NEET 2024 Question Paper with Answers and Solutions PDF T6 in English

Step 1: Calculate the angular frequency \(\omega\) of the needle's oscillations.

Given that the needle completes 20 oscillations in 5 seconds, the frequency \(f\) is: \[ f = \frac{20 \, oscillations}{5 \, s} = 4 \, Hz \]
The angular frequency \(\omega\) is given by: \[ \omega = 2\pi f = 2\pi \times 4 = 8\pi \, rad/s \]

Step 2: Using the formula for the period \(T\) of a magnetic dipole in a magnetic field.

The period \(T\) is given by: \[ T = \frac{2\pi}{\omega} \]
Since \(\omega = 8\pi\), \[ T = \frac{2\pi}{8\pi} = \frac{1}{4} \, s \]

Step 3: Relate the moment of inertia and magnetic moment to the angular frequency.

From the equation for the oscillation of a magnetic dipole: \[ \omega^2 = \frac{\mu B}{I} \]
where \(\mu\) is the magnetic moment, \(B\) is the magnetic field, and \(I\) is the moment of inertia. Plugging in the values we get: \[ (8\pi)^2 = \frac{\mu \times 0.049}{9.8 \times 10^{-6}} \] \[ \mu = \frac{(8\pi)^2 \times 9.8 \times 10^{-6}}{0.049} \]

Step 4: Calculate \(\mu\).
\[ \mu = \frac{64\pi^2 \times 9.8 \times 10^{-6}}{0.049} \approx 1280 \pi^2 \times 10^{-5} \, Am^2 \]
Therefore, \(x = 1280\pi^2\). Quick Tip: For oscillations of a magnetic dipole in a magnetic field, remember that the angular frequency \(\omega\) is directly proportional to the square root of the magnetic moment \(\mu\) and inversely proportional to the square root of the moment of inertia \(I\).

Step 1: Understanding the formula for the potential at an axial point.

The potential at an axial point (distance \( r \) from the center of the dipole) is given by the formula:
\[ V = \frac{1}{4 \pi \epsilon_0} \cdot \frac{2P}{r^3}. \]
Here, \( P \) is the dipole moment and \( r \) is the distance from the center of the dipole.

Step 2: Substituting the given values.

We are given \( P = 4 \times 10^{-6} \) C m, \( r = 2 \) m, and \( \frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \) SI units. Substituting these values into the equation:
\[ V = \frac{9 \times 10^9 \times 2 \times 4 \times 10^{-6}}{(2)^3} = \frac{9 \times 10^9 \times 8 \times 10^{-6}}{8} = 9 \times 10^3 V. \]
Thus, Assertion A is true.

Step 3: Verifying the reason.

The formula given in the reason is incorrect. The correct formula for the potential at an axial point is \( V = \frac{1}{4 \pi \epsilon_0} \cdot \frac{2P}{r^3} \), not \( V = \pm \frac{2P}{4 \pi \epsilon_0 r^2} \). Hence, Reason R is false. Quick Tip: For dipoles, remember the correct formula for the potential at an axial point is \( V = \frac{1}{4 \pi \epsilon_0} \cdot \frac{2P}{r^3} \). This is crucial in avoiding errors in calculations.

Step 1: Identify the transitions and corresponding wavelengths

The spectral lines of hydrogen for transitions from higher energy levels \( n_2 \) to \( n_1 = 2 \) correspond to the following wavelengths:
\[ For n_2 = 3 to n_1 = 2, \, \lambda = 656.3 \, nm \quad (H_\alpha) \] \[ For n_2 = 4 to n_1 = 2, \, \lambda = 486.1 \, nm \quad (H_\beta) \] \[ For n_2 = 5 to n_1 = 2, \, \lambda = 434.1 \, nm \quad (H_\gamma) \] \[ For n_2 = 6 to n_1 = 2, \, \lambda = 410.2 \, nm \quad (H_\delta) \]

Step 2: Match the spectral lines with the wavelengths from List II
\( n_2 = 3 to n_1 = 2 \) corresponds to the wavelength 656.3 nm (\( H_\alpha \)), so it matches with option III.
\( n_2 = 4 to n_1 = 2 \) corresponds to 486.1 nm (\( H_\beta \)), so it matches with option IV.
\( n_2 = 5 to n_1 = 2 \) corresponds to 434.1 nm (\( H_\gamma \)), so it matches with option II.
\( n_2 = 6 to n_1 = 2 \) corresponds to 410.2 nm (\( H_\delta \)), so it matches with option I.


Thus, the correct matching is: \[ A-III, B-IV, C-II, D-I. \] Quick Tip: For hydrogen spectral lines, use the Balmer series formula to identify the wavelength corresponding to transitions from higher to lower energy levels.

N/A Quick Tip: When dealing with problems on material properties like elasticity, converting all units to match (e.g., converting meters to millimeters) before performing calculations is crucial to ensure accuracy.

Conservation of momentum in inelastic collision.

Momentum before collision:
\[ m v_1 + 0 = m v_2 \]
Momentum after collision: \[ (m + m) v_2 = 2m v_2 \]
Thus, \(v_2 = \frac{v_1}{2}\) and the ratio \(v_1 : v_2\) is \(2:1\). Quick Tip: In completely inelastic collisions, remember that the bodies stick together, and the velocity of the system is the same for both bodies after the collision. Use the conservation of momentum to find the final velocity.

Understanding the vernier scale.

One main scale division (MSD) is 0.1 mm. Since (N+1) vernier divisions (VD) coincide with N MSDs, each VD is:
\[ VD = \frac{N \times 0.1 \, mm}{N+1} \]
Calculate the vernier constant.

The vernier constant (VC) is the difference between one MSD and one VD:
\[ VC = 0.1 \, mm - \frac{N \times 0.1 \, mm}{N+1} = \frac{0.1 \, mm}{N+1} \]
Converting to cm: \[ VC = \frac{0.1 \, cm}{10(N+1)} = \frac{1}{100(N+1)} \, cm \] Quick Tip: To calculate the Vernier constant, subtract the value of one Vernier scale division from one main scale division. The result gives the precision of the Vernier caliper.

N/A Quick Tip: For cycles involving different thermodynamic processes, it's essential to consider the type of each segment to determine overall work done and energy changes. Recall that the area enclosed by a process path on a P-V diagram represents the work done during the cycle.

N/A Quick Tip: Remember in rolling motion, points on the rim of the wheel have varying speeds depending on their position relative to the point of contact with the ground. The highest point moves the fastest, while the point in contact with the ground has zero speed.

N/A Quick Tip: Remember that the tension needed for circular motion is directly proportional to the square of the angular velocity. Doubling the speed quadruples the required centripetal force.

Step 1: Analyze the Geometry
The prism is right-angled at B. The light ray enters at P with an angle of incidence 30°, travels parallel to BC, and exits at Q along AC. Let the angle at A be \(\theta\).

Step 2: Apply Snell's Law at P
Let the refractive index of the prism be \(n\). Applying Snell's Law at point P:
\[ 1 \cdot \sin(30^\circ) = n \cdot \sin(r_1) \]
where \(r_1\) is the angle of refraction at P.

Step 3: Angles and Relationships
Since the ray travels parallel to BC, the angle of refraction at P, \(r_1\), is related to the angle at A (\(\theta\)) by:
\[ r_1 + \theta = 90^\circ \Rightarrow r_1 = 90^\circ - \theta \]

Step 4: Apply Snell's Law at Q
At point Q, the angle of incidence is \(\theta\) and the angle of refraction is 90°. Applying Snell's Law:
\[ n \cdot \sin(\theta) = 1 \cdot \sin(90^\circ) = 1 \]

Step 5: Combine and Solve
From the equation at P: \[ \sin(30^\circ) = n \sin(r_1) = n \sin(90^\circ - \theta) = n \cos(\theta) \]
Since \(\sin(30^\circ) = \frac{1}{2}\), we have: \[ \frac{1}{2} = n \cos(\theta) \]
From the equation at Q: \[ n \sin(\theta) = 1 \]
We can write \(\cos(\theta) = \sqrt{1 - \sin^2(\theta)}\). Substituting this into the equation from P: \[ \frac{1}{2} = n \sqrt{1 - \sin^2(\theta)} \]
Substitute \(\sin(\theta) = \frac{1}{n}\) from the equation at Q: \[ \frac{1}{2} = n \sqrt{1 - \frac{1}{n^2}} \] \[ \frac{1}{4} = n^2 \left(1 - \frac{1}{n^2}\right) = n^2 - 1 \] \[ n^2 = \frac{5}{4} \] \[ n = \frac{\sqrt{5}}{2} \]

% Quick tip
\begin{quicktipbox
For a ray to travel parallel inside a prism, the refractive index is calculated using Snell’s law at the first interface.
\end{quicktipbox Quick Tip: For a ray to travel parallel inside a prism, the refractive index is calculated using Snell’s law at the first interface.

Understanding magnetic susceptibility \( \chi \)

Diamagnetic materials have small negative susceptibility (\( \chi < 0 \)).
Ferromagnetic materials have very high positive susceptibility.
Paramagnetic materials have small positive susceptibility.
Non-magnetic materials have zero susceptibility. Quick Tip: Magnetic materials are classified based on their susceptibility \( \chi \):
- \( \chi < 0 \) for diamagnetic,
- \( 0 < \chi < \epsilon \) for paramagnetic,
- \( \chi \gg 1 \) for ferromagnetic.

Step 1: Identifying amplitude and time period
- The given equation is in standard SHM form: \[ x = A \sin (\omega t + \phi) \]
Comparing with \( x = 5 \sin \left( \pi t + \frac{\pi}{3} \right) \), we identify: \[ A = 5 m, \quad \omega = \pi \]
Time period is given by: \[ T = \frac{2\pi}{\omega} = \frac{2\pi}{\pi} = 2 s \] Quick Tip: The amplitude is the coefficient of the sine function, and the time period is given by \( T = \frac{2\pi}{\omega} \).

Step 1: Analyzing Statement A (Solar Cell I-V Characteristics).
- A solar cell operates by converting light energy into electrical energy.
- The I-V characteristics of a solar cell are plotted in the IV quadrant of the voltage-current graph because it supplies power instead of consuming it.
- Thus, Statement A is correct.

Step 2: Analyzing Statement B (Reverse Bias in \( pn \)-Junction Diode).
In reverse bias, the current that flows is minority carrier current (not majority carriers).
Majority carriers do not contribute significantly in reverse bias, and the current measured in \( \mu A \) is due to minority carriers.
Thus, Statement B is incorrect.

Step 3: Conclusion.
Since A is correct and B is incorrect, the correct answer is Option (3). Quick Tip: - A solar cell works as a power-generating device, so its I-V characteristics appear in the IV quadrant. - In a reverse-biased \( pn \)-junction diode, the current is mainly due to minority carriers.

We have four capacitors in the circuit, each of \( 2 \, \mu F \). First, we simplify the capacitors in series and parallel.

