^+\) to form Cu\(^{2+}\)) is very high. Based on IE alone, Cu\(^+\) should be more stable than Cu\(^{2+}\). So, option (1) is incorrect as a reason for Cu\(^{2+}\) stability; it's actually a factor that opposes it.

Hydration Enthalpy: Hydration enthalpy is the energy released when one mole of gaseous ions is dissolved in water. It depends strongly on the charge density of the ion (charge/size ratio). The formula for hydration energy is roughly proportional to the square of the charge (\(q^2\)) and inversely proportional to the radius (\(r\)).
\[ \Delta_{hyd}H \propto -\frac{q^2}{r} \]
The Cu\(^{2+}\) ion has a greater charge (+2) and a smaller ionic radius compared to the Cu\(^+\) ion (+1). Consequently, the hydration enthalpy of Cu\(^{2+}\) is much more negative (i.e., much more energy is released) than that of Cu\(^+\).

Conclusion: This large release of hydration energy for Cu\(^{2+}\) more than compensates for the high energy input required for the second ionization. The overall energy change is more favorable for the formation of Cu\(^{2+}\) in an aqueous solution, making it more stable.



Step 4: Final Answer:

The primary factor responsible for the greater stability of Cu\(^{2+}\) in aqueous solution is its high hydration energy. This corresponds to option (4).
Quick Tip: When comparing the stability of different oxidation states of an ion *in aqueous solution*, always consider the hydration enthalpy. It's often the deciding factor, especially for ions with higher charges, as hydration energy increases significantly with charge.

Step 1: Understanding the Question:

The question asks for the mathematical relationship between the total number of possible values for the magnetic quantum number (\(m\)), denoted as n\(_m\), and the azimuthal quantum number (\(l\)).


Step 2: Key Formula or Approach:

The rules for quantum numbers state that for a given value of the azimuthal quantum number, \(l\), the magnetic quantum number, \(m_l\) (or simply \(m\)), can take any integer value from \( -l \) to \( +l \), including zero.

Possible values of \(m\) are: \( -l, (-l+1), ..., 0, ..., (l-1), +l \).


Step 3: Detailed Explanation:

To find the total number of these values (n\(_m\)), we can count them. The number of values is given by:
\[ n_m = (last value) - (first value) + 1 \] \[ n_m = (l) - (-l) + 1 \] \[ n_m = l + l + 1 \] \[ n_m = 2l + 1 \]
Now we have the relationship \(n_m = 2l + 1\). The question provides options where this relationship is rearranged. We need to find the option that is equivalent to our derived formula. Let's rearrange our formula to solve for \(l\):
\[ n_m - 1 = 2l \] \[ l = \frac{n_m - 1}{2} \]

Step 4: Final Answer:

This rearranged formula matches option (2). Therefore, the correct relation is given in option (2).
Quick Tip: For any subshell `l`, the number of orbitals is always \(2l+1\). For s-subshell (\(l=0\)), number of orbitals = \(2(0)+1 = 1\). For p-subshell (\(l=1\)), number of orbitals = \(2(1)+1 = 3\). For d-subshell (\(l=2\)), number of orbitals = \(2(2)+1 = 5\). This formula is fundamental to understanding electron configurations.

Step 1: Understanding the Question:

This is a stoichiometry problem. We need to calculate the mass of carbon dioxide produced from the thermal decomposition of an impure sample of limestone (CaCO\(_3\)).


Step 2: Key Formula or Approach:

1. Calculate the mass of the pure reactant (CaCO\(_3\)) in the sample.
2. Use the balanced chemical equation to establish the molar relationship between the reactant and the product.
3. Convert the mass of the pure reactant to moles.
4. Use the mole ratio to find the moles of the product (CO\(_2\)).
5. Convert the moles of the product to mass.


Step 3: Detailed Explanation:

1. Mass of pure CaCO\(_3\):

Total mass of limestone sample = 20 g.

Purity = 20%.

Mass of pure CaCO\(_3\) = 20 g \(\times \frac{20}{100} = 4\) g.


2. Molar Masses:

Molar mass of CaCO\(_3\) = 40 (Ca) + 12 (C) + 3 \(\times\) 16 (O) = 100 g/mol.

Molar mass of CO\(_2\) = 12 (C) + 2 \(\times\) 16 (O) = 44 g/mol.


3. Stoichiometric Calculation:

The balanced equation is: \[ CaCO_3 (s) \rightarrow CaO (s) + CO_2 (g) \]
The mole ratio between CaCO\(_3\) and CO\(_2\) is 1:1.

This means that 1 mole of CaCO\(_3\) produces 1 mole of CO\(_2\).

In terms of mass, 100 g of CaCO\(_3\) produces 44 g of CO\(_2\).


4. Calculate the mass of CO\(_2\):

We can use a simple ratio to find the mass of CO\(_2\) produced from 4 g of pure CaCO\(_3\):
\[ Mass of CO_2 = (Mass of CaCO_3) \times \frac{Molar mass of CO_2}{Molar mass of CaCO_3} \] \[ Mass of CO_2 = 4 \, g \times \frac{44 \, g/mol}{100 \, g/mol} \] \[ Mass of CO_2 = \frac{4 \times 44}{100} = \frac{176}{100} = 1.76 \, g \]

Step 4: Final Answer:

The mass of CO\(_2\) produced is 1.76 g.
Quick Tip: In stoichiometry problems involving impure samples, always perform calculations based on the mass of the pure substance. The impurities are assumed to be inert and do not participate in the reaction.

Step 1: Understanding the Question:

The question consists of an Assertion (A) and a Reason (R) related to the thermodynamic properties of an electrochemical cell. We need to determine if A and R are true and if R correctly explains A.


Step 2: Detailed Explanation:

Analysis of Assertion A:

The equation for the Gibbs free energy change of a cell reaction is \(\Delta_r G = -nFE_{cell}\), where:

- \(\Delta_r G\) is the Gibbs free energy change.

- \(n\) is the number of moles of electrons transferred in the balanced redox reaction.

- \(F\) is the Faraday constant (charge per mole of electrons).

- \(E_{cell}\) is the cell potential.

From the equation, it is clear that \(\Delta_r G\) is directly proportional to \(n\). Therefore, the value of \(\Delta_r G\) depends on \(n\). Assertion A is true.


Analysis of Reason R:

- An intensive property does not depend on the amount of matter in the system (e.g., temperature, pressure, density, cell potential \(E_{cell}\)). The potential of a cell is the same regardless of its size.

- An extensive property depends on the amount of matter (e.g., mass, volume, energy, Gibbs free energy \(\Delta_r G\)). The total energy released depends on how many reactants are converted.

So, the statement that \(E_{cell}\) is an intensive property and \(\Delta_r G\) is an extensive property is correct. Reason R is true.


Connecting Reason and Assertion:

The reason why \(\Delta_r G\) depends on \(n\) (the amount of reaction) is precisely because \(\Delta_r G\) is an extensive property. The equation \(\Delta_r G = -nFE_{cell}\) shows how the extensive property (\(\Delta_r G\)) is calculated from the intensive property (\(E_{cell}\)) by multiplying it by factors related to the amount of substance (\(nF\)). Thus, R provides the correct fundamental explanation for A.


Step 3: Final Answer:

Both Assertion A and Reason R are true, and Reason R is the correct explanation for Assertion A. Therefore, option (2) is the correct answer.
Quick Tip: Remember: Intensive properties are intrinsic to the substance (like its color or melting point), while extensive properties depend on how much of it you have (like its mass or total energy). Energy is almost always extensive. Potentials (like voltage) are intensive.

Step 1: Understanding the Question:

We are given the rate law for a reaction and asked how the initial rate of reaction changes when the concentration of one reactant is changed.


Step 2: Key Formula or Approach:

The rate law is given as: Rate = k[A]\(^2\)[B].
We need to compare the initial rate with the new rate after changing the concentration of A.

Let the initial rate be \(r_1\) and the new rate be \(r_2\).


Step 3: Detailed Explanation:

Initial State:

Let the initial concentrations be [A] and [B].
The initial rate is: \[ r_1 = k[A]^2[B] \]

New State:

The concentration of A is tripled, so the new concentration is [A'] = 3[A].
The concentration of B is kept constant, so [B'] = [B].
The new rate is: \[ r_2 = k[A']^2[B'] = k(3[A])^2[B] \] \[ r_2 = k(9[A]^2)[B] = 9(k[A]^2[B]) \]
By substituting \(r_1 = k[A]^2[B]\), we get: \[ r_2 = 9 \times r_1 \]
This means the new rate is 9 times the initial rate. The rate increases by a factor of nine.


Step 4: Final Answer:

The initial rate would increase by a factor of nine.
Quick Tip: To quickly determine the effect of a concentration change on the reaction rate, look at the order of the reaction with respect to that reactant. If the concentration is changed by a factor of 'x', the rate will change by a factor of 'x\(^n\)', where 'n' is the order with respect to that reactant. Here, [A] is tripled (x=3) and the order is 2, so the rate changes by a factor of 3\(^2\) = 9.

Step 1: Understanding the Question:

The question asks for the mass of two moles of the organic product formed from a specific chemical reaction: the decarboxylation of sodium ethanoate.


Step 2: Key Formula or Approach:

1. Identify the reaction. Heating a sodium salt of a carboxylic acid with soda-lime (a mixture of NaOH and CaO) is a standard method for preparing alkanes, known as decarboxylation.
2. Write the balanced chemical equation to identify the organic product.
3. Calculate the molar mass of the product.
4. Calculate the total mass for two moles of the product using the formula: Mass = moles \(\times\) Molar Mass.


Step 3: Detailed Explanation:

The reaction is the decarboxylation of sodium ethanoate (CH\(_3\)COONa) using soda-lime (NaOH + CaO). The CaO does not directly participate but helps to keep the NaOH dry and facilitates the reaction.

The reaction is: \[ \underset{Sodium ethanoate}{CH_3COONa} + NaOH \xrightarrow{CaO, \Delta} \underset{Methane}{CH_4} + Na_2CO_3 \]
The organic compound obtained is methane (CH\(_4\)).


Now, we need to find the weight of two moles of methane.
First, calculate the molar mass of methane (CH\(_4\)):
Molar Mass (CH\(_4\)) = (Atomic mass of C) + 4 \(\times\) (Atomic mass of H)
Molar Mass (CH\(_4\)) = 12.0 g/mol + 4 \(\times\) 1.0 g/mol = 16.0 g/mol.


Finally, calculate the weight of two moles:
Weight = 2 moles \(\times\) 16.0 g/mol = 32 g.


Step 4: Final Answer:

The weight of two moles of the organic product (methane) is 32 g.
Quick Tip: Soda-lime decarboxylation is a step-down reaction for alkanes. The alkane produced has one less carbon atom than the parent carboxylic acid salt. The -COONa group is effectively replaced by an -H atom.

Step 1: Understanding the Question:

The question requires us to identify the incorrect statements from a list of five statements about the properties of transition metals and their oxides.


Step 2: Detailed Explanation:

Let's analyze each statement:

A. All the transition metals except scandium form MO oxides which are ionic. This is a broad generalization and is not entirely correct. While many MO oxides (like FeO, MnO) are ionic, others (like ZnO) have significant covalent character. Also, higher oxides are covalent. So, this statement is factually weak but let's evaluate others which might be more clearly incorrect.

B. The highest oxidation number corresponding to the group number in transition metal oxides is attained in Sc\(_2\)O\(_3\) to Mn\(_2\)O\(_7\). This statement is correct. Sc (Group 3) shows +3. Ti (Group 4) shows +4 (in TiO\(_2\)). V (Group 5) shows +5 (in V\(_2\)O\(_5\)). Cr (Group 6) shows +6 (in CrO\(_3\)). Mn (Group 7) shows +7 (in Mn\(_2\)O\(_7\)). This trend holds true up to manganese.

C. Basic character increases from V\(_2\)O\(_3\) to V\(_2\)O\(_4\) to V\(_2\)O\(_5\). This statement is incorrect. The acidic character of metal oxides increases with an increase in the oxidation state of the metal. For vanadium oxides: V\(_2\)O\(_3\) (+3) is basic, V\(_2\)O\(_4\) (+4) is amphoteric, and V\(_2\)O\(_5\) (+5) is acidic. Thus, the basic character decreases, not increases.

D. V\(_2\)O\(_4\) dissolves in acids to give VO\(^{3+}\) salts. This statement is incorrect. V\(_2\)O\(_4\) contains vanadium in the +4 oxidation state. When it dissolves in acid, it forms the vanadyl ion, which is VO\(^{2+}\), not VO\(^{3+}\). The VO\(^{3+}\) ion would correspond to a +5 oxidation state.

E. CrO is basic but Cr\(_2\)O\(_3\) is amphoteric. This statement is correct. Following the trend with oxidation states, CrO (+2) is basic, Cr\(_2\)O\(_3\) (+3) is amphoteric, and CrO\(_3\) (+6) is acidic.



Step 3: Final Answer:

The incorrect statements are C and D. Therefore, the correct option is (4).
Quick Tip: A crucial trend for metal oxides is that their acidity increases with the oxidation state of the metal. Low oxidation states yield basic oxides, intermediate states yield amphoteric oxides, and high oxidation states yield acidic oxides.

Step 1: Understanding the Question:

The question asks for the fundamental thermodynamic relationship between enthalpy change (\(\Delta\)H) and internal energy change (\(\Delta\)U).


Step 2: Key Formula or Approach:

The definition of enthalpy (H) is given by the equation: \[ H = U + PV \]
where U is the internal energy, P is the pressure, and V is the volume.


Step 3: Detailed Explanation:

For a change in the state of the system, the change in enthalpy (\(\Delta\)H) can be written as: \[ \Delta H = \Delta U + \Delta (PV) \]
For a process occurring at constant pressure, this simplifies to: \[ \Delta H = \Delta U + P\Delta V \]
For chemical reactions involving gases, we often assume they behave ideally. According to the ideal gas law: \[ PV = nRT \]
If the reaction involves a change in the number of moles of gas, \(\Delta n_g\), at constant temperature (T) and pressure (P), then the change in volume is related to the change in moles of gas: \[ P\Delta V = (\Delta n_g)RT \]
Here, \(\Delta n_g\) = (total moles of gaseous products) - (total moles of gaseous reactants).
Substituting this back into the enthalpy equation gives the desired relationship: \[ \Delta H = \Delta U + \Delta n_g RT \]

Step 4: Final Answer:

Comparing this derived equation with the given options, we find that option (3) is the correct relation.
Quick Tip: A simple way to remember the sign in the \(\Delta H = \Delta U + \Delta n_g RT\) equation is to think of enthalpy as the "total heat content". It includes the internal energy (\(\Delta U\)) plus the work the system has to do on the surroundings to make space for itself (\(P\Delta V \approx \Delta n_g RT\)). So, you add the work term to the internal energy.