1. The two capacitors in series on the left-hand side (each of \( 2 \, \mu F \)) have an equivalent capacitance \( C_1 \) given by: \[ \frac{1}{C_1} = \frac{1}{2 \, \mu F} + \frac{1}{2 \, \mu F} = 1 \, \mu F \quad \Rightarrow \quad C_1 = 1 \, \mu F. \]

2. The two capacitors in series on the right-hand side (each of \( 2 \, \mu F \)) also have an equivalent capacitance \( C_2 \) given by: \[ \frac{1}{C_2} = \frac{1}{2 \, \mu F} + \frac{1}{2 \, \mu F} = 1 \, \mu F \quad \Rightarrow \quad C_2 = 1 \, \mu F. \]

3. Now, the two series capacitors \( C_1 \) and \( C_2 \) are in parallel with each other, so the total equivalent capacitance \( C_{eq} \) is: \[ C_{eq} = C_1 + C_2 = 1 \, \mu F + 1 \, \mu F = 2 \, \mu F. \]

Thus, the equivalent capacitance between terminals A and B is \( 2 \, \mu F \). Quick Tip: For capacitors in series, the equivalent capacitance is found using \( \frac{1}{C_{eq}} = \sum \frac{1}{C_i} \), and for capacitors in parallel, \( C_{eq} = \sum C_i \).

Understanding Transformer Voltage Relation
The voltage transformation equation for an ideal transformer is: \[ \frac{V_S}{V_P} = \frac{N_S}{N_P} \]
- Given \( \frac{N_P}{N_S} = \frac{1}{2} \), we take the reciprocal: \[ \frac{N_S}{N_P} = 2 \]
Thus, \[ \frac{V_S}{V_P} = 2 \]
\(\Rightarrow\) The voltage ratio \( V_S : V_P \) is \( 2:1 \).

% Quick tip
\begin{quicktipbox
In an ideal transformer, voltage is directly proportional to the turns ratio: \[ \frac{V_S}{V_P} = \frac{N_S}{N_P} \]
More secondary turns result in a step-up transformer, while fewer secondary turns result in a step-down transformer.
\end{quicktipbox Quick Tip: In an ideal transformer, voltage is directly proportional to the turns ratio: \[ \frac{V_S}{V_P} = \frac{N_S}{N_P} \] More secondary turns result in a step-up transformer, while fewer secondary turns result in a step-down transformer.

N/A Quick Tip: When calculating forces related to surface tension, always ensure the linear measurement (such as circumference in this case) is calculated accurately to get precise results. The surface tension force calculation essentially involves a linear force distribution along the periphery of the object in contact with the fluid.

N/A Quick Tip: Remember, when combining resistances in series and parallel, always ensure calculations follow the correct formula for each type of connection. This will ensure accurate results in complex circuits.

N/A Quick Tip: Solid angle, like planar angle and strain, is a dimensionless quantity. When identifying dimensionally consistent quantities, always consider the fundamental nature of each quantity's measurement and units.

Step 1: Applying Lenz’s Law.

According to Lenz’s Law, the induced current in a solenoid opposes the change in magnetic flux.

Step 2: Determining the current direction in Solenoid-1.

Since the bar magnet is moving towards Solenoid-2, the flux through Solenoid-1 is decreasing.
To oppose this decrease, the solenoid induces a current that maintains the original north polarity near the magnet.
Using the right-hand rule, the current in Solenoid-1 circulates in the direction \( AB \).

Step 3: Determining the current direction in Solenoid-2.

The approaching magnet increases the flux in Solenoid-2.

To oppose this increase, Solenoid-2 induces a current to create a south pole at the nearest end.

Using the right-hand rule, the induced current flows in the \( DC \) direction.


Step 4: Conclusion.
Thus, the correct directions of the induced currents are \( AB \) in Solenoid-1 and \( DC \) in Solenoid-2. Quick Tip: Lenz’s Law states that the induced current opposes the change in magnetic flux. Always apply the right-hand rule to determine the direction of induced current.

The total resistance in the circuit is the sum of the internal resistance \( r = 1 \, \Omega \) and the external resistance \( R = 4 \, \Omega \), so the total resistance is \( R_{total} = 5 \, \Omega \).

The current \( I \) in the circuit is given by: \[ I = \frac{emf}{R_{total}} = \frac{10}{5} = 2 \, A. \]

The terminal voltage \( V_{terminal} \) is the emf minus the voltage drop across the internal resistance: \[ V_{terminal} = emf - I \times r = 10 - 2 \times 1 = 8 \, V. \] Quick Tip: To calculate the terminal voltage, subtract the voltage drop across the internal resistance from the emf of the battery: \( V_{terminal} = emf - I \times r \).

N/A Quick Tip: Understanding the relationship between kinetic energy and wavelength is crucial in quantum mechanics and helps in analyzing particle behavior at quantum scales.

Understanding the photon properties.

Statement A is correct because photon energy is given by \( E = h\nu \).

Statement B is correct because photons travel at the speed of light \( c \).

Statement C is correct as photon momentum is given by \( p = \frac{h\nu}{c} \).

Statement D is correct as photon-electron collisions obey conservation of energy and momentum.

Statement E is incorrect because photons are neutral (no charge).


Final Answer: A, B, C, and D are correct. Quick Tip: Photon has no mass, always moves at speed \( c \), and follows energy-momentum relations.

N/A Quick Tip: In nuclear reactions, each type of decay affects the atomic and mass numbers differently. Alpha decay reduces both by 2 and 4, respectively. Positron emission and electron capture lower the atomic number by 1, while beta minus decay increases it by 1.

Step 1: Finding velocity

Velocity is given by differentiating displacement:

\[ v = \frac{dx}{dt} = \frac{d}{dt} (2t - 1) = 2 m/s \]

Step 2: Calculating Instantaneous Power

Power is given by:
\[ P = F \cdot v \]
Substituting \( F = 5 \) N and \( v = 2 \) m/s,
\[ P = 5 \times 2 = 10 W \]

% Quick tip
\begin{quicktipbox
Instantaneous power is calculated as \( P = Fv \), where \( v \) is the instantaneous velocity of the particle.
\end{quicktipbox Quick Tip: Instantaneous power is calculated as \( P = Fv \), where \( v \) is the instantaneous velocity of the particle.

N/A Quick Tip: Always check units and conversion factors when dealing with physical constants and measurements. This ensures the accuracy of the calculated results.

Step 1: Understanding the Effect of White Light in Young’s Experiment

When a monochromatic source is used, the fringes are of uniform colour and spacing.

When white light is used, all wavelengths interfere, resulting in a central bright white fringe.

Surrounding fringes have different colours because different wavelengths interfere at different positions.


Step 2: Elimination of Incorrect Options

Option (2) is incorrect because different wavelengths result in different fringe widths.

Option (3) is incorrect because interference still occurs but with colour variation.

Option (4) is incorrect because the central fringe remains bright, not dark.
Quick Tip: For Young’s experiment with white light, remember:
The central fringe is always white.
Coloured fringes appear due to wavelength-dependent interference.
The fringe width varies for different wavelengths.

Step 1: Understanding the properties of a charged spherical shell

A charged spherical shell behaves such that the electric potential inside and on the surface of the shell remains constant.

The potential at any point inside the shell is the same as the potential on its surface and is given by:

\[ V = \frac{1}{4\pi\epsilon_0} \cdot \frac{q}{R} \]

Step 2: Applying the concept to given points \( C \) and \( P \)

Both \( C \) (center) and \( P \) (on the shell) experience the same potential.
Since the potential difference is given by:

\[ V_C - V_P \]
and \( V_C = V_P \), we conclude:

\[ V_C - V_P = 0 \]

% Quick tip
\begin{quicktipbox
For a charged conducting spherical shell, the electric potential remains constant everywhere inside the shell, meaning the potential difference between any two internal points is always zero.
\end{quicktipbox Quick Tip: For a charged conducting spherical shell, the electric potential remains constant everywhere inside the shell, meaning the potential difference between any two internal points is always zero.

Step 1: Identifying the logic gates in the circuit.

The first gate is a NAND gate with both inputs as \( A \), giving:
\[ Output = \neg(A \cdot A) = \neg A. \]

The second gate is an OR gate with inputs \( A \) and \( B \), giving:
\[ Output = A + B. \]

The final gate is an AND gate that takes the outputs of the previous two gates as inputs:
\[ Y = (\neg A) \cdot (A + B). \]

Step 2: Simplifying the expression.

Expanding using Boolean algebra:
\[ Y = \neg A \cdot A + \neg A \cdot B. \]

Since \( A \cdot \neg A = 0 \), we get:
\[ Y = \neg A \cdot B. \]

This is the same as an AND gate operation where one input is \( B \) and the other is \( \neg A \), meaning the circuit functions like an AND gate. Quick Tip: When analyzing logic circuits, break them down into individual logic gates and use Boolean algebra to simplify step by step.

N/A Quick Tip: In logic circuits, if the output is dependent on one variable and independent of another, analyzing the truth table closely can simplify finding the correct logical expression.

N/A Quick Tip: The formula \( B = \frac{\mu_0 N I}{2R} \) is applicable for a tightly wound coil, ensuring uniform field distribution.

Step 1: Calculate the radius of the planet.

Given that the diameter of the planet is half that of Earth, the radius of the planet \( r \) will also be half of Earth's radius \( R \). If \( R \) is the radius of Earth, then:
\[ r = \frac{R}{2} \]

Step 2: Use the formula for gravitational acceleration.

The formula for gravitational acceleration \( g \) on the surface of a planet is given by:
\[ g = \frac{G M}{r^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( r \) is the radius of the planet. The gravitational acceleration on Earth \( g_{Earth} \) is \( 9.8 \, m/s^{-2} \).

Step 3: Apply the ratios of mass and radius to the formula.
The mass of the planet is \( \frac{1}{10} \) of Earth's mass \( M_{Earth} \), and the radius is \( \frac{R}{2} \). Plugging these into the formula for \( g \) gives: \[ g_{planet} = \frac{G \left(\frac{1}{10} M_{Earth}\right)}{\left(\frac{R}{2}\right)^2} = \frac{G \left(\frac{1}{10} M_{Earth}\right)}{\frac{R^2}{4}} = \frac{4G \left(\frac{1}{10} M_{Earth}\right)}{R^2} = \frac{4}{10} g_{Earth} \] \[ g_{planet} = \frac{4}{10} \times 9.8 \, m/s^{-2} = 3.92 \, m/s^{-2} \] Quick Tip: When calculating gravitational acceleration on other planets, remember to account for both changes in mass and radius, as gravity is sensitive to both these factors. This problem highlights the square law nature of gravitational force relative to distance (radius).

Understanding circular motion

Velocity changes because the direction changes continuously.

Acceleration also changes due to the changing direction of velocity (centripetal acceleration).
Quick Tip: In uniform circular motion, speed is constant but velocity and acceleration continuously change due to direction changes.

N/A Quick Tip: Always isolate the component or part of the system you are analyzing to simplify force calculations in systems with multiple objects.

Understanding Brewster’s angle
At Brewster’s angle, the reflected light is completely polarised perpendicular to the plane of incidence, while the refracted light remains partially polarised. Quick Tip: At Brewster’s angle \( \theta_B \), the reflected light is completely polarised perpendicular to the incident plane. The refracted light is not fully polarised.

N/A Quick Tip: Understanding atomic structure and behavior involves knowing when and why atoms emit light. The emission of light in the form of a spectrum generally indicates a transition between energy levels, not the inherent stability of atoms.

N/A Quick Tip: Understanding how changes in physical dimensions of a capacitor affect its electrical properties is crucial in both theoretical and practical applications in circuits.

Step 1: Understanding the Graph

The equation governing isobaric processes is Charles’s Law:
\[ V \propto T \quad (for constant pressure) \]
Higher pressure results in a less steep slope in the \( T \)-\( V \) graph.


Step 2: Identifying the Correct Order

Since \( P_1 \) corresponds to the least steep curve, it represents the highest pressure.

\( P_3 \) has the steepest slope, indicating the lowest pressure.

Thus, the correct order is:

\[ P_1 > P_2 > P_3 \]

% Quick tip
\begin{quicktipbox
For an ideal gas, the steeper the isobaric curve in a \( T \)-\( V \) graph, the lower the pressure.
\end{quicktipbox Quick Tip: For an ideal gas, the steeper the isobaric curve in a \( T \)-\( V \) graph, the lower the pressure.

Analysis of Circuits:


In a balanced Wheatstone bridge:
\[ \frac{R_1}{R_2} = \frac{R_3}{R_4} \]
where \( R_1 \) and \( R_2 \) are resistances on one side of the bridge, and \( R_3 \) and \( R_4 \) are on the opposite side.


Circuit Examination:


Circuit 1: \( \frac{10\Omega}{15\Omega} \neq \frac{15\Omega}{10\Omega} \)
Circuit 2: \( \frac{10\Omega}{15\Omega} \neq \frac{15\Omega}{10\Omega} \)
Circuit 3: \( \frac{15\Omega}{10\Omega} = \frac{15\Omega}{10\Omega} \), satisfying the balance condition.
Circuit 4: \( \frac{10\Omega}{15\Omega} \neq \frac{15\Omega}{10\Omega} \)


Conclusion:

Circuit 3 is correctly configured to achieve a bridge balance, as the ratios of the resistances on opposite arms are equal. Thus, Circuit 3 is the correct answer, aligning with the solution provided. Quick Tip: For a Wheatstone bridge to be balanced, the ratio of the resistances in one diagonal must equal the ratio in the other diagonal.

If the sheet is magnetic, it will experience attraction towards the magnetic pole, requiring a force to hold it in place (Statement A is correct).

If the sheet is conducting, eddy currents are induced, which generate a repulsive force moving the sheet away from the pole (Statement C is correct).

A non-magnetic sheet does not experience attraction, so Statement B is incorrect.

A non-conducting, non-polar sheet does not interact with the field significantly, so Statement D is incorrect.



% Quick tip
\begin{quicktipbox
Eddy currents in a conductor generate a repulsive force when placed in a time-varying magnetic field.
\end{quicktipbox Quick Tip: Eddy currents in a conductor generate a repulsive force when placed in a time-varying magnetic field.

N/A Quick Tip: Displacement current is crucial for the symmetry of Maxwell's equations and explains phenomena like electromagnetic waves where no actual charge carriers are present.

N/A Quick Tip: The magnifying power of a telescope increases with the increase of the objective lens's focal length or the decrease of the eyepiece's focal length. This is crucial for achieving high-resolution images of distant celestial objects.

From the given velocity-time graph:

Initially, velocity increases linearly, implying constant acceleration.

Then, velocity remains constant, meaning zero acceleration.

Finally, velocity decreases linearly, implying constant negative acceleration.


Thus, the correct acceleration-time graph will have a positive constant value initially, followed by zero acceleration, and then a negative constant value. Quick Tip: Acceleration is the slope of the velocity-time graph. A constant velocity means zero acceleration, while a linear velocity change means constant acceleration.

N/A Quick Tip: When dealing with thermal expansion and stress calculations, always verify that units are consistent across all terms to avoid calculation errors.

The capacitive reactance is given by: \[ X_C = \frac{1}{\omega C} = \frac{1}{2 \pi f C}. \]
Substituting \( f = 50 \) Hz and \( C = 10 \mu F = 10^{-5} \) F: \[ X_C = \frac{1}{2 \times 3.14 \times 50 \times 10^{-5}} \approx 318 \Omega. \]
The peak current is given by: \[ I_0 = \frac{V_0}{X_C} = \frac{210 \times \sqrt{2}}{318} \approx 0.93 A. \] Quick Tip: For AC circuits with capacitors, use \( X_C = \frac{1}{2\pi f C} \) and \( I_0 = \frac{V_0}{X_C} \) to find peak current.

N/A Quick Tip: When calculating power in series and parallel circuits, remember to first find equivalent resistances and then use the power formula accordingly. This helps avoid calculation errors in complex circuits.

N/A Quick Tip: When determining dimensionality in physical equations, match units across terms carefully to identify dimensionless combinations.

N/A Quick Tip: Recall that electromagnetic waves are generated by accelerating charges, not just moving charges. The change in acceleration is key to changing the electromagnetic fields.

N/A Quick Tip: When dealing with vector quantities like magnetic moments, the angle between components and the method of their vector addition play critical roles in determining the resultant vector's magnitude and direction.

N/A Quick Tip: In mechanics, understanding the dependencies of oscillatory motion such as the pendulum's period on length but not on mass is crucial. This conceptual clarity helps in problem-solving across various topics in physics.

N/A Quick Tip: In launching satellites, it is crucial to consider both the energy required to overcome Earth's gravitational pull to the desired altitude and the kinetic energy needed to achieve orbital velocity. Both energies significantly contribute to the total launch energy requirement.

Step 1: Analyze the chemical properties and trends within Group 16.
Oxygen (\( O \)) is known for almost exclusively showing a \(-2\) oxidation state due to its high electronegativity.

Sulfur (\( S \)), Selenium (\( Se \)), and Tellurium (\( Te \)) can also commonly exhibit the \(-2\) oxidation state, reflecting their non-metallic nature and ability to gain electrons.

Polonium (\( Po \)), however, being a metalloid and more metallic than its predecessors, does not typically show a \(-2\) oxidation state. Polonium is more likely to exhibit positive oxidation states, such as \(+2\) or \(+4\), which are typical for metals and metalloids.
Quick Tip: Understanding periodic trends, such as increasing metallic character and decreasing electronegativity from oxygen to polonium, helps in predicting typical oxidation states of elements.

Step 1: Convert each option to the number of atoms using Avogadro's number.


4 g of helium: The molar mass of helium is approximately 4 g/mol.


Thus, 4 g of helium is 1 mol, which contains \(6.022 \times 10^{23}\) atoms.


2.271098 L of helium at STP: At STP, 1 mol of any gas occupies 22.4 L. Therefore, \(2.271098 \, L\) of helium corresponds to \(\frac{2.271098}{22.4} \approx 0.1014\) mol, which is about \(6.1 \times 10^{22}\) atoms.

4 mol of helium: 4 mol of helium contains \(4 \times 6.022 \times 10^{23} = 2.409 \times 10^{24}\) atoms.


4 u of helium: Since 4 u corresponds to 1 mol of helium (4 g/mol), it represents \(6.022 \times 10^{23}\) atoms.


Step 2: Conclusion.

The option with the highest number of helium atoms is 4 mol of helium, containing \(2.409 \times 10^{24}\) atoms. Quick Tip: Remember that Avogadro's number (\(6.022 \times 10^{23}\) atoms/mol) is key to converting moles to atoms.

Statement I: The boiling points of alkanes generally decrease with increased branching because branched molecules have less surface area for intermolecular forces to act upon, which lowers the boiling point. Thus, \(n\)-pentane (straight-chain) has the highest boiling point, followed by isopentane (branched), and neopentane (more highly branched) has the lowest boiling point.

Statement II: This is correct because, as branching increases, the molecule becomes more spherical in shape, leading to reduced surface area and weaker intermolecular forces (van der Waals forces), thus lowering the boiling point.
Quick Tip: When comparing boiling points, remember that more branching usually leads to lower boiling points due to decreased surface area and weaker intermolecular forces.

Understand the Arrhenius Equation.

The Arrhenius plot, specifically \(\ln k\) vs \( \frac{1}{T} \), provides critical insights into the reaction mechanism:

Slope (\(-\frac{E_a}{R}\)): This value directly indicates the activation energy required to initiate the reaction. A steeper negative slope implies a higher activation energy, indicating a more temperature-sensitive process.

Intercept (\(\ln A\)): The intercept on the \( \ln k \) axis provides the natural logarithm of the pre-exponential factor, which is a measure of the frequency of collisions and the orientation factor conducive to the reaction.


These parameters are crucial for chemists to understand the temperature dependence of reaction rates and to estimate the feasibility and speed of chemical reactions under different conditions.
Quick Tip: Always remember that a plot of \( \ln k \) vs \( \frac{1}{T} \) in the context of the Arrhenius equation should be a straight line decreasing as \( \frac{1}{T} \) increases, due to the negative relationship.

Analyze the stability of the carbocations based on structural features.

The most stable carbocation typically has the most alkyl groups attached to the positively charged carbon, providing greater electron-donating effects through hyperconjugation and inductive effects. Among the given options, the structure with extensive alkyl substitution around the carbocation (as described for option (2)) would be the most stable. Quick Tip: Carbocations are stabilized by alkyl groups through hyperconjugation and inductive effects. The more alkyl groups directly attached, the more stable the carbocation.

N/A Quick Tip: When analyzing redox couples and their electrode potentials, consider how the electron configurations before and after the redox change contribute to the overall stability and energy of the ions. This can significantly affect the observed \( E^\circ \) values.

Understanding the trend in ionization energies across the periodic table.

The first ionization energy generally increases across a period from left to right due to increasing nuclear charge and decreasing atomic radius. However, there are exceptions due to electron configuration stability:

Li (1s\(^2\) 2s\(^1\)) has the lowest ionization energy because it has only one electron in the outermost shell.

B (1s\(^2\) 2s\(^2\) 2p\(^1\)) is lower than Be due to the p-orbital having a higher energy level than the s-orbital.

Be (1s\(^2\) 2s\(^2\)) has higher ionization energy due to a fully filled s-orbital, which is more stable.

C (1s\(^2\) 2s\(^2\) 2p\(^2\)) and N (1s\(^2\) 2s\(^2\) 2p\(^3\)) have progressively higher ionization energies, with nitrogen being the highest due to half-filled p-orbital stability.
Quick Tip: Recall the effects of electronic configurations and orbital types (s vs. p) on ionization energies to predict trends accurately.

A. [Co(NH\(_3\))\(_5\)(NO\(_2\))]Cl\(_2\): The NO\(_2\) ligand can bind through either nitrogen (nitro) or oxygen (nitrito), exhibiting linkage isomerism. Match: A - II


B. [Co(NH\(_3\))\(_5\)(SO\(_4\))]Br: The SO\(_4^{2-}\) ion can be inside or outside the coordination sphere, leading to ionization isomerism. Match: B - III


C. [Co(NH\(_3\))\(_6\)][Cr(CN)\(_6\)]: This complex has two complex ions. The ligands on the metal centers can be exchanged, resulting in coordination isomerism. Match: C - IV


D. [Co(H\(_2\)O)\(_6\)]Cl\(_3\): This complex can exist as different hydrates, with varying numbers of water molecules coordinated to the cobalt ion. This is a form of solvate isomerism (specifically, hydrate isomerism). Match: D - I


Therefore, the correct matching is A-II, B-III, C-IV, and D-I. The correct answer is (3). Quick Tip: Review each complex's structure and the nature of its ligands to identify the possible isomerism types effectively.