Step 1: Understanding the Question:

The question asks for the contribution of a single octahedral void located at the edge center of a face-centered cubic (fcc) unit cell to that specific unit cell.


Step 2: Key Formula or Approach:

In a crystal lattice, atoms or voids located at different positions (corners, faces, edges, body center) are shared by multiple adjacent unit cells. The contribution of a particle at a specific position to a single unit cell is given by:

Corner: shared by 8 cells, contribution = 1/8
Face center: shared by 2 cells, contribution = 1/2
Edge center: shared by 4 cells, contribution = 1/4
Body center: shared by 1 cell, contribution = 1


Step 3: Detailed Explanation:

In a face-centered cubic (fcc) lattice, there are octahedral voids at two types of locations:
1. One void at the body center of the cube.
2. One void at the center of each of the 12 edges of the cube.

The question specifically asks about an "edge centred octahedral void". An edge of a cubic unit cell is shared by four other unit cells. Therefore, a void located at the center of an edge is also shared by those four unit cells.

The fraction of this void that lies within one particular unit cell is \(\frac{1}{4}\).


Step 4: Final Answer:

The contribution of one edge-centered octahedral void to a single unit cell is \(\frac{1}{4}\). This corresponds to option (4).
Quick Tip: To easily remember contributions, visualize the unit cell in a 3D lattice. An edge is a line shared by the 4 cubes that meet at that line. Anything in the middle of that edge must also be shared by those 4 cubes.

Step 1: Understanding the Question:

The question asks to evaluate two statements about eutrophication, a process of water pollution, and determine their correctness.


Step 2: Detailed Explanation:

Analysis of Statement I:

"The nutrient deficient water bodies lead to eutrophication."

This statement is incorrect. Eutrophication is the process of nutrient enrichment of a water body. It is caused by an excess of nutrients, particularly nitrates and phosphates, from sources like agricultural runoff (fertilizers) and sewage. Nutrient deficiency describes an oligotrophic water body, which is the opposite of a eutrophic one.


Analysis of Statement II:

"Eutrophication leads to decrease in the level of oxygen in the water bodies."

This statement is correct. The excess nutrients in a eutrophic water body cause a massive growth of algae and other aquatic plants, a phenomenon known as an algal bloom. When this large mass of algae dies, it sinks to the bottom and is decomposed by aerobic bacteria. This decomposition process consumes large amounts of dissolved oxygen from the water. The resulting depletion of oxygen (hypoxia or anoxia) can lead to the death of fish and other aquatic organisms.


Step 3: Final Answer:

Based on the analysis, Statement I is incorrect, and Statement II is true. This corresponds to option (1).
Quick Tip: Remember that "eu-" is a prefix meaning "good" or "well," and "trophic" relates to nutrition. So, eutrophication literally means "well-nourished," implying an excess of nutrients, not a deficiency. This excess leads to a cascade of negative effects, including oxygen depletion.

Step 1: Understanding the Question:

The question asks to identify which of the given alcohols will undergo dehydration most easily (i.e., at the fastest rate) under acidic conditions.


Step 2: Key Formula or Approach:

The acid-catalyzed dehydration of alcohols typically proceeds via an E1 mechanism. The mechanism involves three steps:
1. Protonation of the hydroxyl group.
2. Loss of a water molecule to form a carbocation intermediate. This is the rate-determining step.
3. Deprotonation of an adjacent carbon to form an alkene.

The rate of the reaction is determined by the stability of the carbocation intermediate formed in the second step. The more stable the carbocation, the lower the activation energy, and the faster the reaction.


Step 3: Detailed Explanation:

Let's analyze the stability of the carbocation formed from each alcohol:

(1) Dehydration of 4-nitrohexan-2-ol forms a secondary carbocation. The electron-withdrawing nitro (-NO\(_2\)) group is far away (at C4), so its destabilizing inductive effect is weak.
(2) Dehydration of 3-methyl-4-nitropentan-2-ol forms a secondary carbocation. Here, the electron-withdrawing -NO\(_2\) group is closer (at C4), exerting a stronger destabilizing inductive (-I) effect on the carbocation at C2.
(3) Dehydration of butane-2,3-diol. Let's consider the dehydration of the OH at C2. It forms a secondary carbocation at C2: CH\(_3\)-C\(^+\)H-CH(OH)-CH\(_3\). This carbocation is exceptionally stable because the lone pair of electrons on the oxygen atom of the adjacent hydroxyl group can donate electron density through resonance, which is a very powerful stabilizing effect.
(4) Dehydration of 1-nitropropan-2,3-diol. Dehydration of the secondary OH at C2 would form a carbocation directly adjacent to the carbon bearing the very strong electron-withdrawing -NO\(_2\) group. This would be an extremely unstable carbocation.

Comparing the stabilities, the carbocation formed from alcohol (3) is by far the most stable due to resonance stabilization from the adjacent -OH group. Therefore, this alcohol will be dehydrated most readily.


Step 4: Final Answer:

Alcohol (3), butane-2,3-diol, forms the most stable carbocation intermediate, and thus will be dehydrated most readily. This corresponds to option (3).
Quick Tip: When assessing carbocation stability, remember the order of stabilizing effects: Resonance > Hyperconjugation > Inductive effect. A carbocation adjacent to an atom with a lone pair (like O or N) is significantly stabilized by resonance. Conversely, a carbocation near a strong electron-withdrawing group (like -NO\(_2\)) is significantly destabilized.

Step 1: Understanding the Question:

The question asks to identify the most stable coordination compound from the given list. The stability of a complex is a key concept in coordination chemistry.


Step 2: Key Formula or Approach:

A major factor contributing to the stability of coordination complexes is the chelate effect. The chelate effect states that complexes formed by polydentate ligands (ligands that can bind to the central metal ion through more than one donor atom, forming a ring) are significantly more stable than complexes with analogous monodentate ligands.


Step 3: Detailed Explanation:

Let's analyze the ligands present in each complex:

(1) [Co(NH\(_3\))\(_6\)]\(_2\)(SO\(_4\))\(_3\): The ligand is ammonia (NH\(_3\)), which is a monodentate ligand. It does not form a chelate ring.

(2) [Co(NH\(_3\))\(_4\)(H\(_2\)O)Br](NO\(_3\))\(_2\): The ligands are ammonia (NH\(_3\)), water (H\(_2\)O), and bromide (Br\(^-\)). All are monodentate ligands. No chelation occurs.

(3) [Co(NH\(_3\))\(_3\)(NO\(_3\))\(_3\): The ligands are ammonia (NH\(_3\)) and nitrate (NO\(_3\)\(^{-}\)). Both are acting as monodentate ligands here. No chelation occurs.

(4) [CoCl\(_2\)(en)\(_2\)]NO\(_3\): The ligands are chloride (Cl\(^-\)) and ethylenediamine (en). Ethylenediamine (H\(_2\)N-CH\(_2\)-CH\(_2\)-NH\(_2\)) is a bidentate ligand. It binds to the cobalt ion through its two nitrogen atoms, forming a stable five-membered chelate ring.

Because the complex in option (4) contains a chelating ligand (ethylenediamine), it benefits from the chelate effect, which leads to a large increase in thermodynamic stability compared to the other complexes which only contain monodentate ligands.


Step 4: Final Answer:

The complex [CoCl\(_2\)(en)\(_2\)]NO\(_3\) is the most stable due to the chelate effect. This corresponds to option (4).
Quick Tip: When asked to compare the stability of complexes, the first thing to look for is the presence of chelating (polydentate) ligands like 'en', 'edta', or 'ox'. Complexes with these ligands are almost always more stable than those without them.

Step 1: Understanding the Question:

The question requires matching four oxoacids of sulfur with the correct description of the chemical bonds present in their structures.


Step 2: Key Formula or Approach:

To solve this, one needs to know the molecular structures of the given oxoacids.


Step 3: Detailed Explanation:

Let's determine the structure and bonds for each acid in List-I:

A. Peroxodisulphuric acid (H\(_2\)S\(_2\)O\(_8\)): Also known as Marshall's acid. Its structure is HO-SO\(_2\)-O-O-SO\(_2\)-OH. It contains a peroxide linkage (-O-O-). Counting the bonds:

Two S-OH single bonds.
Four S=O double bonds.
One S-O-O-S linkage.

This matches with description III in List-II. So, A-III.
B. Sulphuric acid (H\(_2\)SO\(_4\)): Its structure is HO-SO\(_2\)-OH. Counting the bonds:

Two S-OH single bonds.
Two S=O double bonds.

This matches with description IV in List-II. So, B-IV.
C. Pyrosulphuric acid (H\(_2\)S\(_2\)O\(_7\)): Also known as oleum. Its structure is HO-SO\(_2\)-O-SO\(_2\)-OH. It contains an S-O-S linkage. Counting the bonds:

Two S-OH single bonds.
Four S=O double bonds.
One S-O-S linkage.

This matches with description I in List-II. So, C-I.
D. Sulphurous acid (H\(_2\)SO\(_3\)): Its structure is HO-SO-OH, with a lone pair on the sulfur atom. Counting the bonds:

Two S-OH single bonds.
One S=O double bond.

This matches with description II in List-II. So, D-II.


Step 4: Final Answer:

The correct set of matches is A-III, B-IV, C-I, D-II. This corresponds to option (3).
Quick Tip: Being able to draw the structures of common oxoacids (of sulfur, phosphorus, chlorine) is a very useful skill for competitive exams. Key structural features to remember are peroxide linkages (-O-O-) in "peroxo" acids and direct M-O-M linkages in "pyro" acids.

Step 1: Understanding the Question:

The question asks for the major product of the reaction between the given reactant and Tollens' reagent ([Ag(NH\(_3\))\(_2\)]\(^+\)) in a basic medium, followed by heating. The reactant shown is 1,3-indandione, which has an active methylene group at position 2.


Step 2: Key Formula or Approach:

The reaction appears to be a non-standard oxidation. Tollens' reagent is a mild oxidizing agent. 1,3-Dicarbonyl compounds, like the reactant, have a highly acidic methylene group (\(-CH_2-\)) between the two carbonyls. In the presence of a base (OH\(^-\)), this group can be deprotonated to form an enolate. This enolate might be susceptible to oxidative cleavage. The product shown in the correct option (4) is the salt of 2-acetylbenzoic acid. This product indicates that the five-membered ring has been opened oxidatively.


Step 3: Detailed Explanation:

The question as presented has some inconsistencies, as the conversion of 1,3-indandione to 2-acetylbenzoic acid with Tollens' reagent is not a standard textbook reaction. However, we must deduce the transformation based on the given answer.

Reactant: 1,3-Indandione.
Reagent: Tollens' reagent ([Ag(NH\(_3\))\(_2\)]\(^+\) / OH\(^-\)), which is an oxidizing agent.
Product (from answer key): Salt of 2-acetylbenzoic acid.

This transformation requires the cleavage of one of the C-C bonds in the five-membered ring and rearrangement. Let's assume the active methylene group is first oxidized to a carbonyl group, forming 1,2,3-indantrione. This highly reactive intermediate, in the presence of a strong base, can undergo a benzilic acid-type rearrangement and further cleavage.
The overall transformation can be rationalized as an oxidative ring cleavage of the active methylene compound. The C1-C2 bond breaks, and the C2 and C3 atoms are rearranged and oxidized to form an acetyl group (\(-COCH_3\)) and a carboxylate group (\(-COO^-\)), respectively, attached to the benzene ring.
Although the mechanism is complex and not straightforward, the net result is the conversion of the fused ring system into the ortho-substituted benzene derivative shown in option (4).


Step 4: Final Answer:

Given the options, the reaction represents an oxidative cleavage of the 1,3-dicarbonyl system, leading to the formation of the carboxylate salt of 2-acetylbenzoic acid. This corresponds to structure (4).
Quick Tip: In complex organic reactions where the mechanism is not immediately obvious, analyze the change in the carbon skeleton and functional groups between the reactant and product. Here, a fused ring system is converted to a single ring with two functional groups, indicating a ring-opening reaction, which in this case is oxidative.

Step 1: Understanding the Question:

The question asks to identify which of the given chemical reactions does not occur in the higher temperature zone (900 K - 1500 K) of a blast furnace used for iron extraction.


Step 2: Key Formula or Approach:

The blast furnace has different temperature zones, and specific reactions occur in each zone.

Lower Temperature Zone (500 K - 800 K): This is the upper part of the furnace. Here, the iron oxides are reduced by carbon monoxide.
Higher Temperature Zone (900 K - 1500 K): This is the lower part of the furnace. Here, the final reduction of iron oxide occurs, and slag is formed. Also, the reducing agent CO is regenerated.


Step 3: Detailed Explanation:

Let's analyze the reactions based on the temperature zones:

(1) CaO + SiO\(_2\) \(\rightarrow\) CaSiO\(_3\): This is the formation of slag (calcium silicate). The limestone (CaCO\(_3\)) decomposes to CaO at high temperatures, which then reacts with the silica (SiO\(_2\)) impurity. This process occurs at about 1200 K, which is within the 900 K - 1500 K range.

(2) Fe\(_2\)O\(_3\) + CO \(\rightarrow\) 2FeO + CO\(_2\): This is one of the initial reduction steps of hematite ore. This reaction takes place in the upper part of the furnace at lower temperatures, typically around 500 K - 800 K. Therefore, it does NOT occur in the 900 K - 1500 K range.

(3) FeO + CO \(\rightarrow\) Fe + CO\(_2\): This is the final reduction step where iron(II) oxide is reduced to molten iron. This occurs at higher temperatures, around 1075 K, well within the specified range.

(4) C + CO\(_2\) \(\rightarrow\) 2CO: This is the Boudouard reaction, where hot coke reduces carbon dioxide to produce carbon monoxide, the main reducing agent. This reaction is favored at high temperatures (above 1075 K) and occurs in the lower part of the furnace.



Step 4: Final Answer:

The reaction Fe\(_2\)O\(_3\) + CO \(\rightarrow\) 2FeO + CO\(_2\) occurs at lower temperatures and not in the 900 K - 1500 K range. Hence, option (2) is the correct answer.
Quick Tip: Remember the temperature gradient in a blast furnace is hottest at the bottom and cooler at the top. The initial reduction of the ore (Fe\(_2\)O\(_3\)) happens at the top (cooler region), while the final reduction to Fe and slag formation happens at the bottom (hotter region).

Step 1: Understanding the Question:

The question provides equilibrium concentrations for a reaction and asks to calculate the standard Gibbs free energy change (\(\Delta\)G\(^\circ\)).