Option (1): This is a redox reaction where hydrogen and chlorine are both reduced and oxidized, respectively.

Option (3): This is a redox reaction where zinc is oxidized and copper is reduced.

Option (4): This is a redox reaction where potassium chlorate is reduced and iodine is oxidized.


Option (2) is a double displacement reaction, where no change in oxidation states occurs. Therefore, it is not a redox reaction. Quick Tip: In redox reactions, the oxidation states of elements change. In non-redox reactions, there is no change in oxidation states.

Step 1: Analyze Statement I.

Aniline, being an aromatic amine, has a lone pair of electrons on the nitrogen atom that can interact with Lewis acid catalysts used in the Friedel-Crafts alkylation, leading to complex formation that deactivates the catalyst. Thus, it does not effectively undergo Friedel-Crafts reactions.

Step 2: Analyze Statement II.

Gabriel synthesis is typically used for synthesizing primary amines from primary alkyl halides. Aniline, which involves an aromatic ring, cannot be prepared using Gabriel synthesis because it requires conditions that would lead to the destruction of the aromatic ring. Quick Tip: Understanding the chemical properties and reactivity of compounds helps in predicting their behavior in different chemical reactions.

NH3 (A) has a trigonal pyramidal shape due to the lone pair on nitrogen, corresponding to I.

BrF5 (B) has a square pyramidal shape due to 5 bonding pairs and 1 lone pair on the central atom, matching with IV.

XeF4 (C) has a square planar shape due to 4 bonding pairs and 2 lone pairs, aligning with II.

SF6 (D) has an octahedral shape with 6 bonding pairs, corresponding to III.
Quick Tip: For molecular geometries, consider the number of bonding pairs and lone pairs around the central atom using VSEPR theory.

Lucas reagent, a mixture of zinc chloride and concentrated hydrochloric acid, is employed to classify low molecular weight alcohols based on their reactivity. The nature of the reaction is influenced by the stability of the carbocation formed during the substitution process:


Primary alcohols (like n-butyl alcohol) form a primary carbocation, which is least stable and reacts very slowly, often showing no visible change at room temperature.
Secondary alcohols (like isobutyl alcohol) form a more stable secondary carbocation and react at a moderate rate.
Tertiary alcohols (like tert-butyl alcohol), form a highly stable tertiary carbocation almost instantaneously, leading to a rapid reaction evident by the quick formation of a cloudy solution or precipitate. Quick Tip: Lucas reagent is used to classify alcohols based on their reactivity. Tertiary alcohols react instantaneously, secondary alcohols react within minutes, and primary alcohols do not react significantly at room temperature.

According to Henry's Law, solubility is inversely proportional to the Henry’s law constant (\(K_H\)). A higher value of \(K_H\) corresponds to a lower solubility. Based on the given constants:

Gas A has the highest \(K_H\), so it has the lowest solubility.

Gas C has a medium \(K_H\) value, so its solubility is intermediate.

Gas B has the lowest \(K_H\), so it has the highest solubility.


Thus, the order of solubility is \(B > C > A\). Quick Tip: For gases, solubility in a liquid is inversely proportional to the Henry’s law constant. A higher constant means lower solubility.

Tollen’s reagent (A) and Schiff’s reagent (B) react with aldehydes; glucose has an aldehyde group and thus reacts with both.

HCN (C) reacts with glucose, forming cyanohydrin.

NH2OH (D) reacts with glucose to form an oxime derivative.

NaHSO3 (E) reacts with aldehydes, but glucose does not typically react with sodium bisulfite under normal conditions.
Quick Tip: Glucose, with its aldehyde group, can react with reagents like Tollen's and Schiff's, but not all reagents with aldehyde-specific reactions will work with glucose.

N/A Quick Tip: Understanding the sequence and specificity of reagent actions in organic synthesis is crucial for designing effective synthetic routes and achieving the desired chemical transformations.

Step 1: Understanding Fehling's solution.

Fehling's solution is used as a chemical test to differentiate between water-soluble carbohydrate and ketone functional groups, and more specifically, it detects the presence of aldehyde groups which are oxidizable.

Step 2: Composition of Fehling's solution.

Fehling's solution is typically made up of two parts:

Fehling's 'A': This part contains aqueous copper(II) sulphate, which provides the Cu\(^{2+}\) ions necessary for the oxidation reaction.

Fehling's 'B': This part contains an alkaline solution of sodium potassium tartrate (Rochelle salt), which acts to keep the copper ions in solution.


Since the correct answer for Fehling's solution 'A' is aqueous copper sulphate, the best choice is (3). Quick Tip: Remember the specific roles of each component in Fehling's solution: 'A' supplies the copper ions, and 'B' maintains these ions in an alkaline and complexed form.

Step 1: Analyze each quantum number and its information.

\(m_l\) (A): Magnetic quantum number, which indicates the orientation of the orbital in space, thus matching with III.

\(m_s\) (B): Spin quantum number, indicating the orientation of the electron's spin, which is IV.

\(l\) (C): Azimuthal quantum number, related to the shape of the orbital, thus matching with I.

\(n\) (D): Principal quantum number, which influences the size and energy level of the orbital, corresponding to II. Quick Tip: Memorize the functions of each quantum number: principal (\(n\)), azimuthal (\(l\)), magnetic (\(m_l\)), and spin (\(m_s\)) to easily solve quantum mechanics problems.

Step 1: Calculate the moles of HCl used.

The molarity (M) of HCl is 0.75 M, and the volume (V) used is 25 mL (or 0.025 L). The moles of HCl used are given by:
\[ n = M \times V = 0.75 \, mol/L \times 0.025 \, L = 0.01875 \, mol \]
Step 2: Stoichiometry of the reaction.

The reaction between sodium hydroxide (NaOH) and hydrochloric acid (HCl) is:
\[ NaOH + HCl \rightarrow NaCl + H_2O \]
This reaction is a 1:1 molar reaction. Therefore, 0.01875 mol of NaOH will react with 0.01875 mol of HCl.


Step 3: Calculate the mass of NaOH that reacts.

The molar mass of NaOH is approximately 40 g/mol. Therefore, the mass of NaOH that reacts is:
\[ Mass = 0.01875 \, mol \times 40 \, g/mol = 0.75 \, g \]
Step 4: Calculate the mass of NaOH left unreacted.

Initially, there was 1 gram of NaOH. The amount of NaOH left unreacted after the reaction is: \[ 1 \, g - 0.75 \, g = 0.25 \, g = 250 \, mg \] Quick Tip: When dealing with stoichiometry problems, always check the mole ratio in the balanced chemical equation to ensure correct calculations for reactants and products.

N/A Quick Tip: Sublimation is particularly useful for purifying substances that are unstable or decompose at temperatures near their melting points. It can also be used to separate volatile substances from non-volatile impurities.

N/A Quick Tip: Entropy, a measure of disorder or randomness in a system, tends to increase in processes where gases form from liquids or solids, or where more particles are produced from fewer, especially at higher temperatures or when bonds are broken.

Determine the correct conditions for each process.

Isothermal process (A) occurs at a constant temperature, matching with II.

Isochoric process (B) occurs at constant volume, matching with III.

Isobaric process (C) occurs at constant pressure, matching with IV.

Adiabatic process (D) occurs with no heat exchange, matching with I.
Quick Tip: Associate each thermodynamic process with its defining characteristic to quickly identify correct matching in problems.

Understanding the trend in electronegativity across the periodic table.


Electronegativity generally increases across a period from left to right and decreases down a group. Therefore, among the given elements:


Silicon (Si) being a group 14 element and further down the period compared to carbon, has the lowest electronegativity.


Carbon (C) has a higher electronegativity than Si.


Nitrogen (N) is more electronegative than C due to its position in the periodic table and having a smaller atomic radius.


Oxygen (O) is more electronegative than N, attributed to its higher effective nuclear charge.


Fluorine (F), being the most electronegative element in the periodic table, ranks highest among the options.
Quick Tip: Remember that electronegativity helps predict how atoms share electrons in chemical bonds, with higher values indicating stronger attraction for bonding electrons.

Statement I: Both \([Co(NH_3)_6]^{3+}\) and \([CoF_6]^{3-}\) are octahedral complexes with similar coordination numbers but differ in their magnetic properties.


Statement II: \([Co(NH_3)_6]^{3+}\) is diamagnetic due to the low-spin configuration of Co\((^{3+})\) (\(3d^6\)), while \([CoF_6]^{3-}\) is paramagnetic due to the weak field fluoride ligands, leading to unpaired electrons.
Quick Tip: Check the type of ligands and the electron configuration of the central metal ion to determine the magnetic behavior of coordination complexes.

Matching molecules with bond types.

Ethane (A) has a single \(\sigma\)-bond between carbons.

Ethene (B) has one \(\sigma\)-bond and one \(\pi\)-bond.

The carbon molecule \( C_2 \) (C) typically has two \(\pi\)-bonds.

Ethyne (D) has one \(\sigma\)-bond and two \(\pi\)-bonds.
Quick Tip: Visualizing the molecular structure and counting the types of bonds between atoms are crucial for correctly identifying molecular bonding.

The activation energy \( E_a \) can be calculated using the Arrhenius equation, which involves the rate constants at two different temperatures: \[ k = A \exp\left(-\frac{E_a}{RT}\right) \]
where \( k \) is the rate constant, \( A \) is the frequency factor, \( E_a \) is the activation energy, \( R \) is the gas constant, and \( T \) is temperature. Quick Tip: Use the Arrhenius equation to relate the rate constant at two different temperatures to determine the activation energy.

Understand the relationship between \( K_p \) and \( K_c \).

The equilibrium constants \( K_p \) and \( K_c \) are related by the equation \( K_p = K_c(RT)^{\Delta n} \) where \( \Delta n \) is the change in moles of gas.

In Reaction (3), \( \Delta n = 1 - 2 = -1 \), therefore \( K_p \neq K_c \). Quick Tip: When \( \Delta n \neq 0 \), \( K_p \) and \( K_c \) will differ due to the contribution of the gas constant and temperature to the equation.