Step 2: Key Formula or Approach:

1. First, calculate the equilibrium constant (\(K_c\)) from the given concentrations.
2. Then, use the thermodynamic relationship between \(\Delta\)G\(^\circ\) and \(K_c\):
\[ \Delta G^\circ = -RT \ln K_c \]
This can also be written as:
\[ \Delta G^\circ = -2.303 RT \log_{10} K_c \]

Step 3: Detailed Explanation:

Calculation of Equilibrium Constant (\(K_c\)):

The reaction is: A + B \(\rightleftharpoons\) C + D

The equilibrium concentrations are:

[A] = 2 mol L\(^{-1}\)

[B] = 3 mol L\(^{-1}\)

[C] = 10 mol L\(^{-1}\)

[D] = 6 mol L\(^{-1}\)

The expression for \(K_c\) is: \[ K_c = \frac{[C][D]}{[A][B]} \] \[ K_c = \frac{(10)(6)}{(2)(3)} = \frac{60}{6} = 10 \]

Calculation of \(\Delta\)G\(^\circ\):

Given values are:
R = 2 cal / mol K
T = 300 K
K\(_c\) = 10

Using the formula: \[ \Delta G^\circ = -RT \ln K_c \] \[ \Delta G^\circ = -(2 cal / mol K) \times (300 K) \times \ln(10) \]
We know that \(\ln(10) \approx 2.303\). \[ \Delta G^\circ = -600 \times 2.303 cal/mol \] \[ \Delta G^\circ = -1381.8 cal/mol \]

Step 4: Final Answer:

The standard Gibbs free energy change for the reaction is -1381.80 cal. This corresponds to option (4).
Quick Tip: Remember the relationship between the sign of \(\Delta G^\circ\) and the value of K. If K > 1, then ln(K) is positive, and \(\Delta G^\circ\) is negative (reaction is spontaneous under standard conditions). If K < 1, then ln(K) is negative, and \(\Delta G^\circ\) is positive. Here K=10, so we expect a negative \(\Delta G^\circ\).

Step 1: Understanding the Question:

The question asks to classify pumice stone based on the type of colloidal system it represents. A colloidal system is defined by its dispersed phase and dispersion medium.


Step 2: Key Formula or Approach:

Let's define the components of pumice stone and the different types of colloids listed:

Pumice stone: It is a porous volcanic rock formed when super-heated, highly pressurized rock is rapidly ejected from a volcano. The porous texture is due to gas bubbles being trapped in the rock as it cooled. So, the dispersed phase is a gas, and the dispersion medium is a solid.
Foam: Dispersed phase = Gas, Dispersion medium = Liquid. (e.g., whipped cream).
Sol: Dispersed phase = Solid, Dispersion medium = Liquid. (e.g., paint).
Gel: Dispersed phase = Liquid, Dispersion medium = Solid. (e.g., cheese, jelly).
Solid Sol: This term can be used for two types of colloids: Solid in Solid (e.g., colored glass) or Gas in Solid. The latter is more precisely called a "solid foam".


Step 3: Detailed Explanation:

Based on its structure, pumice stone is a system where a gas is dispersed in a solid medium. This type of colloid is technically called a solid foam. However, among the given options, "solid sol" is often used in textbooks as a classification that can include gas-in-solid systems. Given the choices, "solid sol" is the intended answer representing the dispersion of gas bubbles within the solid rock matrix.


Step 4: Final Answer:

Pumice stone is an example of a gas dispersed in a solid. In the context of the given options, this is classified as a solid sol. Therefore, option (4) is correct.
Quick Tip: Memorize the eight types of colloidal systems by creating a table of dispersed phase vs. dispersion medium. Remember key examples for each type. Pumice stone and styrofoam are classic examples of a gas-in-solid colloid, which may be called either a solid foam or a solid sol depending on the classification system used.

Step 1: Understanding the Question:

The question asks us to count how many of the given seven species are aromatic based on Hückel's rule.


Step 2: Key Formula or Approach:

Hückel's rule states that for a species to be aromatic, it must satisfy four conditions:
1. It must be cyclic.
2. It must be planar.
3. It must be completely conjugated (every atom in the ring must have a p-orbital).
4. It must contain (4n + 2) \(\pi\) electrons, where n is a non-negative integer (0, 1, 2, ...).


Step 3: Detailed Explanation:

Let's analyze each species:


i. Naphthalene: It is cyclic, planar, and fully conjugated. It has 10 \(\pi\) electrons. For 4n + 2 = 10, 4n = 8, so n = 2. It obeys Hückel's rule. (Aromatic)

ii. Cyclopentadienyl anion: It is cyclic, planar, and fully conjugated. It has 6 \(\pi\) electrons (4 from double bonds, 2 from the lone pair/negative charge). For 4n + 2 = 6, 4n = 4, so n = 1. It obeys Hückel's rule. (Aromatic)

iii. Cyclopropenyl cation: It is a three-membered ring with a positive charge. It is cyclic, planar, and fully conjugated. It has 2 \(\pi\) electrons. For 4n + 2 = 2, 4n = 0, so n = 0. It obeys Hückel's rule. (Aromatic)

iv. Bicyclo[1.1.0]butane: This is a bicyclic, non-planar molecule. It is not aromatic. (Non-aromatic)

v. Cyclopropenyl cation: This appears to be the same as species iii. Assuming it's a distinct species intended, like cyclobutadiene, which has 4\(\pi\) electrons (anti-aromatic), or some other non-aromatic species, it doesn't add to the count. Let's assume it is just a repeated structure.

vi. Cyclooctatetraene (COT): It is cyclic and has 8 \(\pi\) electrons (a 4n system, where n=2). To avoid the instability of being anti-aromatic, it adopts a non-planar, tub-like shape. Since it's not planar, it is not aromatic. (Non-aromatic)

vii. Anthracene: It is cyclic, planar, and fully conjugated. It has 14 \(\pi\) electrons. For 4n + 2 = 14, 4n = 12, so n = 3. It obeys Hückel's rule. (Aromatic)



Step 4: Final Answer:

The species that are aromatic are i, ii, iii, and vii. Counting these, we find there are 4 aromatic species. This corresponds to option (2).
Quick Tip: When applying Hückel's rule, remember to check all four criteria: Cyclic, Planar, Conjugated, and (4n+2) \(\pi\) electrons. A species failing even one criterion is not aromatic. Common non-aromatic examples include molecules with sp\(^3\) carbons in the ring or non-planar systems like Cyclooctatetraene.

Step 1: Understanding the Question:

The question asks to find the stoichiometric coefficients (a, b, and c) for the given redox reaction in an acidic medium.


Step 2: Key Formula or Approach:

We will use the ion-electron (or half-reaction) method to balance the equation.


Step 3: Detailed Explanation:

1. Identify and balance the half-reactions:

Reduction half-reaction: Dichromate(VI) is reduced to Chromium(III).
\[ Cr_2O_7^{2-} \rightarrow Cr^{3+} \]
- Balance Cr atoms: \( Cr_2O_7^{2-} \rightarrow 2Cr^{3+} \)
- Balance O atoms with H\(_2\)O: \( Cr_2O_7^{2-} \rightarrow 2Cr^{3+} + 7H_2O \)
- Balance H atoms with H\(^+\): \( Cr_2O_7^{2-} + 14H^+ \rightarrow 2Cr^{3+} + 7H_2O \)
- Balance charge with electrons (e\(^-\)): L.H.S charge = (-2) + (+14) = +12. R.H.S charge = 2(+3) = +6. Add 6e\(^-\) to L.H.S.
\[ Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O \quad \cdots(i) \]
Oxidation half-reaction: Sulfite(IV) is oxidized to Sulfate(VI).
\[ SO_3^{2-} \rightarrow SO_4^{2-} \]
- S atoms are balanced.
- Balance O atoms with H\(_2\)O: \( SO_3^{2-} + H_2O \rightarrow SO_4^{2-} \)
- Balance H atoms with H\(^+\): \( SO_3^{2-} + H_2O \rightarrow SO_4^{2-} + 2H^+ \)
- Balance charge with electrons (e\(^-\)): L.H.S charge = -2. R.H.S charge = (-2) + (+2) = 0. Add 2e\(^-\) to R.H.S.
\[ SO_3^{2-} + H_2O \rightarrow SO_4^{2-} + 2H^+ + 2e^- \quad \cdots(ii) \]

2. Combine the half-reactions:
To make the electrons equal, multiply equation (ii) by 3. \[ 3SO_3^{2-} + 3H_2O \rightarrow 3SO_4^{2-} + 6H^+ + 6e^- \]
Now add this to equation (i): \[ Cr_2O_7^{2-} + 14H^+ + 6e^- + 3SO_3^{2-} + 3H_2O \rightarrow 2Cr^{3+} + 7H_2O + 3SO_4^{2-} + 6H^+ + 6e^- \]
3. Simplify the final equation:
Cancel species that appear on both sides (6e\(^-\), 6H\(^+\), 3H\(_2\)O). \[ Cr_2O_7^{2-} + 3SO_3^{2-} + 8H^+ \rightarrow 2Cr^{3+} + 3SO_4^{2-} + 4H_2O \]

Step 4: Final Answer:

Comparing the balanced equation with the given format `a \(Cr_2O_7^{2-}\) + b \(SO_3^{2-}\) + c H+ ...`, we find:
a = 1
b = 3
c = 8
These coefficients correspond to option (2).
Quick Tip: Always double-check your final balanced redox equation by ensuring that both the atoms of each element and the total charge are balanced on both sides of the reaction. This final check can catch simple arithmetic errors.

Step 1: Understanding the Question:

The question asks to predict the products (A and B) of the reaction between benzyl phenyl ether and hydrogen iodide (HI) with heat. This is a classic ether cleavage reaction.


Step 2: Key Formula or Approach:

The cleavage of ethers by hydrogen halides (like HI or HBr) follows specific rules:

1. The oxygen atom of the ether is first protonated by the acid.

2. The halide ion (I\(^-\)) then acts as a nucleophile and attacks one of the carbon atoms attached to the oxygen, displacing the other part of the molecule.

3. The crucial rule for mixed ethers (with different alkyl/aryl groups) is that an aryl-oxygen bond (C\(_aryl\)-O) is very strong due to resonance and the sp\(^2\) character of the carbon, and it does not break. The cleavage always occurs at the alkyl-oxygen bond.


Step 3: Detailed Explanation:

The starting material is benzyl phenyl ether: \( C_6H_5-O-CH_2-C_6H_5 \).


The oxygen is bonded to a phenyl group (\(C_6H_5\)) and a benzyl group (\(-CH_2-C_6H_5\)).
As per the rule, the C\(_phenyl\)-O bond will not break.
Therefore, cleavage must occur at the O-\(CH_2\) (benzyl) bond.
The iodide ion (I\(^-\)) will attack the benzylic carbon (\(-CH_2-\)).
The other fragment will be the phenoxide ion, which will be protonated by the acid to form phenol.

The overall reaction is: \[ C_6H_5-O-CH_2-C_6H_5 + HI \xrightarrow{\Delta} C_6H_5-OH + I-CH_2-C_6H_5 \]
So, the products A and B are phenol and benzyl iodide.


Step 4: Final Answer:

Matching the products with the given options, we find that option (4) correctly identifies A as benzyl iodide (\(CH_2I\) attached to a benzene ring) and B as phenol (OH attached to a benzene ring).
Quick Tip: A key rule for ether cleavage with HX: the bond between an sp\(^2\) carbon (from a phenyl or vinyl group) and the ether oxygen is never broken. This means that if a phenyl group is present, phenol will always be one of the products.

Step 1: Understanding the Question:

The question presents a multi-step reaction sequence and asks for the structure of the final product [D].


Step 2: Key Formula or Approach:

We need to identify the product of each step in the sequence. The key reactions are reduction of an aldehyde, dehydration of an alcohol, addition of HBr to an alkene, and a Wurtz-Fittig reaction.


Step 3: Detailed Explanation:


Step 1: CH\(_3\)CHO \(\xrightarrow{i) LiAlH_4 ii) H_3O^+}\) [A]

Lithium aluminium hydride (LiAlH\(_4\)) is a strong reducing agent that reduces the aldehyde ethanal (CH\(_3\)CHO) to a primary alcohol.
\[ [A] is CH_3CH_2OH (Ethanol) \]
Step 2: [A] \(\xrightarrow{H_2SO_4, \Delta}\) [B]

Ethanol is dehydrated by concentrated sulfuric acid upon heating to form an alkene.
\[ [B] is CH_2=CH_2 (Ethene) \]
Step 3: [B] \(\xrightarrow{HBr}\) [C]

Ethene undergoes electrophilic addition with hydrogen bromide.
\[ [C] is CH_3CH_2Br (Bromoethane) \]
Step 4: [C] + Bromobenzene \(\xrightarrow{Na/dry ether}\) [D]

This is a Wurtz-Fittig reaction. Bromoethane (an alkyl halide) reacts with bromobenzene (an aryl halide) and sodium metal in dry ether to form an alkylbenzene. The ethyl group from bromoethane attaches to the phenyl ring from bromobenzene.
\[ CH_3CH_2Br + C_6H_5Br + 2Na \xrightarrow{dry ether} C_6H_5-CH_2CH_3 + 2NaBr \]
\[ [D] is Ethylbenzene \]


Step 4: Final Answer:

The final product [D] is ethylbenzene, which is represented by the structure in option (2).
Quick Tip: Pay close attention to the layout of reaction schemes. The reaction of an alkyl halide and an aryl halide with sodium in ether is a specific named reaction called the Wurtz-Fittig reaction, which is used to synthesize alkylbenzenes.

Step 1: Understanding the Question:

This Assertion-Reason question deals with the characteristics of late wood (autumn wood) and the activity of the cambium.


Step 2: Detailed Explanation:

Analysis of Assertion A:

In temperate regions, trees form annual rings due to seasonal variations in the activity of the vascular cambium.

Early wood (Spring wood): Formed during spring when conditions are favorable. It has a larger number of xylary elements (vessels, tracheids), and the vessels have wider cavities.
Late wood (Autumn wood): Formed during autumn or winter when conditions are less favorable. It has fewer xylary elements, and the vessels are narrow and thick-walled. This wood is denser.

So, the statement "Late wood has fewer xylary elements with narrow vessels" is true.


Analysis of Reason R:

The activity of the vascular cambium is regulated by physiological and environmental factors, including hormones and climate. In spring, the cambium is very active. In contrast, during the unfavorable conditions of winter, the cambium becomes less active or dormant. So, the statement "Cambium is less active in winters" is true.


Conclusion:

The reduced activity of the cambium during winter (Reason R) is the direct cause for the formation of late wood with its characteristic features of fewer and narrower vessels (Assertion A). Therefore, Reason R is the correct explanation for Assertion A.