A. 1 mol of H\(_2\)O to O\(_2\): The balanced equation is 2H\(_2\)O \(\rightarrow\) O\(_2\) + 4H\(^+\) + 4e\(^-\). 2 moles of H\(_2\)O require 4 moles of electrons, so 1 mole of H\(_2\)O requires 2 moles of electrons (2F). Match: A - II


B. 1 mol of MnO\(_4^-\) to Mn\(^{2+}\): The balanced equation is MnO\(_4^-\) + 8H\(^+\) + 5e\(^-\) \(\rightarrow\) Mn\(^{2+}\) + 4H\(_2\)O. 1 mole of MnO\(_4^-\) requires 5 moles of electrons (5F). Match: B - IV


C. 1.5 mol of Ca from molten CaCl\(_2\): The balanced equation is Ca\(^{2+}\) + 2e\(^-\) \(\rightarrow\) Ca. 1 mole of Ca requires 2 moles of electrons. 1.5 moles of Ca require 1.5 \(\times\) 2 = 3 moles of electrons (3F). Match: C - I


D. 1 mol of FeO to Fe\(_2\)O\(_3\): The balanced equation is 2FeO + \(\frac{1}{2}\)O\(_2\) \(\rightarrow\) Fe\(_2\)O\(_3\). The oxidation state of Fe changes from +2 in FeO to +3 in Fe\(_2\)O\(_3\). For 1 mole of FeO, the change is 1 mole of electrons (1F). Match: D - III


Therefore, the correct matching is A-II, B-IV, C-I, and D-III. The correct answer is (3). Quick Tip: In electrochemical processes, one Faraday (F) of charge corresponds to the charge of one mole of electrons, about \( 96,485 \) coulombs.

N/A Quick Tip: When analyzing whether a reaction will proceed in the forward or reverse direction, always calculate and compare the reaction quotient \(Q_c\) to the equilibrium constant \(K_c\). This will indicate whether the system needs to adjust toward more products or more reactants to achieve equilibrium.

Identify structures capable of intramolecular hydrogen bonding.

Intramolecular hydrogen bonding typically occurs when hydrogen-bonding groups are positioned to form a stable five- or six-membered ring. Option (3) allows for such a configuration, with the OH and NO2 on the same carbon.
Quick Tip: Intramolecular hydrogen bonds are stronger and more stable when they form smaller rings, typically five- or six-membered.

Evaluate the substrate stability for \( S_N1 \) reactions.


Compound 1: Benzyl Bromide - Forms a benzyl carbocation, which is stabilized by resonance with the phenyl ring.
Compound 2: para-Methylbenzyl Bromide - Forms a para-methylbenzyl carbocation, which is further stabilized by electron-donating effects of the methyl group through resonance, making it more stable than the benzyl carbocation.
Compound 3: Secondary Alkyl Bromide - Forms a secondary carbocation, which lacks the additional resonance stabilization seen in benzyl carbocations.
Compound 4: Cyclohexyl Bromide - Forms a cyclohexyl carbocation, which is not resonance stabilized and is generally less stable.


Conclusion:
Compound 2 will undergo an \( S_N1 \) reaction the fastest due to the enhanced stability of the carbocation intermediate by both resonance and inductive effects from the methyl group. The correct answer is 2. Quick Tip: Benzyl and allyl halides are excellent substrates for \( S_N1 \) reactions due to carbocation stabilization.

Let's analyze and match List I (Reactions) with List II (Reagents/Conditions):


Reaction A: The reaction shows the cleavage of a benzyl group into two benzaldehyde molecules. This is characteristic of ozonolysis using \(O_3/Zn-H_2O\).
\(\Rightarrow\) Matches with IV

Reaction B: The reaction results in the formation of a benzophenone-like structure. This suggests a Friedel-Crafts acylation reaction, which requires benzoyl chloride and anhydrous AlCl\(_3\).
\(\Rightarrow\) Matches with I

Reaction C: The oxidation of a benzylic alcohol to a ketone suggests the use of chromium trioxide (CrO\(_3\)).
\(\Rightarrow\) Matches with II

Reaction D: The conversion of an alkyl benzene into a carboxylate group suggests a strong oxidizing agent like KMnO\(_4\)/KOH, \(\Delta\).
\(\Rightarrow\) Matches with III


Correct Answer: (1) A-IV, B-I, C-II, D-III Quick Tip: When identifying reagents for specific organic reactions, consider the common oxidation agents (CrO\(_3\), KMnO\(_4\)) and electrophilic substitution reagents like AlCl\(_3\).

Step 1: Calculate the 'spin only' magnetic moment for each ion.

The formula for the 'spin only' magnetic moment (\(\mu\)) is \(\mu = \sqrt{n(n+2)}\) Bohr magnetons, where \(n\) is the number of unpaired electrons.

Ti\(^{3+}\) has 1 unpaired electron, \(\mu = \sqrt{1(1+2)} = 1.73\) Bohr magnetons.

Cr\(^{2+}\) has 4 unpaired electrons, \(\mu = \sqrt{4(4+2)} = 4.90\) Bohr magnetons.

Mn\(^{2+}\) has 5 unpaired electrons, \(\mu = \sqrt{5(5+2)} = 5.92\) Bohr magnetons.

Fe\(^{2+}\) also has 4 unpaired electrons, \(\mu = \sqrt{4(4+2)} = 4.90\) Bohr magnetons.

Sc\(^{3+}\) has no unpaired electrons, \(\mu = \sqrt{0(0+2)} = 0\) Bohr magnetons.


Step 2: Match the ions with the same magnetic moments.

Cr\(^{2+}\) and Fe\(^{2+}\) both have the same magnetic moment of 4.90 Bohr magnetons. Quick Tip: Use the number of unpaired electrons to quickly estimate the magnetic properties of transition metal ions.

N/A Quick Tip: When comparing the energies of electrons in hydrogen-like ions across different atomic numbers and principal quantum numbers, the square of the charge \( Z^2 \) and the inverse square of the principal quantum number \( n^2 \) are key to understanding the energy scaling.

Step 1: Analyze Statement I.

Statement I is true as \( H_2O \) has a higher boiling point than other Group 16 hydrides due to hydrogen bonding, which is stronger than in \( H_2Te \), \( H_2Se \), and \( H_2S \).

Step 2: Analyze Statement II.

Statement II is also true. While the molecular mass of \( H_2O \) is lower compared to other hydrides, which suggests a lower boiling point, the extensive hydrogen bonding in \( H_2O \) significantly raises its boiling point above those of its heavier congeners. Quick Tip: Remember that hydrogen bonding can significantly affect physical properties like boiling points beyond what would be expected based purely on molecular mass.

To calculate the mass of copper deposited, we can use the formula for electrolysis:
\[ Mass of Cu = \frac{I \times t \times M}{n \times F} \]

Where:

\( I = 9.6487 \, A \) (current),
\( t = 100 \, seconds \) (time),
\( M = 63 \, g/mol \) (molar mass of copper),
\( n = 2 \) (valency of copper, \( Cu^{2+} \)),
\( F = 96487 \, C/mol \) (Faraday's constant).


Now, substituting the values into the formula:
\[ Mass of Cu = \frac{9.6487 \times 100 \times 63}{2 \times 96487} \]

Thus, the mass of copper deposited is \( 0.315 \, grams \). Quick Tip: To calculate the mass of a metal deposited during electrolysis, use the formula: \[ Mass = \frac{I \times t \times M}{n \times F} \] where \( I \) is the current, \( t \) is the time, \( M \) is the molar mass, \( n \) is the valency of the metal, and \( F \) is Faraday's constant.

To determine the empirical formula, we calculate the moles of A, B, and C based on their percentages:

Moles of A = \( \frac{32}{64} = 0.5 \) mol

Moles of B = \( \frac{20}{40} = 0.5 \) mol

Moles of C = \( \frac{48}{32} = 1.5 \) mol


Now, dividing each by the smallest number of moles (0.5), the ratio of A : B : C is 1 : 1 : 3. Therefore, the empirical formula is ABC\(_3\). Quick Tip: To find the empirical formula, first convert the mass percentages into moles, then simplify the ratios to get the simplest whole-number formula.

Let the initial concentration of NO be 0.1 M, and \(\alpha\) be the degree of dissociation. At equilibrium, the concentration of NO will be:
\[ [NO] = 0.1 - 2\alpha \]
The concentration of \(N_2\) and \(O_2\) formed will be \(\alpha\). Using the equilibrium concentrations and applying the reaction stoichiometry, we can set up the equilibrium expression and solve for \(\alpha\). After solving, we find that \(\alpha = 0.717\). Quick Tip: For dissociation problems, use the stoichiometry of the reaction to set up the concentration changes and solve for \(\alpha\).

The work done during reversible isothermal expansion is given by: \[ W = -nRT \ln \left( \frac{P_1}{P_2} \right) \]

Given:

\( n = 1 \, mol \)
\( R = 2.0 \, cal K^{-1} mol^{-1} \)
\( T = 25^\circ C = 298 \, K \)
\( P_1 = 20 \, atm \)
\( P_2 = 10 \, atm \)


Step 1: Substitute the values into the formula. \[ W = - (1 \, mol) \cdot (2.0 \, cal K^{-1} mol^{-1}) \cdot (298 \, K) \cdot \ln \left( \frac{20 \, atm}{10 \, atm} \right) \]

Step 2: Simplify the pressure ratio. \[ \frac{20 \, atm}{10 \, atm} = 2 \]

\subsection{Step 3: Calculate the natural logarithm. \[ \ln(2) \approx 0.6931 \]

Step 4: Calculate the work done. \[ W = - (1) \cdot (2.0) \cdot (298) \cdot (0.6931) \] \[ W = - 413.14 \, cal \]

The negative sign indicates that work is done by the system (gas) during expansion.

Final Answer: \[ \boxed{-413.14 \, calories} \] Quick Tip: For isothermal expansion, use the equation \(W = -nRT \ln \left(\frac{P_2}{P_1}\right)\) to calculate work done by the gas.

In the first reaction with PCl\(_3\), alcohol reacts with phosphorus trichloride (PCl\(_3\)) to produce alkyl chloride (RCl) and phosphorous acid (H\(_3\)PO\(_3\)) as product A.

In the second reaction, alcohol reacts with phosphorus pentachloride (PCl\(_5\)) to produce alkyl chloride (RCl), hydrogen chloride (HCl), and POCl\(_3\) as product B.

Thus, the products are H\(_3\)PO\(_3\) and POCl\(_3\). Quick Tip: Phosphorus trichloride reacts with alcohol to form phosphorous acid, while phosphorus pentachloride yields POCl\(_3\) and releases HCl.

\([Co(NH_3)_6]^{3+}\) is a homoleptic complex because it contains only one type of ligand (NH\(_3\)).

\([Co(NH_3)_4Cl_2]^+\) is a heteroleptic complex because it contains two different types of ligands (NH\(_3\) and Cl\(^-\)).


Thus, both statements are true. Quick Tip: A homoleptic complex contains only one type of ligand, while a heteroleptic complex contains more than one type of ligand.

Step 1: Reaction analysis and product identification.

First, the alkyl iodide (\(CH_3 - CH_2 - CH_2 - I\)) reacts with sodium cyanide (\(NaCN\)) to replace the iodide with a cyano group, forming nitrile \(A\) (\(CH_3 - CH_2 - CH_2 - CN\)).

Step 2: Hydrolysis of the nitrile.

Partial hydrolysis of the nitrile by \(OH^-\) converts \(A\) to the corresponding amine, \(B\) (\(CH_3 - CH_2 - CH_2 - NH_2\)), rather than fully hydrolyzing it to the carboxylic acid.

Step 3: Reaction with bromine.

Treatment of \(B\) with \(NaOH\) and \(Br_2\) does not lead to further functional group transformation but is intended to ensure the bromination reaction does not occur. Thus, \(C\) remains as propylamine, confirming that no further transformation of the amine occurs under these conditions. Quick Tip: Partial hydrolysis of nitriles can strategically be stopped at the amine stage by controlling the reaction conditions, avoiding full conversion to carboxylic acids.

During the preparation of Mohr's salt (ferrous ammonium sulfate), dilute sulfuric acid is added to prevent the hydrolysis of Fe\(^{2+}\) ions, as it does not form insoluble hydroxides at the pH conditions used. Quick Tip: To prevent hydrolysis of metal ions, choose acids that do not increase the pH significantly.

\(CO_3^{2-}\) (carbonate ion) has three equivalent resonance structures.
The other statements are incorrect:

NF\(_3\) has a dipole moment smaller than NH\(_3\) due to the different electronegativities of nitrogen and fluorine.

Ozone (\(O_3\)) has two main resonance structures, not three.

BF\(_3\) is a symmetrical molecule and has no dipole moment.
Quick Tip: For resonance structures, draw all possible equivalent forms that contribute to the overall electronic distribution.

In qualitative inorganic analysis, the cations are grouped according to their precipitation behavior and solubility in various reagents:

Group I: Alkali metals and ammonium (not relevant here).

Group II: Alkaline earth metals (e.g., Ba\(^{2+}\) and Mg\(^{2+}\)).

Group III: Transition metals like Cu\(^{2+}\).

Group IV: Cobalt (Co\(^{2+}\)).

Group V: Aluminum (Al\(^{3+}\)).


Thus, the correct order is: \( Cu^{2+} \) (B), \( Al^{3+} \) (A), \( Co^{2+} \) (D), \( Ba^{2+} \) (C), \( Mg^{2+} \) (E). Quick Tip: Familiarize yourself with the qualitative analysis chart to quickly identify group numbers of cations based on their solubility and precipitation behavior.

Oxidation Mechanism:

Potassium permanganate (\(KMnO_4\)) in acidic condition (\(H^+\)) is a strong oxidizing agent. When it reacts with alkenes, the typical initial product is a diol (vicinal diol). However, the reaction can proceed further, especially under the reaction conditions provided (\(KMnO_4\) with acid), leading to the cleavage of the carbon-carbon double bond.

Given Substrate:

The substrate shown in the problem is a cyclohexene derivative with a carbon-carbon double bond.

Reaction Pathway:

In this case, the oxidation does not stop at diol formation due to the strength of the oxidizing agent and the acidic environment. Instead, the alkene is likely cleaved, and the oxidation progresses to form carboxylic acids at the positions originally involved in the double bond.

Product Formation:

Given the structure of the alkene and the conditions:

The double bond carbons are fully oxidized to carboxylic acids, leading to the formation of a dicarboxylic acid compound as the major product.


Why Option (4) is Correct:

The product \( P \) corresponds to the chemical structure of a dicarboxylic acid derivative, where each of the carbon atoms originally involved in the double bond is now part of a carboxyl group (\(-COOH\)). This transformation is typical of strong oxidative conditions applied to alkenes. Quick Tip: Always consider the strength and conditions of the oxidizing agent when predicting reaction products from complex organic substrates.

Gd\(^{3+}\) and Eu\(^{3+}\): Gd\(^{3+}\) has an unpaired electron in the 4f orbital, making it paramagnetic. Eu\(^{3+}\) also has unpaired electrons and is paramagnetic.


Pm\(^{3+}\) and Sm\(^{3+}\): Both Pm\(^{3+}\) and Sm\(^{3+}\) have unpaired electrons, making them paramagnetic.


Ce(\(^{4+}\)): Ce(\(^{4+}\)) has a 4f\(^0\) configuration, which is diamagnetic.


Yb(\(^{2+}\)): Yb(\(^{2+}\)) has a 4f\(^{14}\) configuration, which is also diamagnetic.


Thus, the pair Ce\(^{4+}\) and Yb\(^{2+}\) are diamagnetic because they both have no unpaired electrons.
Quick Tip: To determine if an ion is diamagnetic, check if it has no unpaired electrons. Ions with all electrons paired are diamagnetic.

Step 1: Reaction with \( PBr_3 \)


The reaction of an alcohol with \( PBr_3 \) leads to the formation of an alkyl bromide, through a mechanism where the hydroxyl group of the alcohol is substituted by a bromide ion.
In this case, the hydroxyl group attached to the cyclohexane ring in the given molecule is replaced by bromine, forming the brominated product \( A \).


Step 2: Reaction with Alcoholic KOH


Alcoholic KOH generally promotes elimination reactions, particularly \( E2 \) mechanism.
The reaction likely involves the removal of a hydrogen atom from a carbon adjacent to the carbon bearing the bromine atom, leading to the formation of a double bond as the bromine leaves, forming product \( B \).


Identification of Products:


\( A \) is the alkyl bromide where the OH group was replaced by Br.
\( B \) is the alkene formed by the elimination of \( HBr \) from \( A \).


Why Option (3) is Correct:


\( A \) in option (3) shows the bromine correctly positioned at the site of the original hydroxyl group.
\( B \) shows a double bond resulting from an elimination reaction, consistent with the action of alcoholic KOH. Quick Tip: When converting alcohols to more reactive halides or preparing alkenes via elimination, reagents like \( PBr_3 \) and bases like KOH are commonly used.

Calculate the temperature using the ideal gas law relation for osmotic pressure.

Using the formula \(\Pi = cRT\), and given the slope as \( R \times T \): \[ T = \frac{25.73 bar mol^{-1}}{0.083 L bar mol^{-1} K^{-1}} \approx 310 K \]
Converting to Celsius: \( 310 K - 273.15 = 36.85 \)°C, which rounds to 37°C. Quick Tip: Always check your units and conversions, especially when working with temperature and pressure in gas laws.

The Arrhenius equation is given by:
\[ k_2 = k_1 \exp\left(\frac{-E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)\right) \]
Since the rate quadruples, \(\frac{k_2}{k_1} = 4\). Taking the natural logarithm of both sides:
\[ \ln 4 = \frac{-E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \]
Substituting the given values:
\[ \log 4 = 0.6021, \, T_1 = 27^\circ C = 300 \, K, \, T_2 = 57^\circ C = 330 \, K \] \[ \ln 4 = 0.6021 \quad \Rightarrow \quad \frac{E_a}{R} = \frac{\ln 4}{\left(\frac{1}{T_1} - \frac{1}{T_2}\right)} \] \[ E_a = \left( 8.314 \times \frac{\ln 4}{\frac{1}{300} - \frac{1}{330}} \right) \]
Calculating gives \(E_a = 38.04 \, kJ/mol\). Quick Tip: Use the Arrhenius equation to relate rate changes with temperature and calculate the activation energy.

In the classification of fungi, characteristics such as mode of spore formation, the structure of fruiting bodies, and the morphology of mycelium are commonly used. However, mode of nutrition, while variable among different organisms, is not a primary criterion for classification within fungi. Fungi are heterotrophic organisms that absorb nutrients, and this characteristic is general to all fungi rather than specific to different groups within fungi. Quick Tip: When classifying fungi, focus on reproductive and structural characteristics, not on general traits like nutrition which are common to the entire kingdom.

A. Nucleolus is the site for the active synthesis of ribosomal RNA, so it matches with III.


B. Centriole is known for its organization like the cartwheel structure, so it matches with II.


C. Leucoplasts are involved in storing nutrients, especially starch, oils, and proteins, so it matches with IV.


D. Golgi apparatus is involved in the formation of glycolipids and processing lipids, so it matches with I.


Thus, the correct match is A-III, B-II, C-IV, D-I. Quick Tip: The organelles in the cell have specialized functions like synthesizing ribosomal RNA, storing nutrients, or forming glycolipids. Familiarize yourself with their roles.

Step 1: Evaluate each statement for its contribution to species richness.


A (Correct): The stability over millions of years in tropical regions has allowed extensive speciation and diversification.


B (Incorrect): Contrary to the statement, tropical environments are less seasonal, which contributes to more stable conditions and supports biodiversity.


C (Correct): The availability of abundant solar energy supports higher primary productivity, which in turn supports a wider range of species through increased food availability.

\
D (Correct): Stable, predictable environments allow species to specialize and occupy specific niches, promoting diversity through niche differentiation.


E (Correct): The predictability and constancy of the environment reduce seasonal stress on species and allow for year-round growth and reproduction.


Step 2: Identify the correct combination of factors.

Based on the analysis, the correct combination of factors that contribute to species richness in the tropics includes A, C, D, and E. Quick Tip: Stability and constancy in environmental conditions, along with high solar input, play critical roles in promoting high biodiversity in tropical ecosystems.

Step 1: Understand flower structure and terminology.

Hypogynous: The ovary is positioned above the other parts of the flower, making it superior.


Perigynous: The ovary is located centrally, and other flower parts (calyx, corolla, androecium) form a cup around it but do not attach.


Epigynous: The ovary is below the other parts, which appear to arise from the top of the ovary, making it inferior.


Step 2: Analyze the description or figures provided in the question (assuming general descriptions as figures are not visible here).


For both (a) and (b) described as perigynous, the flower parts surround but do not attach to the ovary, indicating that neither is superior nor inferior but rather centrally located with respect to the other floral parts. Quick Tip: Visualize or reference flower cross-sections to identify ovary positions accurately; this will aid in distinguishing between hypogynous, perigynous, and epigynous arrangements effectively.

Malonate is a competitive inhibitor of succinic dehydrogenase. It competes with succinate (the substrate) for binding to the active site of the enzyme, thereby inhibiting the enzyme's activity. In competitive inhibition, the inhibitor binds to the active site, preventing the substrate from binding. Quick Tip: In competitive inhibition, the inhibitor competes with the substrate for binding to the enzyme's active site, reducing enzyme activity.

Step 1: Identify typical cell types with described characteristics.

Cells with thin outer walls and highly thickened inner walls are typical of certain specialized cells like sclerenchyma fibers or xylem vessels, which are involved in support and water transport, respectively.


Step 2: Match descriptions to typical plant cell components.

Assuming 'C' refers to such cells, it would be characteristic of a component designed for structural support or conduction, such as sclerenchyma.
Quick Tip: Identifying cell types based on wall thickness can help in understanding their roles in plant structure and function, such as support, conduction, or storage.

The process described in the question refers to the conservation of species by taking them out of their natural habitat and placing them in a controlled environment where they can be protected and nurtured. This type of conservation is known as ex-situ conservation.

Understanding the different types of conservation.

In-situ conservation refers to the conservation of species in their natural habitats, i.e., protecting ecosystems and the species within them in their natural environment.

Ex-situ conservation is when the species are taken out of their natural habitat and placed in special settings like zoos, botanical gardens, or aquariums where they can receive special care. This method involves human intervention to provide a safe environment for endangered species.

Since the given description refers to taking the species out of their natural habitat, the correct answer is related to Biodiversity conservation (the broad field under which both in-situ and ex-situ methods are included). The specific term for this process is ex-situ conservation. However, since the correct answer choice is (4) "Biodiversity conservation", we conclude this as the intended answer.