Step 3: Final Answer:

Both Assertion A and Reason R are true, and R is the correct explanation of A.
Quick Tip: Remember the contrast: Spring = Active Cambium \(\rightarrow\) Wide Vessels (Early Wood). Winter = Less Active Cambium \(\rightarrow\) Narrow Vessels (Late Wood). The reason (cambium activity) directly explains the result (wood structure).

Step 1: Understanding the Question:

The question asks for the definition of pleiotropism (or pleiotropy).


Step 2: Detailed Explanation:

Let's analyze the definition and compare it with the options.

Pleiotropy is a condition in which a single gene influences two or more seemingly unrelated phenotypic traits. A mutation in a pleiotropic gene can have a range of effects on the organism.

A classic example is phenylketonuria (PKU), an inherited disorder in humans. A single gene that codes for the enzyme phenylalanine hydroxylase is mutated. This single genetic defect leads to multiple phenotypic consequences, including mental retardation, reduced hair and skin pigmentation, and eczema.

Now let's examine the given options:


(A) more than two genes affecting a single character: This describes polygenic inheritance, the opposite of pleiotropy.
(B) presence of several alleles of a single gene controlling a single crossover: This describes multiple alleles, which relates to the variety of alleles for a single gene in a population, not the number of traits a single gene affects. Crossover is a separate meiotic process.
(C) presence of two alleles, each of the two genes controlling a single trait: This is a confusing statement but does not describe pleiotropy.
(D) a single gene affecting multiple phenotypic expression: This is the correct definition of pleiotropy.


Step 3: Final Answer:

The phenomenon of pleiotropism refers to a single gene affecting multiple phenotypic expressions.
Quick Tip: To avoid confusion, remember: \textbf{Pleiotropy:} One gene \(\rightarrow\) Many traits (e.g., PKU, sickle-cell anemia). \textbf{Polygenic Inheritance:} Many genes \(\rightarrow\) One trait (e.g., skin color, height). They are inverse concepts.

Step 1: Understanding the Question:

The question asks for the standard unit of measurement for the thickness of the atmospheric ozone layer.


Step 2: Detailed Explanation:

Let's examine the units provided:


Kilobase (kb): A unit of length for DNA or RNA sequences, equal to 1000 base pairs. Not used for atmospheric measurements.

Dobson units (DU): This is the standard unit used to measure the total amount of ozone in a vertical column of air in the atmosphere. One Dobson Unit is the number of molecules of ozone that would be required to create a layer of pure ozone 0.01 millimeters thick at a temperature of 0 degrees Celsius and a pressure of 1 atmosphere.

Decibels (dB): A logarithmic unit used to measure the intensity of sound.

Decameter (dam): A unit of length equal to 10 meters. While a unit of thickness, it is not the specialized unit used for the ozone layer.


The correct unit specifically for ozone layer thickness is the Dobson unit.


Step 3: Final Answer:

The thickness of the ozone layer is measured in Dobson Units.
Quick Tip: Associate "Ozone layer" with "Dobson Units". This is a key factual point in environmental science and ecology topics. Remember that the other units are from completely different fields of science.

Step 1: Understanding the Question:

The question asks to identify a sequence of structures from a fertilized embryo sac in angiosperms that are haploid (n), diploid (2n), and triploid (3n), respectively.


Step 2: Detailed Explanation:

In angiosperms, double fertilization occurs. The embryo sac (female gametophyte) contains several cells with different ploidy levels before and after fertilization.


Haploid (n) structures: Before fertilization, the egg cell, synergids, and antipodal cells are all haploid. After fertilization, the synergids and antipodals degenerate, but they are still considered haploid structures of the embryo sac.

Diploid (2n) structure: One male gamete (n) fuses with the egg cell (n) to form the zygote, which is diploid (2n). The zygote develops into the embryo.

Triploid (3n) structure: The other male gamete (n) fuses with the central cell, which contains two polar nuclei (n + n). This fusion results in the Primary Endosperm Nucleus (PEN), which is triploid (3n). The PEN develops into the endosperm, a nutritive tissue.


Now let's check the options for the sequence: haploid, diploid, triploid.


(A) Synergids (n), antipodals (n), Polar nuclei (n+n, not a single structure after fertilization) - Incorrect sequence.
(B) Synergids (n), Primary endosperm nucleus (3n), zygote (2n) - Incorrect order.
(C) Antipodals (n), synergids (n), primary endosperm nucleus (3n) - Incorrect sequence.
(D) Synergids (n), Zygote (2n), and Primary endosperm nucleus (3n) - Correct sequential order of ploidy (n, 2n, 3n).



Step 3: Final Answer:

The correct sequence of haploid, diploid, and triploid structures is Synergids (n), Zygote (2n), and Primary endosperm nucleus (3n).
Quick Tip: Memorize the products of double fertilization: 1. Sperm (n) + Egg (n) \(\rightarrow\) Zygote (2n) 2. Sperm (n) + 2 Polar Nuclei (n+n) \(\rightarrow\) Primary Endosperm Nucleus (3n) This will help you quickly identify the ploidy of key structures in a fertilized embryo sac.

Step 1: Understanding the Question:

The question presents two statements about transpiration and asks to evaluate their correctness.


Step 2: Detailed Explanation:

Analysis of Statement I:

This statement refers to the transpiration pull, which is the main driver of water movement in the xylem according to the cohesion-tension theory. The combined forces of cohesion (water molecules sticking together), adhesion (water molecules sticking to xylem walls), and the tension created by evaporation from leaves (transpiration) can generate a very strong negative pressure potential. This force is sufficient to pull water up to the tops of the tallest trees, such as the coastal redwood (\textit{Sequoia sempervirens), which can exceed 100 meters. Therefore, lifting water over 130 meters is within the capability of this mechanism. Statement I is correct.


Analysis of Statement II:

Transpiration is the process of water evaporating from the leaf surface. Evaporation is a cooling process because the water molecules with the highest kinetic energy (i.e., the "hottest" molecules) are the ones that escape as vapor, leaving behind the cooler molecules. This process of evaporative cooling can significantly lower the temperature of the leaf surface, often by 10 to 15 degrees Celsius, protecting the leaves from overheating in direct sunlight. Statement II is correct.


Step 3: Final Answer:

Since both statements accurately describe phenomena related to transpiration, both Statement I and Statement II are correct.
Quick Tip: For questions with two statements, analyze each one independently first. Determine if it's true or false on its own before looking at the combined options. This prevents confusion.

Step 1: Understanding the Question:

The question asks who first used the frequency of genetic recombination to create a map of genes on a chromosome.


Step 2: Detailed Explanation:

Let's look at the contributions of the individuals mentioned:


Henking: Discovered the X chromosome.

Thomas Hunt Morgan: Working with \textit{Drosophila melanogaster (fruit flies), he established the chromosomal theory of inheritance, demonstrating that genes are located on chromosomes. He also discovered concepts like linkage and recombination.

Sutton and Boveri: Independently proposed the Chromosomal Theory of Inheritance, which states that chromosomes are the carriers of genetic material.

Alfred Sturtevant: He was an undergraduate student in T.H. Morgan's lab. In 1913, he had a crucial insight: he reasoned that the frequency of recombination between two linked genes is proportional to the physical distance between them on the chromosome. Based on this principle, he used recombination data from fruit fly crosses to construct the very first genetic map, showing the linear arrangement of genes on a chromosome.


While Morgan's work laid the foundation, it was his student, Sturtevant, who first applied the concept of recombination frequency to gene mapping.


Step 3: Final Answer:

Alfred Sturtevant was the first to use recombination frequency to map the position of genes on a chromosome.
Quick Tip: Remember the relationship: Morgan was the professor who discovered linkage and recombination. Sturtevant was his student who had the brilliant idea to use the recombination frequencies to make the first gene map.

Step 1: Understanding the Question:

The question asks to identify what the term 'R' represents in the ecological equation relating Gross Primary Productivity (GPP) and Net Primary Productivity (NPP).


Step 2: Detailed Explanation:

Let's define the terms in the equation:


Gross Primary Productivity (GPP): This is the total rate at which solar energy is captured by producers (like plants) during photosynthesis to create organic matter. It represents the total amount of food produced.

Net Primary Productivity (NPP): This is the rate at which producers create biomass that is actually available to the next trophic level (herbivores). It is the energy that remains after the producers have met their own metabolic needs.

R (Respiration): Producers use a significant portion of the energy they capture (GPP) for their own life processes, such as cellular respiration, growth, and maintenance. This energy consumed by the producer is known as respiratory loss.


The equation GPP - R = NPP means that the Net Primary Productivity is what's left over from the Gross Primary Productivity after subtracting the energy lost to respiration.


Step 3: Final Answer:

In the given equation, R represents the energy lost by the producers through respiration, i.e., Respiratory loss.
Quick Tip: Think of GPP as the 'gross salary' of an ecosystem's producers. 'R' is the 'tax' or 'living expenses' (energy used for respiration). NPP is the 'net salary' or 'take-home pay'—the energy available to be saved (as biomass) or spent (by consumers).

Step 1: Understanding the Question:

The question asks to identify the most significant cause of species extinction from the four major causes known as 'The Evil Quartet'.


Step 2: Detailed Explanation:

'The Evil Quartet' is a term used to describe the four main causes of biodiversity loss:


Habitat loss and fragmentation: This involves the destruction or division of natural habitats due to activities like deforestation for agriculture, urbanization, and mining. When an organism's home is destroyed or broken into small, isolated patches, its populations decline, and it faces a higher risk of extinction. This is widely recognized by ecologists as the single greatest threat to biodiversity worldwide.

Over-exploitation: This refers to harvesting species from the wild at rates faster than natural populations can recover. Overfishing, overhunting, and excessive logging are examples.

Alien species invasions: When non-native species are introduced into an ecosystem, they can outcompete native species for resources, introduce diseases, or alter the habitat, leading to the decline and extinction of native organisms.

Co-extinctions: This occurs when the extinction of one species causes the extinction of another species that depended on it, such as a plant and its specific pollinator.


While all four are significant threats, habitat loss and fragmentation is considered the most important and primary driver of species extinction.


Step 3: Final Answer:

Among 'The Evil Quartet', habitat loss and fragmentation is the most important cause of species extinction.
Quick Tip: Remember that all other causes of extinction are often exacerbated by habitat loss. If an animal has nowhere to live, it cannot survive, making this the most fundamental threat.

Step 1: Understanding the Question:

The question asks for the total number of ATP and NADPH molecules required to produce one molecule of glucose via the Calvin cycle. (Note: \(NADPH_{2}\) is often used to represent NADPH + H\(^+\)).


Step 2: Key Formula or Approach:

The synthesis of one molecule of glucose (\(C_{6}H_{12}O_{6}\)) requires 6 molecules of \(CO_{2}\) to be fixed. This means the Calvin cycle must turn 6 times. We need to calculate the energy requirement for a single turn and then multiply it by 6.


Step 3: Detailed Explanation:

The Calvin cycle has three main stages: Carboxylation, Reduction, and Regeneration.


Carboxylation: \(CO_{2}\) is fixed to RuBP. No energy is consumed here.

Reduction: The product of carboxylation (3-PGA) is reduced to triose phosphate. For each molecule of \(CO_{2}\) fixed:

2 molecules of ATP are used.
2 molecules of NADPH are used.

Regeneration: The starting molecule, RuBP, is regenerated from triose phosphate. For each molecule of \(CO_{2}\) fixed:

1 molecule of ATP is used.


Total requirement per turn (for 1 \(CO_{2}\)):
\[ ATP = 2 (from reduction) + 1 (from regeneration) = 3 ATP \] \[ NADPH = 2 (from reduction) = 2 NADPH \]
Total requirement for 6 turns (for 1 glucose):
\[ Total ATP = 3 ATP/turn \times 6 turns = 18 ATP \] \[ Total NADPH = 2 NADPH/turn \times 6 turns = 12 NADPH \]

Step 4: Final Answer:

Therefore, the synthesis of one molecule of glucose requires 18 ATP and 12 NADPH.
Quick Tip: Remember the "3-2-1" rule for the Calvin cycle per \(CO_{2}\) fixed: 3 ATPs are used (2 in reduction, 1 in regeneration) and 2 NADPHs are used (both in reduction). Since glucose has 6 carbons, multiply these numbers by 6.

Step 1: Understanding the Question:

The question presents two statements about the arrangement of xylem in plants (endarch and exarch) and asks for an evaluation of their correctness.


Step 2: Detailed Explanation:

Analysis of Statement I:

The terms "endarch" and "exarch" describe the pattern of development of primary xylem, not secondary xylem. These terms refer to the position of the first-formed primary xylem (protoxylem) relative to the later-formed primary xylem (metaxylem).


Endarch: Protoxylem is located towards the center (pith), and metaxylem is towards the periphery. This is characteristic of stems.

Exarch: Protoxylem is located towards the periphery, and metaxylem is towards the center. This is characteristic of roots.


Secondary xylem is formed by the vascular cambium during secondary growth and does not follow this developmental pattern. Therefore, Statement I is incorrect.


Analysis of Statement II:

As explained above, the exarch arrangement of primary xylem (protoxylem on the outside, metaxylem on the inside) is the defining characteristic of the vascular bundles in the roots of vascular plants. Therefore, Statement II is true.


Step 3: Final Answer:

Based on the analysis, Statement I is incorrect, and Statement II is true.
Quick Tip: Use a mnemonic to remember the xylem arrangement: "Exarch in roots, Endarch in stems." The 'x' in exarch can remind you of the 'x' shape xylem often forms in a root cross-section. Also, remember these terms apply only to primary xylem.

Step 1: Understanding the Question:

The question asks to identify the characteristics of stamens that are specific to the family Fabaceae when compared to Solanaceae and Liliaceae.


Step 2: Detailed Explanation:

Let's compare the stamen characteristics of the three families:


Family Fabaceae (Legume family): The androecium typically consists of ten stamens. A key feature is that the filaments are often fused. The most common arrangement is diadelphous, where they are fused into two groups, usually nine in one bundle and one free ((9)+1). The anthers are dithecous (two-lobed).

Family Solanaceae (Potato family): There are five stamens. They are epipetalous, meaning the filaments are attached to the petals. The anthers are dithecous.

Family Liliaceae (Lily family): There are six stamens, arranged in two whorls of three (3+3). They are often epiphyllous or epitepalous, meaning the filaments are attached to the tepals (undifferentiated petals and sepals). The anthers are dithecous.


Now let's evaluate the options based on this comparison:

(A) Epiphyllous condition is characteristic of Liliaceae.