Thus, the correct answer is option (4) Biodiversity conservation. Quick Tip: Ex-situ conservation is crucial for species that can no longer be protected in their natural habitat or where natural habitats have been destroyed.

Mendel's Law of Dominance states that when two different alleles are present, one may mask the expression of the other. The dominant allele will express its trait, while the recessive allele will not show unless both alleles are recessive.

A is correct: One allele is dominant, the other recessive.

C is correct: Alleles occur in pairs in diploid organisms.

D is correct: The discrete unit controlling a particular character is called a "factor" (now known as a gene).

E is correct: In a monohybrid cross, only the dominant character is typically expressed in the F1 generation.

B is incorrect because it describes incomplete dominance, not dominance.


Thus, the correct answer is (4). Quick Tip: Mendel's Law of Dominance explains how one allele can dominate the expression of another allele in a heterozygote.

Auxins, particularly synthetic ones like 2,4-D, are used to kill dicot weeds by causing abnormal growth. However, they have minimal effect on monocot plants like grasses. This is because monocots are less sensitive to auxins compared to dicots. In grasses, auxin does not cause abnormal growth, so it doesn't harm them, but it can effectively target dicot weeds. Quick Tip: Auxins are more effective on dicot plants, as monocots (like grasses) are less sensitive to them.

Step 1: Understand the function of each component of a transcription unit.

Promoter: The region where RNA polymerase binds to initiate transcription.

Structural gene: The actual segment of DNA that is transcribed into RNA.

Terminator: The sequence that signals the end of transcription.


Step 2: Relate the components to the transcription process.

These three regions define the start, the body, and the end of the gene transcription process, which fits the description in option (2). Quick Tip: A clear understanding of gene structure and function helps in deciphering how genes are expressed and regulated in cells.

In bacterial cells, particularly in the model organism E. coli, lactose uptake is facilitated by a protein called lactose permease. This protein is integral to the cell membrane and functions by transporting lactose from the environment into the cell through a process called facilitated diffusion. This process is specific to lactose and crucial for bacteria to utilize lactose as an energy source. The gene encoding lactose permease is part of the lac operon, which is regulated by the presence or absence of lactose. Quick Tip: Lactose permease is crucial for the lactose metabolism in E. coli and other bacteria, enabling the transport of this disaccharide into cells where it can be processed by enzymes like beta-galactosidase.

Step 1: Understand the concept of a test cross.

A test cross is used to determine the genotype of an individual exhibiting a dominant phenotype by crossing it with a homozygous recessive individual.

Step 2: Apply the test cross to the scenario.

To determine whether the black-seeded plant is homozygous dominant (BB) or heterozygous (Bb), it should be crossed with a white-seeded plant, which has a known homozygous recessive genotype (bb).

Step 3: Interpret possible outcomes.

If any of the offspring from this cross are white, the black-seeded parent must be heterozygous (Bb). If all offspring are black, the parent is likely homozygous dominant (BB).
Quick Tip: Using a test cross (crossing with a homozygous recessive individual) is a standard method in genetics to determine the genotype of an individual exhibiting a dominant phenotype.

Totipotency is a fundamental characteristic of plant cells that allows each cell to regenerate into a complete plant. This capacity underlies techniques like tissue culture, where cells from a single plant can be used to generate new plants, thereby ensuring genetic uniformity and rapid propagation. It highlights the ability of plant cells to differentiate and organize into complete structures, a trait utilized in biotechnological applications. Quick Tip: Totipotency is a key concept in plant biotechnology, enabling the regeneration and cloning of plants from single cells.

Over-exploitation (A): Unsustainable hunting, fishing, and resource extraction lead to species decline.

Co-extinction (B): When a species goes extinct, dependent species also face extinction.

Habitat loss and fragmentation (D): Deforestation, urbanization, and agriculture destroy ecosystems.

Mutation (C) and Migration (E) are natural processes and do not directly cause biodiversity loss.


% Quick tip
\begin{quicktipbox
The biggest threats to biodiversity are habitat destruction, climate change, pollution, and invasive species.
\end{quicktipbox Quick Tip: The biggest threats to biodiversity are habitat destruction, climate change, pollution, and invasive species.

Step 1: Understanding the Logistic Growth Equation

The logistic growth model described by this equation represents how the growth rate of a population (dN/dt) evolves over time in a limited environment. The rate of change of the population size \(N\) is dependent on both the current population size and the carrying capacity \(K\).


Step 2: Role of \(K\) in the Equation

\(K\) represents the carrying capacity of the environment, which is the maximum population size that the environment can sustain indefinitely under given conditions. As \(N\) approaches \(K\), the factor \(\left[\frac{K - N}{K}\right]\) approaches zero, which slows down the growth rate, indicating that the environment cannot support a higher number of individuals without depleting resources.


Step 3: Explanation of Other Terms

Population density refers to the number of individuals per unit area and is not specifically represented by \(K\) in this model.

Intrinsic rate of natural increase (r) is represented by \(r\) in the equation, which is the per capita rate of increase assuming no resource limitation.

Biotic potential is the maximum reproductive capacity of an organism under optimal conditions, not directly addressed in this logistic model.
Quick Tip: Carrying capacity (\( K \)) can change over time due to environmental changes and is crucial for managing wildlife and conservation efforts.

Statement I: During the leptotene stage of meiosis, chromosomes start to condense and become gradually visible under a light microscope, which is true.

Statement II: The diplotene stage of meiosis is indeed marked by the dissolution of the synaptonemal complex, facilitating the separation of homologous chromosomes. This statement is also true.
Quick Tip: Understanding the stages of meiosis is crucial for identifying when chromosomal structures appear and behave in specific ways.

Dedifferentiation refers to the process where mature, differentiated cells regain the ability to divide and become meristematic (able to form new tissue). In this case, parenchyma cells in the vascular bundles undergo dedifferentiation to form interfascicular cambium, which will participate in secondary growth. Quick Tip: Dedifferentiation allows mature cells to revert to a meristematic state and contribute to growth.

Carboxypeptidase is a metalloenzyme, which means it requires a metal ion as a cofactor for its enzymatic activity. Specifically, carboxypeptidase uses zinc as its cofactor. The zinc ion plays a crucial role in stabilizing the transition state and activating the water molecule that hydrolyzes the peptide bond. This process allows the enzyme to cleave amino acids from the carboxyl end of proteins and peptides efficiently. Quick Tip: Knowing the class of enzymes and their cofactors can help predict their function and mechanism, as seen in metalloenzymes like carboxypeptidase.

Step 1: Define the role of bulliform cells.

Bulliform cells are large, thin-walled cells located on the upper epidermis of leaves, particularly in grasses.


Step 2: Link the physiological function of bulliform cells with leaf movements.

These cells swell and shrink in response to changes in water availability, leading to the curling of leaves during dry conditions. This action reduces the surface area exposed to the sun and minimizes water loss. Quick Tip: Bulliform cells are an excellent example of structural adaptation in plants to environmental stress, particularly in reducing water loss during drought.

Step 1: Understanding the Microorganisms and Their Products

A. Clostridium butylicum: Known for producing butyric acid through the fermentation process. Hence, it matches with III. Butyric acid.

B. Saccharomyces cerevisiae: Commonly known as brewer's yeast or baker's yeast, this microorganism is famous for its ability to produce ethanol by fermenting sugars. Thus, it matches with I. Ethanol.

C. Trichoderma polysporum: This fungus is noted for its ability to produce cyclosporin A, an immunosuppressive drug, not streptokinase. Therefore, it corresponds to IV. Cyclosporin-A.

D. Streptococcus sp.: Certain species within this genus are utilized for the production of streptokinase, an enzyme used in medical settings to break down blood clots. This fits with II. Streptokinase.


Step 2: Matching Each Organism with the Correct Product

Given the roles and products associated with each microorganism, the matches are:

A-III: Clostridium butylicum produces butyric acid.

B-I: Saccharomyces cerevisiae is used for ethanol production.

C-IV: Trichoderma polysporum is known for cyclosporin-A production.

D-II: Streptococcus sp. produces streptokinase.
Quick Tip: Each microorganism has a unique metabolic pathway that can be harnessed biotechnologically for specific product synthesis.

Step 1: Understand the genetics of Snapdragon flower color.

Snapdragon flowers exhibit incomplete dominance, where the heterozygote (Rr) results in a pink phenotype, different from the homozygous red (RR) or white (rr) flowers.


Step 2: Predict the offspring from a cross between pink (Rr) and red (RR).

Crossing a pink (Rr) Snapdragon with a red (RR) Snapdragon results in the following genotypes:

50% Red flowered (RR)

50% Pink flowered (Rr)


Step 3: Conclude the expected phenotypes.
The progeny will exhibit both red and pink flowers, as the genotypes RR and Rr correspond to these phenotypes, respectively. Quick Tip: \textbf{Incomplete Dominance}: Neither allele is completely dominant, and heterozygous individuals show an intermediate phenotype. Example: Snapdragon flower color.

Statement I: Bt toxins, encoded by cry genes such as cry IAc, are indeed insect group-specific. These toxins target specific insect larval stages, affecting only those that are susceptible to the particular cry protein.

Statement II: The Bt toxin is an inactive protoxin within the bacterial spore. However, it is activated in the alkaline pH of the insect's gut, not acidic pH. The statement is false because it incorrectly states the pH condition. Quick Tip: Bt toxin activation is pH-dependent and occurs in the alkaline environment of many insect guts. This specificity is crucial for its role in biological control.

Step 1: Assess each statement for accuracy regarding botanical characteristics and pollination strategies.

Statement A: Vallisneria flowers are small and not colorful; they also do not produce nectar. Instead, they are adapted for hydrophily, a type of pollination involving water. Thus, statement A is incorrect.

Statement B: Water lilies are indeed pollinated by insects, not by water, making this statement correct.

Statement C: True for many hydrophytes, where pollen grains have adaptations to prevent wetting, supporting underwater pollination.

Statement D: Correct, as seen in hydrophytes like seagrasses, where pollen grains can be long and ribbon-like to be carried by water.

Statement E: Also true, particularly in water-pollinated plants, where pollen grains are passively transported by water.


Step 2: Determine the correct combination of true statements.

The correct set based on the analysis is B, C, D, and E. Quick Tip: Understanding adaptations in plant reproduction, especially concerning their pollination mechanisms, can clarify which features are associated with specific environments like aquatic settings.

Recall the stages of mitosis.
During metaphase, chromosomes are lined up at the metaphase plate, and spindle fibers attach to the kinetochores located on the chromosomes. Quick Tip: Understanding the sequence of mitosis stages is crucial for recognizing the timing of cellular events.

Understand the terms.

Allele: Alternative forms of a gene.

Test cross: Cross between F1 progeny and homozygous recessive parent.

Back cross: Cross between F1 progeny and one of the parents.

Ploidy: Number of sets of chromosomes.
Quick Tip: Remember, test cross is used to determine the genotype of an individual by crossing with a homozygous recessive individual.

Step 1: Identifying Organisms and Their Characteristics


Rhizopus: Known as black bread mold, Rhizopus is a genus of common saprophytic fungi on plants and specialized parasites on animals. It is best known as bread mold, hence matches with III (Bread mould).