(B) Diadelphous condition is a hallmark of Fabaceae and is not found in Solanaceae or Liliaceae. Dithecous anthers are common to all three, but the combination is specific.

(C) Polyadelphous (filaments in many bundles) is found in families like Rutaceae (Citrus). Epipetalous condition is found in Solanaceae.

(D) Monoadelphous (filaments in one bundle) and monothecous anthers are characteristic of the family Malvaceae (China rose), not Fabaceae.


Step 3: Final Answer:

The diadelphous condition of stamens is the specific characteristic of family Fabaceae among the given options.
Quick Tip: To differentiate plant families, focus on unique floral characteristics. For Fabaceae, remember the "pea flower" structure: papilionaceous corolla and diadelphous stamens ((9)+1). For Solanaceae, remember epipetalous stamens. For Liliaceae, remember parts in threes and epiphyllous stamens.

Step 1: Understanding the Question:

The question asks to identify a pair where both members are heterosporous pteridophytes.


Step 2: Detailed Explanation:

Pteridophytes can be classified based on the types of spores they produce:


Homosporous: These plants produce only one type of spore, which develops into a bisexual (monoecious) gametophyte bearing both antheridia and archegonia. The majority of pteridophytes are homosporous. Examples include Psilotum, \textit{Lycopodium, and \textit{Equisetum.

Heterosporous: These plants produce two distinct types of spores: small microspores that develop into male gametophytes, and large megaspores that develop into female gametophytes. Heterospory is an important evolutionary step towards the seed habit. Key examples of heterosporous pteridophytes are \textit{Selaginella, \textit{Salvinia, \textit{Azolla, and \textit{Marsilea.


Let's analyze the given pairs:

(A) \textit{Equisetum (homosporous) and \textit{Salvinia (heterosporous).

(B) \textit{Lycopodium (homosporous) and \textit{Selaginella (heterosporous).

(C) \textit{Selaginella (heterosporous) and Salvinia (heterosporous). This pair consists of two heterosporous pteridophytes.

(D) Psilotum (homosporous) and \textit{Salvinia (heterosporous).


Step 3: Final Answer:

The pair of heterosporous pteridophytes is \textit{Selaginella and \textit{Salvinia.
Quick Tip: For competitive exams, it's essential to memorize the key examples of heterosporous pteridophytes: \textit{Selaginella, Salvinia, Azolla, and Marsilea. Most other common pteridophytes you encounter will be homosporous.

Step 1: Understanding the Question:

The question asks to identify the specific phase of the eukaryotic cell cycle during which DNA replication occurs.


Step 2: Detailed Explanation:

The eukaryotic cell cycle is divided into two main stages: Interphase and the M phase (Mitotic phase). Interphase is further subdivided into three phases:


G\(_1\) phase (Gap 1): This is the first growth phase where the cell grows and carries out normal metabolic functions.

S phase (Synthesis phase): This is the phase where the cell synthesizes a complete copy of the DNA in its nucleus. DNA replication occurs during this stage. The amount of DNA in the cell doubles.

G\(_2\) phase (Gap 2): The cell continues to grow and prepares for mitosis. It synthesizes proteins and organelles needed for cell division.

M phase (Mitotic phase): This is the phase of actual cell division, which includes mitosis (nuclear division) and cytokinesis (cytoplasmic division).


Therefore, DNA replication is confined to the S phase of the cell cycle.


Step 3: Final Answer:

In eukaryotes, DNA replication takes place during the S phase of the cell cycle.
Quick Tip: Remember the mnemonic for the cell cycle: Go (G\(_1\)) -> Sally (S) -> Go (G\(_2\)) -> Make (M) -> Cookies (Cytokinesis). The 'S' for Sally stands for Synthesis, which is when DNA replication happens.

Step 1: Understanding the Question:

The question asks to identify the plant hormone responsible for the rapid elongation of internodes or petioles in deep water rice varieties when they are submerged.


Step 2: Detailed Explanation:

Deep water rice has a remarkable adaptation to survive flooding. When the plant is submerged, the reduced oxygen (hypoxia) triggers a significant increase in the synthesis of the gaseous hormone ethylene.


Ethylene accumulates in the submerged parts of the plant because its diffusion out of the tissues is much slower in water than in air.
This high concentration of ethylene triggers a signaling cascade that either reduces the level of abscisic acid (ABA, a growth inhibitor) or increases the sensitivity of the cells to gibberellins (GA).
The ultimate result is a dramatic increase in cell division and elongation in the internodes, causing the stem to grow rapidly and keep the leaves above the water surface for photosynthesis and gas exchange.

While gibberellins (like \(GA_3\)) are the direct promoters of stem elongation, ethylene is the primary hormonal signal that initiates this specific response to submergence in deep water rice. Among the choices given, ethylene is the most appropriate answer for promoting this adaptive elongation.


2, 4-D is a synthetic auxin.
\(GA_3\) is involved, but ethylene is the primary trigger.
Kinetin is a cytokinin, primarily involved in cell division.


Step 3: Final Answer:

Ethylene is the hormone that promotes internode/petiole elongation in deep water rice as an adaptation to flooding.
Quick Tip: For questions about specific plant adaptations, remember the primary trigger. In deep water rice, the trigger for elongation is submergence, which leads to the accumulation of the gaseous hormone Ethylene.

Step 1: Understanding the Question:

The question describes a set of floral characteristics (large, colourful, fragrant, with nectar) and asks to identify the corresponding mode of pollination.


Step 2: Detailed Explanation:

Plants have evolved specific floral traits to attract different types of pollinators. This is known as pollination syndrome. Let's analyze the traits given:


Large and Colourful Flowers: These are visual signals to attract pollinators from a distance. Insects, especially bees and butterflies, have good colour vision.

Fragrant Flowers: Scent is a chemical attractant, especially for insects like moths (at night) and bees.

Nectar: This is a sugary fluid that serves as a food reward for the pollinator.


These characteristics—visual appeal, scent, and a food reward—are classic adaptations for attracting insects. This is called entomophily.

Let's consider other options:


Wind pollinated (anemophilous) plants: Flowers are typically small, inconspicuous, not colourful, lack nectar and scent, and produce large amounts of light, dusty pollen.

Bird pollinated (ornithophilous) plants: Flowers are often large, brightly coloured (especially red or orange), have little to no scent (birds have a poor sense of smell), and produce copious amounts of dilute nectar.

Bat pollinated (chiropterophilous) plants: Flowers are typically large, pale or white, open at night, and have a strong, musty or fruity odour.


The combination of being colourful AND fragrant is most characteristic of insect pollination.


Step 3: Final Answer:

The combination of large, colourful, fragrant flowers with nectar is a hallmark of insect pollinated plants.
Quick Tip: Think of floral characteristics as "advertisements" for pollinators. Bright colors and sweet smells attract insects, just like advertisements attract customers. Wind-pollinated plants don't need to advertise, so their flowers are plain and functional.

Step 1: Understanding the Question:

The question asks to identify the micronutrient that is essential for the photolysis (splitting) of water during the light-dependent reactions of photosynthesis.


Step 2: Detailed Explanation:

The splitting of water molecules (H\(_2\)O) into protons (H\(^+\)), electrons (e\(^-\)), and oxygen (O\(_2\)) occurs in Photosystem II (PS II). This reaction is catalyzed by the Oxygen Evolving Complex (OEC).
\[ 2H_2O \rightarrow 4H^+ + 4e^- + O_2 \]
The core of the OEC is a cluster of four manganese ions (Mn) and one calcium ion (Ca\(^{2+}\)), which are stabilized by proteins. The manganese ions cycle through different oxidation states to catalyze the water-splitting reaction. Therefore, manganese is the essential micronutrient for this process.

Let's look at the other options:


Copper (Cu): A component of plastocyanin, an electron carrier protein in the electron transport chain between PS II and PS I.

Molybdenum (Mo): A component of enzymes like nitrogenase (for nitrogen fixation) and nitrate reductase.

Magnesium (Mg): A macronutrient, not a micronutrient. It is the central atom in the chlorophyll molecule, essential for capturing light energy, but not directly involved in splitting water.



Step 3: Final Answer:

Manganese (Mn) is the crucial micronutrient required for the splitting of water molecules during photosynthesis.
Quick Tip: Remember the key roles of micronutrients in photosynthesis: Mn for water splitting, Cu in plastocyanin, and Mg (a macronutrient) as the core of chlorophyll.

Step 1: Understanding the Question:

The question asks about the appearance of DNA stained with ethidium bromide when it is exposed to ultraviolet (UV) light.


Step 2: Detailed Explanation:

Ethidium bromide (EtBr) is a fluorescent dye commonly used in molecular biology laboratories to visualize DNA in agarose gel electrophoresis.


EtBr works by intercalating, or inserting itself, between the base pairs of the DNA double helix.

When the gel containing the EtBr-stained DNA is placed under a UV transilluminator, the EtBr molecules absorb the UV radiation.

This absorption of energy excites the EtBr molecules, causing them to fluoresce.

The light emitted during this fluorescence is in the orange part of the visible spectrum.


Therefore, the DNA bands on the gel appear as a bright orange colour. The other colours listed are incorrect for EtBr staining.


Step 3: Final Answer:

DNA stained with ethidium bromide fluoresces with a bright orange colour under UV radiation.
Quick Tip: This is a standard technique in molecular biology. Associate Ethidium Bromide (EtBr) + DNA + UV light with "bright orange bands". This is a frequently asked direct recall question.

Step 1: Understanding the Question:

The question asks for the function of "tassels in the corn cob". This phrasing is slightly ambiguous, as the tassel and the cob are separate parts of the corn plant. The tassel is the male inflorescence at the top, and the cob is the ear that bears the female flowers and kernels. The structures emerging from the cob are called silks. It is highly likely the question is referring to the function of these silks.


Step 2: Detailed Explanation:

Let's clarify the botanical terms for a corn plant (maize):


Tassel: This is the male flower cluster located at the apex (top) of the plant. Its function is to produce and disperse pollen grains into the wind. So, option (D) is the function of the actual tassel.

Ear (Cob): This is the female inflorescence located on a side shoot. It develops into the fruit (the "corn on the cob").

Silks: These are the long, thread-like structures that emerge from the tip of the ear. Each silk is a style and stigma. The function of the silks is to trap airborne pollen grains. Each silk is connected to an ovule, and successful pollination of a silk leads to the fertilization of that ovule, which then develops into a corn kernel.


Given the options and the phrasing "in the corn cob", the question is almost certainly asking about the function of the silks, not the tassel at the top of the plant. The function of the silks is to trap pollen grains.


(A) To protect seeds: The husk protects the seeds (kernels).
(B) To attract insects: Corn is wind-pollinated, not insect-pollinated.
(C) To trap pollen grains: This is the function of the silks on the cob.
(D) To disperse pollen grains: This is the function of the tassel at the top of the plant.

Based on the likely intent of the question, option (C) is the correct answer.


Step 3: Final Answer:

Assuming "tassels in the corn cob" refers to the silks emerging from the cob, their function is to trap pollen grains.
Quick Tip: Be aware that exam questions sometimes use common or slightly inaccurate terminology. The Tassel (top) produces pollen, and the Silk (on the cob) receives it. Understanding the function of both parts helps you deduce the intended answer even with confusing wording.

Step 1: Understanding the Question:

The question asks what biological macromolecule precipitates when chilled ethanol is added during a typical DNA purification protocol.


Step 2: Detailed Explanation:

The process of isolating DNA involves several steps:


Lysis: Breaking open the cells to release their contents, including DNA.
Purification: Removing other macromolecules like proteins, RNA, and lipids. This is often done using enzymes like proteases and RNases.
Precipitation: Separating the DNA from the remaining solution. DNA is a polar molecule and is soluble in aqueous solutions. However, it is insoluble in ethanol (or isopropanol).

When cold ethanol is added to the aqueous DNA solution, it disrupts the hydration shell around the DNA molecules. The presence of salt (like sodium acetate) in the buffer neutralizes the negative charge of the phosphate backbone of DNA. This combination causes the DNA molecules to aggregate and precipitate out of the solution. The precipitated DNA can then be seen as a mass of fine white threads, which can be spooled out with a glass rod.


Step 3: Final Answer:

The addition of chilled ethanol causes the precipitation of DNA.
Quick Tip: Remember the final step of most DNA extraction kits: "Add ethanol to precipitate DNA." The cold temperature of the ethanol helps to increase the yield of the precipitate.

Step 1: Understanding the Question:

The question presents an assertion and a reason related to the consumption of ATP during glycolysis. We need to determine if both statements are true and if the reason correctly explains the assertion.


Step 2: Detailed Explanation:

Analysis of Assertion A:

Glycolysis is a 10-step pathway that breaks down glucose. The initial part of the pathway is known as the preparatory or investment phase, where energy (in the form of ATP) is consumed to prepare the glucose molecule for cleavage. In this phase, ATP is indeed used at two specific steps. So, Assertion A is true.


Analysis of Reason R:

The reason specifies the two steps where ATP is used:


Step 1: Glucose is phosphorylated to glucose-6-phosphate. This reaction is catalyzed by the enzyme hexokinase and consumes one molecule of ATP.

\[ Glucose + ATP \rightarrow Glucose-6-phosphate + ADP \]
Step 3: Fructose-6-phosphate is phosphorylated to fructose-1,6-bisphosphate. (Note: "diphosphate" is an older term, "bisphosphate" is preferred but both refer to the same molecule). This reaction is catalyzed by phosphofructokinase-1 (PFK-1) and consumes a second molecule of ATP.

\[ Fructose-6-phosphate + ATP \rightarrow Fructose-1,6-bisphosphate + ADP \]

The reason correctly identifies the exact two steps of ATP consumption in glycolysis. So, Reason R is true.


Conclusion:

Since Assertion A states that ATP is used in two steps, and Reason R correctly lists these two steps, Reason R is the correct explanation for Assertion A.


Step 3: Final Answer:

Both Assertion A and Reason R are true, and R is the correct explanation of A.
Quick Tip: For Assertion-Reason questions, follow a systematic approach: 1. Check if Assertion (A) is true. 2. Check if Reason (R) is true. 3. If both are true, check if R correctly explains A by asking "Is A true because of R?".

Step 1: Understanding the Question:

The question asks for the correct definition of Expressed Sequence Tags (ESTs).


Step 2: Detailed Explanation:

The term "Expressed Sequence Tag" itself gives a clue.


Expressed: This refers to gene expression, which is the process of a gene's information being used to synthesize a functional gene product. The primary step of expression is transcription, where a gene's DNA sequence is copied into an RNA molecule (mRNA, tRNA, rRNA, etc.).

Sequence Tag: This refers to a short subsequence of a DNA sequence.


ESTs are generated by sequencing short fragments of complementary DNA (cDNA). cDNA is synthesized from messenger RNA (mRNA) using the enzyme reverse transcriptase. Since mRNA is the product of gene transcription, cDNA represents the sequences of genes that are being expressed (transcribed into RNA) in a particular cell or tissue at a particular time.

Therefore, ESTs are tags or markers for genes that are expressed as RNA. Option (B) is the most accurate description.


Option (A) is too restrictive; ESTs represent all expressed genes found in the cDNA library, not just "certain important" ones.

Option (C) is incorrect because not all transcribed RNAs are translated into proteins (e.g., rRNA, tRNA). ESTs are derived from all mRNA, regardless of whether it's translated.

Option (D) is incorrect because ESTs only represent expressed genes, not unexpressed genes (which are not transcribed into mRNA).



Step 3: Final Answer:

Expressed Sequence Tags (ESTs) refer to all genes that are expressed as RNA.
Quick Tip: Break down the term: "Expressed" means transcribed into RNA. "Sequence Tag" means a piece of a sequence. So, ESTs are pieces of sequences from transcribed (expressed) genes. This helps in deducing the correct answer.

Step 1: Understanding the Question:

The question asks to identify the transport mechanism responsible for moving ions across a membrane from a region of lower concentration to a region of higher concentration, which is "against their concentration gradient."


Step 2: Detailed Explanation:

Let's analyze the given options:


Active Transport: This process moves substances (like ions) across a cell membrane against their concentration gradient (from low to high concentration). This movement requires energy, typically in the form of ATP (adenosine triphosphate).

Osmosis: This is the movement of water molecules across a selectively permeable membrane from a region of high water potential to a region of low water potential. It does not directly involve the transport of ions against their gradient.

Facilitated Diffusion: This is a type of passive transport where substances move across membranes down their concentration gradient (from high to low concentration) with the help of membrane proteins (channels or carriers). It does not require energy.

Passive Transport: This is the general term for the movement of substances across a membrane without the use of metabolic energy, following the concentration gradient. Simple diffusion and facilitated diffusion are types of passive transport.


The key phrase in the question is "against their concentration gradient," which is the defining characteristic of active transport.


Step 3: Final Answer:

The movement of ions against their concentration gradient is explained by active transport, which requires energy expenditure.
Quick Tip: Remember the key distinction: "Down the gradient" (high to low concentration) = Passive Transport (no energy). "Against the gradient" (low to high concentration) = Active Transport (requires energy).

Step 1: Understanding the Question:

This is an Assertion-Reason question about the life cycle of a moss. We need to evaluate both statements and their relationship.


Step 2: Detailed Explanation:

Analysis of Assertion A:

The life cycle of a moss (a bryophyte) shows a dominant gametophyte generation. When a haploid spore germinates, it first develops into a filamentous, branching, green structure called the protonema. This is the juvenile, or first stage, of the gametophyte. Later, buds arise from the protonema, which develop into the mature, leafy gametophyte (gametophore). Therefore, Assertion A is correct.


Analysis of Reason R:

The sporophyte generation in mosses ends with the capsule (sporangium), which produces haploid spores through meiosis. When these spores are released and land on a suitable substrate, they germinate and grow directly into the protonema. Therefore, the statement that the protonema develops directly from spores produced in the capsule is correct. Reason R is correct.


Conclusion:

Assertion A states that the protonema is the first stage of the gametophyte. Reason R explains how this protonema stage originates (from a spore). Thus, Reason R provides the correct explanation for Assertion A.


Step 3: Final Answer:

Both Assertion A and Reason R are correct, and R is the correct explanation of A.
Quick Tip: To master bryophyte life cycles, draw a simple diagram: Spore (n) \(\rightarrow\) Germination \(\rightarrow\) Protonema (n) \(\rightarrow\) Leafy Gametophyte (n) \(\rightarrow\) Gametes (n) \(\rightarrow\) Fertilization \(\rightarrow\) Zygote (2n) \(\rightarrow\) Sporophyte (2n) \(\rightarrow\) Meiosis in Capsule \(\rightarrow\) Spores (n). This visual aid helps clarify the sequence of stages.

Step 1: Understanding the Question:

The question asks to identify the correct statements related to the process of decomposition from the given list.


Step 2: Detailed Explanation:

Let's analyze each statement:


A. Detrivores perform fragmentation. Detrivores, such as earthworms, break down dead organic matter (detritus) into smaller particles. This process is known as fragmentation. This statement is correct.

B. The humus is further degraded by some microbes during mineralization. Humus is a dark, amorphous substance that is highly resistant to microbial action and decomposes at an extremely slow rate. Microbes degrade humus to release inorganic nutrients, a process called mineralization. This statement is correct.

C. Water soluble inorganic nutrients go down into the soil and get precipitated by a process called leaching. Leaching is the process where water-soluble inorganic nutrients percolate down into the soil horizon and can become unavailable as precipitated salts. This statement is correct.

D. The detritus food chain begins with living organisms. The detritus food chain (DFC) begins with dead organic matter (detritus). The grazing food chain (GFC) begins with living organisms (producers). This statement is incorrect.

E. Earthworms break down detritus into smaller particles by a process called catabolism. Earthworms perform fragmentation. Catabolism is the enzymatic breakdown of detritus into simpler inorganic substances by bacteria and fungi. This statement is incorrect.



Step 3: Final Answer:

Statements A, B, and C are correct. Therefore, the correct option includes only these three statements.
Quick Tip: Remember the three main steps of decomposition in order: Fragmentation (physical breakdown by detritivores), Leaching (washing away of soluble nutrients), and Catabolism (chemical breakdown by microbes). Mineralization and humification occur as part of catabolism.

Step 1: Understanding the Question:

The question asks for the peak absorption wavelength of the reaction center chlorophyll of Photosystem II (PS II).


Step 2: Detailed Explanation:

In higher plants and algae, the light-dependent reactions of photosynthesis involve two photosystems: Photosystem I (PS I) and Photosystem II (PS II). Each photosystem has a light-harvesting complex (antenna molecules) and a reaction center.


Photosystem II (PS II): The reaction center of PS II is a special chlorophyll a molecule designated as P680. This indicates that it has a peak absorption of light at a wavelength of 680 nm. PS II is involved in the splitting of water and is the first photosystem in the Z-scheme of electron transport.

Photosystem I (PS I): The reaction center of PS I is designated as P700, meaning its absorption maximum is at 700 nm.


The other wavelengths listed (780 nm, 660 nm) are not the standard designations for the reaction centers of PS I or PS II.


Step 3: Final Answer:

The reaction centre in PS II has an absorption maximum at 680 nm.
Quick Tip: Remember the order in the Z-scheme: PS II comes first, then PS I. The numbers associated with them are P680 and P700, respectively. Associate the smaller number (680) with the first photosystem (PS II) and the larger number (700) with the second (PS I).

Step 1: Understanding the Question:

The question asks for the specific function of RNA polymerase III in the transcription process within eukaryotic cells.


Step 2: Detailed Explanation:

In eukaryotes, there are three main types of RNA polymerases, each responsible for transcribing different types of RNA:


RNA Polymerase I: Located in the nucleolus, it transcribes most ribosomal RNAs (rRNAs), specifically the 28S, 18S, and 5.8S rRNA genes.

RNA Polymerase II: Located in the nucleoplasm, it transcribes the precursor of messenger RNA (pre-mRNA), which is then processed to form mRNA. It also transcribes most small nuclear RNAs (snRNAs) and microRNAs (miRNAs).

RNA Polymerase III: Located in the nucleoplasm, it transcribes transfer RNA (tRNA), 5S rRNA (a component of the ribosome), and some small nuclear RNAs (snRNAs), such as U6 snRNA.


Based on these functions, option (C) correctly identifies the products of RNA polymerase III transcription.

Option (A) is incorrect because Pol III transcribes more than just snRNAs.

Option (B) is incorrect as these rRNAs are transcribed by RNA Polymerase I.

Option (D) is incorrect as the precursor of mRNA is transcribed by RNA Polymerase II.


Step 3: Final Answer:

Therefore, the role of RNA polymerase III is the transcription of tRNA, 5S rRNA, and some snRNAs.
Quick Tip: Create a simple table to memorize the functions of the three eukaryotic RNA polymerases. For example: Pol I \(\rightarrow\) rRNA (most), Pol II \(\rightarrow\) mRNA, Pol III \(\rightarrow\) tRNA and 5S rRNA. This makes it easy to recall during an exam.

Step 1: Understanding the Question:

The question asks to identify the specific stage of meiosis where the centromeres, which hold sister chromatids together, divide.


Step 2: Detailed Explanation:

Meiosis consists of two successive nuclear divisions, Meiosis I and Meiosis II.


Meiosis I: This is the reductional division. In Anaphase I, homologous chromosomes separate and move to opposite poles. However, the sister chromatids remain attached at their centromeres. The centromeres do not divide.

Meiosis II: This is the equational division, and it is very similar to mitosis.

In Metaphase II, individual chromosomes (each consisting of two sister chromatids) align at the metaphase plate.
In Anaphase II, the centromeres finally divide (split). This allows the sister chromatids to separate and move to opposite poles. Once separated, each chromatid is now considered an individual chromosome.

Telophase and Metaphase I do not involve centromere division.

Therefore, the division of the centromere occurs during Anaphase II.


Step 3: Final Answer:

The division of the centromere in meiosis occurs during Anaphase II.
Quick Tip: Remember this key difference: Anaphase I separates homologous chromosomes. Anaphase II separates sister chromatids. The separation of sister chromatids is only possible after the centromere divides.

Step 1: Understanding the Question:

The question asks to identify the plant hormone (phytohormone) that can be used to speed up the maturation process in young conifer trees, leading to earlier production of seeds.


Step 2: Detailed Explanation:

Let's review the primary functions of the given phytohormones:


Abscisic Acid (ABA): Generally known as a stress hormone, it induces dormancy, stomatal closure, and abscission (shedding of leaves/fruits). It inhibits growth.

Indole-3-butyric Acid (IBA): This is an auxin. Auxins are primarily involved in root initiation, apical dominance, and cell elongation. IBA is commonly used as a rooting hormone for cuttings.

Gibberellic Acid (GA): Gibberellins have a wide range of effects, including promoting stem elongation (bolting), breaking seed dormancy, and inducing flowering and fruit development. A key commercial application is spraying juvenile conifers with GA to hasten maturity and promote early seed cone production for breeding programs.

Zeatin: This is a type of cytokinin. Cytokinins promote cell division (cytokinesis), chloroplast development, and delay senescence (aging). They work in conjunction with auxins to control differentiation.


The specific function of hastening maturity in conifers to induce early seed production is a well-known effect of Gibberellic Acid.


Step 3: Final Answer:

Spraying juvenile conifers with Gibberellic Acid helps in hastening their maturity period, leading to early seed production.
Quick Tip: Associate Gibberellins (GA) with "bolting" (rapid stem elongation before flowering) and overcoming juvenility. This makes it easier to remember their role in promoting flowering and seed production.

Step 1: Understanding the Question:

The question asks for the reason why cellulose does not give a positive iodine test (blue colour), unlike starch.


Step 2: Detailed Explanation:

The iodine test for starch works because of the structure of amylose, a component of starch.


Starch (Amylose): It consists of \(\alpha\)-glucose units linked together, which causes the chain to form a coiled, helical structure. Iodine molecules (\(I_2\)) can fit inside this helix, forming a starch-iodine complex that absorbs light in such a way that it appears blue-black.

Cellulose: It consists of \(\beta\)-glucose units. The linkages between these units result in a straight, linear chain. These chains are packed tightly together via hydrogen bonds to form microfibrils. Cellulose does not form complex helices.


Because cellulose lacks the helical structure necessary to trap iodine molecules, the characteristic blue colour is not produced.

Analyzing the options:

(A) is incorrect. Cellulose does not break down.

(B) is incorrect. Cellulose is a polysaccharide, not a disaccharide.

(C) is incorrect. It is not a helical molecule in the way starch is.

(D) correctly states that cellulose does not have the complex helices required to hold iodine molecules.


Step 3: Final Answer:

Cellulose does not produce a blue colour with iodine because its linear structure lacks the complex helices needed to trap iodine molecules.
Quick Tip: Associate the starch-iodine test with the helical structure of amylose. Remember that cellulose is a straight-chain polymer due to its \(\beta\)-1,4 glycosidic bonds, which makes it structurally different and unable to form the iodine complex.

Step 1: Understanding the Question:

The question describes the process where specialized leaf mesophyll cells are induced to form an undifferentiated mass of cells (callus) in tissue culture and asks for the correct term for this process.


Step 2: Detailed Explanation:

Let's define the relevant terms in plant development:


Differentiation: The process by which cells derived from meristems mature and undergo structural changes to perform specific functions. For example, a meristematic cell differentiating into a mesophyll cell.

Dedifferentiation: The process where already differentiated, mature cells lose their specialization and regain the ability to divide actively. In this case, the specialized, non-dividing mesophyll cells are stimulated by the culture medium to revert to a meristematic state and start dividing to form the undifferentiated callus. This is dedifferentiation.

Redifferentiation: The process where the dedifferentiated cells of the callus once again differentiate to form specialized cells, tissues, and organs like roots and shoots.

Senescence: The process of aging in plants.


The phenomenon described in the question, where specialized cells form an unspecialized callus, is a classic example of dedifferentiation.


Step 3: Final Answer:

The formation of callus from differentiated leaf mesophyll cells is called dedifferentiation.
Quick Tip: Remember the sequence in plant tissue culture: 1. \textbf{Explant} (differentiated cells) 2. \(\rightarrow\) \textbf{Dedifferentiation} \(\rightarrow\) \textbf{Callus} (undifferentiated cells) 3. \(\rightarrow\) \textbf{Redifferentiation} \(\rightarrow\) \textbf{Plantlet} (differentiated organs)

Step 1: Understanding the Question:

The question asks to identify the group of plants that all exhibit axile placentation.


Step 2: Detailed Explanation:

Axile placentation is a type of ovule arrangement where the ovary is divided into two or more chambers (locules) by septa. The ovules are attached to a central column or axis where the septa meet. This is found in multicarpellary, syncarpous (fused carpels) ovaries.

Let's analyze the placentation types in the plants listed in each option:


(A) China rose, Petunia and Lemon:

China rose (Hibiscus): Has axile placentation.
Petunia (Family Solanaceae): Has axile placentation.
Lemon (\textit{Citrus): Has axile placentation.

This option correctly lists three plants with axile placentation.

(B) Mustard, Cucumber and Primrose:

Mustard: Has parietal placentation.
Cucumber: Has parietal placentation.
Primrose: Has free-central placentation.

(C) China rose, Beans and Lupin:

China rose: Has axile placentation.
Beans and Lupin (Family Fabaceae): Have marginal placentation.

(D) Tomato, Dianthus and Pea:

Tomato: Has axile placentation.
Dianthus: Has free-central placentation.
Pea: Has marginal placentation.



Step 3: Final Answer:

The group where all plants show axile placentation is China rose, Petunia, and Lemon.
Quick Tip: Associate placentation types with key examples: \textbf{Marginal: Pea (Legumes) \textbf{Axile:} Tomato, Lemon, China rose \textbf{Parietal:} Mustard, Cucumber \textbf{Free-central:} Dianthus, Primula (Primrose) \textbf{Basal:} Sunflower, Marigold

Step 1: Understanding the Question:

The question asks to identify the scientists who provided the first "unequivocal" or definitive proof that DNA is the genetic material.


Step 2: Detailed Explanation:

Let's review the contributions of the scientists listed:


Frederick Griffith (1928): Conducted the "transforming principle" experiment with Streptococcus pneumoniae. He showed that genetic material could be transferred from heat-killed pathogenic bacteria to living non-pathogenic bacteria, but he did not identify what this material was.

Avery, Macleod, and McCarty (1944): They expanded on Griffith's work and used biochemical methods (enzymes like proteases, RNases, DNases) to show that the transforming principle was DNA. While their evidence was very strong, some in the scientific community remained skeptical, believing protein was a more likely candidate for the complex genetic material.

Alfred Hershey and Martha Chase (1952): They conducted experiments using bacteriophages (viruses that infect bacteria). They radioactively labeled the phage's protein coat with sulfur-35 (\(^{35\)S) and its DNA with phosphorus-32 (\(^{32}\)P). They found that only the \(^{32}\)P (DNA) entered the bacterial cells to direct the synthesis of new phages. This provided clear, direct, and widely accepted proof that DNA, not protein, is the genetic material. Their experiment is considered the unequivocal proof.

Wilkins and Franklin: Their X-ray diffraction images of DNA were crucial for Watson and Crick to determine the double helix structure of DNA, but they did not prove its function as the genetic material.



Step 3: Final Answer:

The unequivocal proof that DNA is the genetic material was provided by the Hershey-Chase experiment.
Quick Tip: Memorize the timeline of discovery: 1. Griffith: "Something" is transferred (Transforming Principle). 2. Avery, Macleod, McCarty: "Something" is DNA (Biochemical proof). 3. Hershey \& Chase: It is definitely the DNA that is transferred (Unequivocal proof using radioactive tracers).

Step 1: Understanding the Question:

The question asks to identify the metal used for making microparticles in the gene gun (biolistics) method of genetic transformation.


Step 2: Detailed Explanation:

The gene gun, or biolistic particle delivery system, is a method for directly transferring genetic material into cells. The procedure involves:


Coating microscopic particles of a heavy metal with the DNA of interest.
Loading these DNA-coated particles onto a carrier sheet.
Propelling the particles at a very high velocity towards the target cells or tissues.

The particles must be dense enough to penetrate the cell walls and membranes without causing excessive damage. They must also be chemically inert so they don't react harmfully with the cell's contents.

The metals commonly used for this purpose are gold (Au) and tungsten (W) due to their high density and chemical inertness. Silver, copper, and zinc are not typically used as they can be toxic to cells.


Step 3: Final Answer:

Microparticles of tungsten or gold are used in the gene gun method.
Quick Tip: Associate "gene gun" or "biolistics" with the precious metals Gold or the heavy metal Tungsten. Think of them as tiny, dense "bullets" used to shoot DNA into cells.

Step 1: Understanding the Question:

The question asks for the year in which the historic "Earth Summit," which led to the Convention on Biological Diversity, was held in Rio de Janeiro.


Step 2: Detailed Explanation:

The United Nations Conference on Environment and Development (UNCED), popularly known as the Earth Summit or the Rio Summit, was a major international conference held in Rio de Janeiro, Brazil.


The summit took place from June 3 to June 14, 1992.

It was a landmark event that brought global attention to environmental issues and sustainable development.

One of the key outcomes of the summit was the signing of the Convention on Biological Diversity (CBD), a multilateral treaty with objectives for conserving biological diversity, the sustainable use of its components, and the fair sharing of benefits from genetic resources.


The other years are incorrect. The World Summit on Sustainable Development was held in Johannesburg in 2002.


Step 3: Final Answer:

The Earth Summit was held in Rio de Janeiro in 1992.
Quick Tip: The 1992 Rio Earth Summit is a critical date in environmental history. Memorize this year along with its key outcomes, such as the Convention on Biological Diversity (CBD) and the Framework Convention on Climate Change (UNFCCC).

Step 1: Understanding the Question:

The question asks to identify the specific substage of Prophase I of meiosis where recombination nodules are observed.


Step 2: Detailed Explanation:

Prophase I is the longest phase of meiosis and is divided into five substages:


Leptotene: Chromosomes start to condense and become visible.
Zygotene: Homologous chromosomes pair up in a process called synapsis, forming bivalents. The synaptonemal complex begins to form.
Pachytene: Synapsis is complete. The paired chromosomes are called bivalents or tetrads. This is the stage where crossing over occurs between non-sister chromatids of homologous chromosomes. The sites where crossing over happens are marked by the appearance of protein complexes called recombination nodules. These nodules contain the enzymes necessary to cut and rejoin the DNA strands.
Diplotene: The synaptonemal complex dissolves, and the homologous chromosomes start to separate, but they remain attached at the sites of crossing over, which are now visible as chiasmata.
Diakinesis: Chromosomes become fully condensed, and the chiasmata terminalize (move towards the ends of the chromatids). The nuclear envelope breaks down.

Therefore, recombination nodules, which are the sites of crossing over, appear during the Pachytene stage.


Step 3: Final Answer:

The appearance of recombination nodules occurs during the Pachytene substage of prophase I.
Quick Tip: Remember the sequence of Prophase I: Leptotene, Zygotene, Pachytene, Diplotene, Diakinesis (LZPDD). A mnemonic like "Lazy Zebra Painted Purple During Day" can help. Associate Pachytene with "Packing together" and "Crossover".

Step 1: Understanding the Question:

The question requires matching essential mineral elements (micronutrients) with their specific physiological roles in plants.


Step 2: Detailed Explanation:

Let's match each nutrient from List I to its function in List II.


A. Iron (Fe): Iron is a crucial component of proteins involved in electron transport, such as cytochromes and ferredoxin. It is also an essential activator of the catalase enzyme. So, A matches with III.

B. Zinc (Zn): Zinc is an activator for many enzymes, especially carboxylases. Critically, it is required for the synthesis of auxin (IAA), as it is involved in the synthesis of the precursor, tryptophan. So, B matches with I.

C. Boron (B): Boron has multiple roles, including pollen germination, membrane function, and carbohydrate translocation. It is also required for cell elongation and cell differentiation. So, C matches with IV.

D. Molybdenum (Mo): Molybdenum is a structural component of several key enzymes involved in nitrogen metabolism, most notably nitrogenase and nitrate reductase. So, D matches with II.



Step 3: Final Answer:

The correct matching is A-III, B-I, C-IV, D-II. This corresponds to option (D).
Quick Tip: For mineral nutrition, focus on the most unique and frequently asked functions: Mo \(\rightarrow\) Nitrate reductase Zn \(\rightarrow\) Auxin synthesis Mg \(\rightarrow\) Chlorophyll center Mn \(\rightarrow\) Photolysis of water Fe \(\rightarrow\) Activator of catalase Memorizing these key pairs will help you solve many matching questions.

Step 1: Understanding the Question:

The question asks to arrange the given steps of creating recombinant DNA in the correct logical order.


Step 2: Detailed Explanation:

The process of recombinant DNA technology follows a specific sequence of steps to create a genetically modified organism. Let's analyze the given steps to find the correct order.


B. Cutting of DNA at specific location by restriction enzyme. The process begins with the source DNA (e.g., a genome) and the vector DNA (e.g., a plasmid). Both must be cut, usually with the same restriction enzyme, to generate compatible ends.

C. Isolation of desired DNA fragment. After cutting the source DNA, which contains many genes, the specific fragment of interest (the gene) must be separated and isolated from the other fragments. This is typically done using gel electrophoresis.

D. Amplification of gene of interest using PCR. Once the desired gene fragment is isolated, the Polymerase Chain Reaction (PCR) is often used to make millions of copies of it. This ensures that there is enough of the gene to ligate into the vector. (This is followed by ligation of the gene into the vector to create the recombinant DNA).

A. Insertion of recombinant DNA into the host cell. Finally, the newly created recombinant DNA (vector containing the gene of interest) is introduced into a suitable host cell (like a bacterium) through a process called transformation.


Therefore, the most logical sequence is B \(\rightarrow\) C \(\rightarrow\) D \(\rightarrow\) A.


Step 3: Final Answer:

The correct sequence of steps for the formation of Recombinant DNA is B, C, D, A.
Quick Tip: Think of recombinant DNA technology like a "cut, copy, and paste" operation. \textbf{Cut (B):} Use restriction enzymes as molecular scissors. \textbf{Isolate (C):} Separate the piece you want. \textbf{Copy (D):} Use PCR to make more copies. \textbf{Paste \& Insert (A):} Ligate the gene into a vector and insert it into a host.

Step 1: Understanding the Question:

The question asks to identify the incorrect statement among the given options related to water pollution and its ecological effects.


Step 2: Detailed Explanation:


(A) This statement describes biomagnification, where the concentration of persistent toxic substances (like DDT or mercury) increases at higher trophic levels in a food chain. This is a well-documented ecological phenomenon. So, this statement is correct.

(B) This statement describes the effect of sewage pollution. The organic matter in sewage serves as food for decomposer microorganisms. Their population explodes, and they consume large amounts of dissolved oxygen (high Biological Oxygen Demand or BOD). This oxygen depletion leads to the death of fish and other aquatic organisms. So, this statement is correct.

(C) This statement claims that algal blooms improve water quality. This is completely false. Algal blooms are caused by eutrophication (excess nutrient enrichment, not just organic matter). These blooms degrade water quality by imparting colour and odour. Furthermore, when the massive amount of algae dies, their decomposition consumes huge quantities of dissolved oxygen, leading to hypoxic or anoxic conditions and mass fish kills. This severely harms, rather than promotes, fisheries. So, this statement is NOT correct.

(D) This statement describes the impact of water hyacinth (\textit{Eichhornia crassipes), an invasive species that thrives in nutrient-rich (eutrophic) waters. Its rapid growth covers the water surface, blocking light and depleting oxygen, which disrupts the aquatic ecosystem. So, this statement is correct.



Step 3: Final Answer:

The statement that is not correct is (C), as algal blooms drastically reduce water quality and are detrimental to fisheries.
Quick Tip: Remember that "eutrophication" and "algal bloom" are negative terms in ecology, associated with pollution and ecosystem damage. Any statement claiming they are beneficial is almost certainly incorrect.

Step 1: Understanding the Question:

The question asks for the total number of different proteins found in a ribosome. As the type of ribosome (prokaryotic vs. eukaryotic) is not specified, we should consider both and see which one fits the options.


Step 2: Detailed Explanation:

Ribosomes are composed of ribosomal RNA (rRNA) and proteins. The number of proteins differs between prokaryotic and eukaryotic ribosomes.


Prokaryotic (70S) Ribosome:

Small subunit (30S): Contains about 21 proteins.
Large subunit (50S): Contains about 31-34 proteins.
Total: Approximately 52-55 proteins.

Eukaryotic (80S) Ribosome:

Small subunit (40S): Contains about 33 proteins.
Large subunit (60S): Contains about 49 proteins.
Total: Approximately 82 proteins.


Comparing these values to the given options:

(A) 20

(B) 80

(C) 60

(D) 40

The value 80 is the closest approximation for the number of proteins in a eukaryotic ribosome. In the context of general biology questions, "ribosome" often implicitly refers to the more complex eukaryotic type, and numbers are often rounded.


Step 3: Final Answer:

A ribosome (typically referring to a eukaryotic 80S ribosome) consists of approximately 80 different proteins.
Quick Tip: When a biology question is general and could apply to both prokaryotes and eukaryotes, check if the options align better with one than the other. The number 80 is a strong indicator that the question is referring to a eukaryotic ribosome.

Step 1: Understanding the Question:

The question asks to match properties related to water transport in plants (List I) with their correct descriptions (List II).


Step 2: Detailed Explanation:

Let's match the terms:


A. Cohesion: This is the intermolecular attraction between like-molecules. In the context of water, it is the mutual attraction among water molecules, due to hydrogen bonding. So, A matches with II.

B. Adhesion: This is the attraction between unlike molecules. In xylem, it is the attraction of water molecules towards polar surfaces of the tracheary elements (xylem walls). So, B matches with IV.

C. Surface tension: This is a property of liquids that allows them to resist an external force, due to the cohesive nature of their molecules. Water molecules are attracted to each other more strongly than to the air above, resulting in a higher attraction for each other in the liquid phase compared to the gaseous phase. So, C matches with I.

D. Guttation: This is the loss of water in the form of liquid droplets from the hydathodes of leaves, which occurs when root pressure is high and transpiration is low. It is a form of water loss in the liquid phase. So, D matches with III.



Step 3: Final Answer:

The correct matching is A-II, B-IV, C-I, D-III, which corresponds to option (B).
Quick Tip: Remember the "C-A-T" of water transport: \textbf{C}ohesion (water-water), \textbf{A}dhesion (water-xylem), and \textbf{T}ension (pull from transpiration). Guttation is different from transpiration as it involves liquid water loss, not vapor.

Step 1: Understanding the Question:

The question asks for the essential components required for the process of chemiosmosis to generate ATP.


Step 2: Detailed Explanation:

Chemiosmosis is the process that couples the energy stored in a proton gradient across a membrane to the synthesis of ATP. According to Peter Mitchell's chemiosmotic theory, the process requires the following key components:


A membrane: An intact membrane (like the inner mitochondrial membrane or the thylakoid membrane) is necessary to create a closed compartment and maintain a separation of charges and concentrations.

A proton pump: This is typically an electron transport chain that uses the energy from electrons to actively pump protons (H\(^+\) ions) from one side of the membrane to the other, against their concentration gradient.

A proton gradient: The pumping of protons creates a high concentration of protons on one side of the membrane. This electrochemical gradient is also known as a proton motive force, which stores potential energy.

ATP synthase: This is a transmembrane enzyme complex. It has a channel that allows protons to flow back across the membrane down their electrochemical gradient. The kinetic energy from this proton flow is used by the enzyme to catalyze the phosphorylation of ADP to ATP.


Option (B) correctly lists all four of these essential components. The other options are incorrect because they mention an "electron gradient" instead of a proton gradient or "NADP synthase" instead of ATP synthase.


Step 3: Final Answer:

The combination required for chemiosmosis is a membrane, a proton pump, a proton gradient, and ATP synthase.
Quick Tip: Think of chemiosmosis like a hydroelectric dam. The \textbf{membrane} is the dam wall. The \textbf{proton pump} is the pump that fills the reservoir. The \textbf{proton gradient} is the water stored at a high level. The \textbf{ATP synthase} is the turbine that generates electricity (ATP) as water flows through it.

Step 1: Understanding the Question:

The question asks to match different types of ecological interactions with their symbolic representations, where '+' indicates benefit, '-' indicates harm, and '0' indicates no effect.


Step 2: Detailed Explanation:

Let's match each interaction with its description:


A. Mutualism: An interaction where both species benefit from each other. The representation is (+, +). So, A matches with IV.

B. Commensalism: An interaction where one species benefits, and the other is neither harmed nor benefited (unaffected). The representation is (+, 0). So, B matches with I.

C. Amensalism: An interaction where one species is harmed, and the other is unaffected. The representation is (-, 0). So, C matches with II.

D. Parasitism: An interaction where one species (the parasite) benefits at the expense of the other (the host), which is harmed. The representation is (+, -). This also applies to predation. So, D matches with III.



Step 3: Final Answer:

The correct set of matches is A-IV, B-I, C-II, D-III. This corresponds to option (C).
Quick Tip: Create a simple table in your notes for all population interactions (+,-,0 notation). This makes it very easy to recall during an exam. Remember that predation, parasitism, and herbivory all fall under the (+, -) category.

Step 1: Understanding the Question:

The question asks which enzyme's activity is inhibited by malonate (spelled here as melonate), thereby inhibiting bacterial growth.


Step 2: Detailed Explanation:

Malonate is a classic example of a competitive inhibitor. This type of inhibition occurs when a molecule that is structurally similar to the enzyme's actual substrate binds to the active site, blocking the substrate from binding.


The enzyme in question is succinic dehydrogenase, which is a key enzyme in the Krebs cycle (citric acid cycle).
The normal substrate for this enzyme is succinate.
The inhibitor, malonate, has a chemical structure very similar to succinate.

Because of this structural similarity, malonate can bind to the active site of succinic dehydrogenase. However, the enzyme cannot act on malonate, so the catalytic reaction is blocked. This inhibition of the Krebs cycle disrupts cellular respiration and ATP production, which in turn inhibits the growth of the bacteria.


Step 3: Final Answer:

Malonate inhibits the activity of succinic dehydrogenase.
Quick Tip: Remember the classic pair: Substrate = Succinate, Enzyme = Succinic Dehydrogenase, Competitive Inhibitor = Malonate. This is a frequently cited example of competitive inhibition in biochemistry.

Step 1: Understanding the Question:

The question asks to identify the correct statements describing Klinefelter's Syndrome from a given list. Klinefelter's Syndrome is a chromosomal disorder with a karyotype of 47, XXY.


Step 2: Detailed Explanation:

Let's evaluate each statement:


A. This disorder was first described by Langdon Down (1866). This is incorrect. Langdon Down described Down's Syndrome. Klinefelter's Syndrome was described by Dr. Harry Klinefelter in 1942.

B. Such an individual has overall masculine development. However, the feminine development is also expressed. This is correct. The presence of a Y chromosome determines the male sex, so the individual is phenotypically male. However, the extra X chromosome leads to the expression of some female characteristics, such as gynecomastia (enlarged breasts).

C. The affected individual is short statured. This is incorrect. Individuals with Klinefelter's Syndrome are often taller than average, with long limbs. Short stature is characteristic of Turner's Syndrome (45, XO).

D. Physical, psychomotor and mental development is retarded. This is an overgeneralization and largely incorrect. While some individuals may have learning disabilities or delayed speech development, severe intellectual disability is not a typical feature.

E. Such individuals are sterile. This is correct. The presence of an extra X chromosome disrupts the development of the testes, leading to small, firm testes that do not produce sperm (azoospermia), resulting in infertility.



Step 3: Final Answer:

The correct statements are B and E. Therefore, the correct option is (D).
Quick Tip: To remember the key features of aneuploidies, create a small comparison table. \textbf{Klinefelter's (XXY):} Tall Male, Gynecomastia, Sterile. \textbf{Turner's (XO):} Short Female, Webbed neck, Sterile. \textbf{Down's (Trisomy 21):} Short stature, Characteristic facial features, Intellectual disability.

Step 1: Understanding the Question:

The question requires matching the phases of the eukaryotic cell cycle (List I) with their corresponding descriptions or events (List II).


Step 2: Detailed Explanation:

Let's perform the matching:


A. M Phase: This is the mitotic phase, where the cell undergoes division. Mitosis is known as equational division because the number of chromosomes in the daughter cells is equal to that of the parent cell. So, A matches with IV.

B. G\(_2\) Phase: This is the second gap phase, which occurs after DNA synthesis (S phase) and before mitosis (M phase). During this phase, the cell continues to grow and synthesizes proteins and organelles needed for mitosis, such as tubulin for the spindle fibers. So, B matches with I (Proteins are synthesized).

C. Quiescent stage (G₀): This is a non-dividing state that cells can enter from the G\(_1\) phase. Cells in G₀ are metabolically active but have exited the cell cycle and are not preparing to divide. It can be considered an inactive phase with respect to cell division. So, C matches with II.

D. G\(_1\) Phase: This is the first gap phase. It represents the interval between the completion of mitosis (M phase) and the initiation of DNA replication (S phase). So, D matches with III.



Step 3: Final Answer:

The correct matching is A-IV, B-I, C-II, D-III, which corresponds to option (D).
Quick Tip: Visualize the cell cycle as a clock: M \(\rightarrow\) G\(_1\) \(\rightarrow\) S \(\rightarrow\) G\(_2\) \(\rightarrow\) M. G\(_1\) is the gap between M and S. G\(_2\) is the gap between S and M. S is for Synthesis (DNA replication). M is for Mitosis (division). G₀ is an exit ramp from G\(_1\).

Step 1: Understanding the Question:

The question requires matching metabolic processes or steps from List I with their associated enzymes, pathways, or systems from List II.


Step 2: Detailed Explanation:

Let's match each item from List I to its correct counterpart in List II.


A. Oxidative decarboxylation: This is a reaction where a carboxyl group is removed as \(CO_{2}\) and the molecule is oxidized. A prime example in cellular respiration is the link reaction, where pyruvate is converted to acetyl-CoA. This reaction is catalyzed by the Pyruvate dehydrogenase complex. So, A matches with II.

B. Glycolysis: This is the pathway for the breakdown of glucose into pyruvate. It is also known by the names of the scientists who elucidated it: Embden, Meyerhof, and Parnas. Hence, it is called the EMP pathway. So, B matches with IV.

C. Oxidative phosphorylation: This is the metabolic pathway in which cells use enzymes to oxidize nutrients, thereby releasing energy which is used to produce ATP. The process is driven by the flow of electrons through the Electron transport system (ETS) located in the inner mitochondrial membrane. So, C matches with III.

D. Tricarboxylic acid (TCA) cycle: Also known as the Krebs cycle. The very first step of this cycle involves the combination of acetyl-CoA (a 2-carbon molecule) with oxaloacetate (a 4-carbon molecule) to form citrate (a 6-carbon molecule). This reaction is catalyzed by the enzyme Citrate synthase. So, D matches with I.


Step 3: Final Answer:

The correct matching is: A-II, B-IV, C-III, D-I. This corresponds to option (A).
Quick Tip: For matching questions on cellular respiration, create a mental flowchart: Glycolysis (EMP pathway) \(\rightarrow\) Pyruvate \(\rightarrow\) Oxidative Decarboxylation (Pyruvate Dehydrogenase) \(\rightarrow\) Acetyl-CoA \(\rightarrow\) TCA Cycle (starts with Citrate Synthase) \(\rightarrow\) Electron Carriers \(\rightarrow\) Oxidative Phosphorylation (Electron Transport System).

Step 1: Understanding the Question:

This is an Assertion-Reason question about the process of pollination in gymnosperms.


Step 2: Detailed Explanation:

Analysis of Assertion A:

Gymnosperms, such as conifers, are characterized by wind pollination (anemophily). Their pollen grains, produced in microsporangia (pollen sacs), are typically light and winged, adapted for dispersal by wind or air currents. This statement is factually correct. So, Assertion A is true.


Analysis of Reason R:

This statement describes what happens after the pollen is carried by air. While air currents do carry the pollen to the ovule, the rest of the statement contains inaccuracies. The pollen grain lands on the micropyle of the ovule, not directly at the mouth of the archegonium. The most significant error is the claim that a "pollen tube is not formed." In all modern gymnosperms, the pollen grain germinates to form a pollen tube. The pollen tube grows towards the archegonium and delivers the male gametes for fertilization. This process is called siphonogamy. Therefore, the statement that a pollen tube is not formed is false. So, Reason R is false.


Step 3: Final Answer:

Assertion A is a true statement, but Reason R is a false statement.
Quick Tip: A key feature of seed plants (gymnosperms and angiosperms) is the formation of a pollen tube (siphonogamy) to deliver non-motile or motile male gametes to the egg. The absence of a pollen tube is characteristic of more primitive plant groups.

Step 1: Understanding the Question:

The question asks to evaluate the correctness of two statements, one about Gause's Competitive Exclusion Principle and the other about the relative impact of competition on carnivores versus herbivores.


Step 2: Detailed Explanation:

Analysis of Statement I:

This statement provides the definition of Gause's 'Competitive Exclusion Principle'. The principle posits that when two species compete for the exact same limited resources in a stable environment, one species will always be more efficient and will eventually eliminate the other. This is the standard and correct definition of the principle. Therefore, Statement I is correct.


Analysis of Statement II:

This statement makes a broad generalization that carnivores are more adversely affected by competition than herbivores. This is not a universally accepted ecological rule. Competition can be extremely intense at any trophic level. For example, herbivores may compete intensely for a specific host plant, while carnivores might compete for territory or a wide range of prey. It is not possible to state definitively that one group is always "more adversely affected" than the other. The intensity of competition depends on factors like resource availability, niche overlap, and population density, not just the trophic level. Therefore, Statement II is considered false.


Step 3: Final Answer:

Based on the analysis, Statement I is a correct definition, while Statement II is an incorrect generalization.
Quick Tip: For statement-based questions in ecology, be wary of absolute generalizations. Ecological interactions are complex and context-dependent. A statement claiming something is "always" or "in general" true for a large, diverse group like "carnivores" should be scrutinized carefully.

Step 1: Understanding the Question:

This is an Assertion-Reason question that requires evaluating the definition of a flower and the developmental changes involved.


Step 2: Detailed Explanation:

Analysis of Assertion A:

The botanical definition of a flower is that it is a highly modified and condensed reproductive shoot. This modification involves the transformation of the shoot apical meristem, which normally produces vegetative structures (leaves and stems), into a floral meristem, which produces the parts of a flower. So, Assertion A is true.


Analysis of Reason R:

The modification from a vegetative shoot to a flower involves several key changes. The axis of the shoot (the future thalamus or receptacle) stops elongating, so the internodes become highly condensed and are not distinguishable. Instead of producing leaves at the nodes, the floral meristem produces modified leaves, which are the floral appendages: sepals, petals, stamens, and carpels, arranged in successive whorls. So, Reason R is also true.


Conclusion:

Reason R perfectly explains the "modification" mentioned in Assertion A. It details the process of how a shoot is modified—by condensing the internodes and producing floral appendages instead of regular leaves. Therefore, Reason R is the correct explanation for Assertion A.


Step 3: Final Answer:

Both Assertion A and Reason R are true, and R is the correct explanation of A.
Quick Tip: Think of a flower as a "telescoped" shoot. The stem is compressed into the thalamus, and the leaves are transformed into the colourful and reproductive parts of the flower. This helps visualize why a flower is considered a modified shoot.

Step 1: Understanding the Question:

The question asks to identify the correct statements about anatomical features of woody stems, particularly bark and related structures.


Step 2: Detailed Explanation:

Let's analyze each statement:


A. Lenticels are the lens-shaped openings permitting the exchange of gases. This is correct. Lenticels are porous tissues on the bark that allow for gas exchange between the internal living cells and the outside atmosphere.

B. Bark formed early in the season is called hard bark. This is incorrect. Bark formed early in the season (spring) is called soft bark, while bark formed late in the season (autumn) is called hard bark.

C. Bark is a technical term that refers to all tissues exterior to vascular cambium. This statement is incorrect because "bark" is generally considered a non-technical term. The description of what it includes is correct, but calling it a "technical term" is botanically imprecise and can be a point of distinction in exam questions.

D. Bark refers to periderm and secondary phloem. This is correct. Bark is composed of two main parts: the periderm (which includes phellogen, phellem, and phelloderm) and the secondary phloem (inner bark). This is a correct description of the components of bark.

E. Phellogen is single-layered in thickness. This is incorrect. Phellogen, or cork cambium, is a meristematic tissue and is typically described as being a couple of layers thick, not strictly a single layer.



Step 3: Final Answer:

The correct statements are A and D. Therefore, the correct option is (C).
Quick Tip: In plant anatomy, precise terminology is key. Remember that "bark" is a common term for everything outside the vascular cambium, which structurally consists of the periderm and secondary phloem. Also, distinguish between early/soft bark and late/hard bark.

Step 1: Understanding the Question:

The question asks to identify the specific type of blood cell in which the Human Immunodeficiency Virus (HIV) replicates.


Step 3: Detailed Explanation:

HIV is a retrovirus that primarily targets cells of the immune system. Its main target is a type of lymphocyte called the helper T cell (or T\(_{H}\) cell). These cells have a specific receptor on their surface called CD4, which the virus uses to gain entry. Once inside the helper T cell, HIV uses an enzyme called reverse transcriptase to convert its RNA genome into DNA. This viral DNA is then integrated into the host cell's genome. The infected cell is then forced to produce new virus particles, becoming a "virus factory". Macrophages are also infected by HIV. The progressive destruction of T\(_{H}\) cells severely weakens the immune system, leading to Acquired Immuno Deficiency Syndrome (AIDS).


Step 4: Final Answer:

The primary site of HIV replication is the helper T cell (T\(_{H}\) cell). Therefore, option (2) is the correct answer.
Quick Tip: Remember the key steps of the HIV life cycle: it infects helper T-cells (CD4+ cells), uses reverse transcriptase to make DNA from its RNA, integrates into the host genome, and replicates, eventually destroying the host cell. This progressive loss of T-cells is the hallmark of AIDS.