Ustilago: Known commonly as smut fungi, these are pathogens of grasses including important crops. Ustilago needs to be matched with II (Smut fungus).


Puccinia: A well-known genus of fungi, responsible for rust diseases on many plants, fits perfectly with IV (Rust fungus).


Agaricus: A genus of mushrooms containing both edible and poisonous species, including the well-known button mushroom, matches with I (Mushroom).


Step 2: Analyzing the Options

Given the characteristics defined above:

A (Rhizopus) matches with III (Bread mould).

B (Ustilago) matches with II (Smut fungus).

C (Puccinia) matches with IV (Rust fungus).

D (Agaricus) matches with I (Mushroom).


Step 3: Selecting the Correct Match

Option (3) states:

A-III, B-II, C-IV, D-I, which corresponds exactly to our analysis.
Quick Tip: Familiarity with the common names of fungi and their characteristics can aid in their identification and understanding of their ecological roles.

Statement I: Parenchyma and collenchyma are both living tissues. Parenchyma is involved in storage and photosynthesis, while collenchyma provides support and flexibility to the plant. Therefore, Statement I is false.

Statement II: Gymnosperms generally lack xylem vessels, which are found in angiosperms. This characteristic aids in efficient water transport in angiosperms, supporting taller growth and greater biomass. Therefore, Statement II is true.
Quick Tip: When studying plant anatomy, it’s important to understand the structure and function of different tissues, especially in how they contribute to the plant's overall physiology and survival strategies.

The Calvin cycle is the process by which plants fix carbon dioxide into glucose. For each molecule of CO\(_2\) fixed:

3 molecules of ATP and 2 molecules of NADPH are consumed during the reduction phase of the Calvin cycle to form glyceraldehyde-3-phosphate (G3P).

This ratio comes from the fact that 3 molecules of CO\(_2\) are incorporated per cycle, requiring 3 ATP molecules for phosphorylation and 2 NADPH molecules for reduction to form the sugar intermediates.
Quick Tip: In the Calvin cycle, ATP is used for phosphorylation and NADPH is used for reduction to synthesize sugar molecules.

Step 1: Identify the organization responsible for monitoring biodiversity.

The International Union for Conservation of Nature (IUCN) is known for publishing the Red List, which assesses the conservation status of species globally.

Step 2: Confirm the role of IUCN.

IUCN's Red List provides detailed analyses on the population status of various species and categorizes them into groups like endangered, vulnerable, and critically endangered. Quick Tip: The IUCN Red List is a critical tool for conservation planning and prioritization, providing detailed information on the status and threats to species across the globe.

Lecithin is a type of phospholipid, which is a major component of cell membranes. It consists of a glycerol backbone, two fatty acid chains, and a phosphate group, making it a phospholipid. Quick Tip: Phospholipids, like lecithin, are crucial in forming the bilayer structure of cell membranes, aiding in membrane fluidity and permeability.

An actinomorphic flower, also known as a radially symmetrical flower, can be divided into two equal halves through multiple planes. Datura is an example of an actinomorphic flower because its petals can be divided symmetrically in multiple directions. In contrast, Pisum (pea) and Sesbania flowers are zygomorphic (bilaterally symmetrical), while Cassia flowers are also zygomorphic. Quick Tip: Actinomorphic flowers are radially symmetrical and can be divided into equal halves through multiple planes.

Step 1: Analyze the potential outcomes for a transgenic piece of DNA.

When a piece of DNA is introduced into an alien organism, it can integrate into the host's genome (B), allowing it to be copied and passed down during cell division (C).

While it can replicate (E), whether it does so independently (A) or not depends on the presence of necessary origin of replication sites and compatibility with the host cell's machinery.


Step 2: Evaluate the statements for biological plausibility.

The DNA piece's integration into the host genome (B) and its subsequent multiplication along with host DNA during cell division (C) are typical events in genetic engineering practices. Quick Tip: When considering the fate of introduced DNA in genetic engineering, focus on integration and inheritance mechanisms, which are critical for stable expression and maintenance of the introduced trait.

Recall the properties of restriction enzymes.

Hind II is a restriction enzyme that recognizes a specific nucleotide sequence to cut DNA, which in this case is 6 base pairs long. Quick Tip: The specific sequence recognized by Hind II is a key detail often asked in molecular biology exams.

The dark reactions of photosynthesis, also known as the Calvin cycle, do not directly require light to proceed, contrary to what the name might suggest. Instead, they rely on the chemical energy stored in ATP and NADPH, which are produced by the light reactions.

C (CO\(_2\)): Carbon dioxide is a critical carbon source for the Calvin cycle, where it is fixed into organic molecules.

D (ATP) and E (NADPH): These molecules are used to convert the fixed carbon dioxide into glucose and other sugars. ATP provides the energy, while NADPH provides the reducing power.

Thus, C, D, and E are essential for the dark reaction, while A (Light) and B (Chlorophyll) are not directly required in this phase of photosynthesis. Quick Tip: Remember, the dark reactions (Calvin cycle) utilize the ATP and NADPH produced by the light-dependent reactions to synthesize organic compounds from CO\(_2\).

In the given seed diagram, the part of the seed that is destined to form the root is the radicle, which is usually marked as part C in seed diagrams.


Understanding the parts of a seed.

Radicle is the part of the embryo that grows into the root during germination.

Plumule is the part of the embryo that grows into the shoot.
The seed coat protects the seed and its embryo.

The cotyledons provide nutrients to the embryo during the early stages of germination.


In the given diagram, C corresponds to the radicle, which forms the root.


Thus, the correct answer is option (1) C. Quick Tip: When studying seed germination, identifying embryonic structures such as the radicle is key to understanding early plant development.

Step 1: Understanding GPP and NPP.

Gross Primary Productivity (GPP) encompasses the total rate of photosynthesis, which includes the energy used for respiration. Net Primary Productivity (NPP) is the leftover energy after plants have respired, calculated as \( NPP = GPP - Respiration \).

Step 2: Estimating energy transfer.

Energy typically transfers at about 10% efficiency between trophic levels. Given \(100x\) at the first level, the second level would theoretically receive \(10x\), considering only about 10% efficiency in energy transfer.

Step 3: Applying the 10% rule to the third level.

The third trophic level would similarly receive 10% of the second level's NPP, suggesting it also handles a GPP of about \(10x\). Quick Tip: When working with ecosystems, always consider the energy efficiency between trophic levels, typically around 10%.

Determine the locations of cellular processes.

Citric acid cycle occurs in the mitochondrial matrix, hence A-II.

Glycolysis takes place in the cytoplasm, making B-I.
Electron transport system is located in the inner mitochondrial membrane, so C-IV.

Proton gradient is formed in the intermembrane space of mitochondria, thus D-III.
Quick Tip: Knowing the locations of key metabolic processes is fundamental in understanding cellular respiration and energy production.

Step 1: Detailed breakdown of stamen types and their examples.

Monoadelphous stamens are fused together by their filaments forming one group. This arrangement is commonly seen in China-rose (Hibiscus).

Diadelphous stamens are fused into two separate groups. The Pea (Pisum sativum) is a classic example where some stamens are fused together forming a bundle distinct from another.

Polyadelphous stamens are fused into multiple groups, as in Citrus, where the stamens form more than two groups.

Epiphyllous stamens are attached directly to the petals. Lily (Lilium) is an example of this arrangement, where stamens arise from the base of the petals.
Quick Tip: Botanical vocabulary related to flower anatomy is crucial in taxonomy and understanding plant diversity.

Examining photosynthesis pathways.
In C\(_3\) plants, RuBisCO does bind O\(_2\) in addition to CO\(_2\), leading to photorespiration which decreases CO\(_2\) fixation efficiency. In C\(_4\) plants, the anatomy and compartmentalization of photosynthesis greatly reduce photorespiration, but it's incorrect to say bundle sheath cells do not show photorespiration; they actually do, but at a very reduced rate due to CO\(_2\) concentration mechanisms. Quick Tip: Understanding the differences between C3 and C4 photosynthesis can aid in botany and ecology, especially in strategies for improving agricultural crop efficiency.

Step 1: Elaborate on the replication mechanism.
During DNA replication in E.coli, the enzyme DNA polymerase III synthesizes new DNA strands by adding nucleotides to the 3' end of the nascent strand, effectively building the new strand in the 5' to 3' direction. This unidirectional synthesis is crucial for the fidelity and regulation of DNA replication. Quick Tip: Memorizing the function and directionality of key enzymes like DNA polymerase helps in understanding fundamental genetic processes.

Verifying the statements about Phaeophyceae.

A: True, asexual reproduction in Phaeophyceae often involves biflagellate zoospores.

B: False, while oogamy is common, it's not the only method of sexual reproduction.

C: True, mannitol and laminarin are typical storage carbohydrates in these algae.

D: True, these pigments are indeed found in Phaeophyceae.

E: True, the cell walls are typically cellulosic with an algin coating.
Quick Tip: Familiarity with the specific biological traits of algae groups like Phaeophyceae enhances understanding in marine biology and botany.

Step 1: Detailed explanation of each term's role.
- GLUT-4 is a glucose transporter type 4 which facilitates glucose uptake into cells, especially in adipose tissue and muscle, responsive to insulin.
- Insulin is a hormone produced by the pancreas, crucial for regulating glucose levels in the blood by enhancing glucose uptake into cells.
- Trypsin is an enzyme that breaks down proteins in the small intestine, crucial for digestion.
- Collagen is the main structural protein found in various connective tissues, forming a scaffold to provide strength and structure. Quick Tip: Linking proteins and molecules to their functions in physiology can help in quickly solving comparative questions in biochemistry and molecular biology.

Associate each individual or team with their discovery.

Frederick Griffith is known for the transformation principle, where genetic material can be transferred between dead and living bacteria.

Francois Jacob and Jacque Monod elucidated the lac operon as a model for gene regulation.

Har Gobind Khorana contributed significantly to the understanding of the genetic code.

Meselson and Stahl demonstrated the semi-conservative mode of DNA replication. Quick Tip: Linking historical scientific figures to their discoveries can aid in remembering their contributions for complex matching questions.

Role of Gibberellin.
Gibberellins are a group of plant hormones that promote growth and elongation of cells. In sugarcane, spraying gibberellin stimulates cell elongation, which results in an increased length of the stem. This increase in stem length directly correlates to a higher yield, as sugarcane's economic value is directly linked to its biomass. Quick Tip: Gibberellins can be used to manipulate plant growth under agricultural settings, especially to maximize crop yields in stem-heavy plants like sugarcane.

Associate each individual with their contribution.

Robert May is known for estimating the global species diversity.

Alexander von Humboldt founded the species-area relationship.

Paul Ehrlich is famous for the rivet popper hypothesis which is an ecological analogy for the importance of biodiversity.

David Tilman conducted long-term ecosystem experiments.
Quick Tip: It helps to remember key contributions of scientists when faced with matching questions in ecology and environmental science.

Characteristics of chloroplast DNA.
Chloroplast DNA is indeed circular and double-stranded, resembling bacterial DNA, which reflects the evolutionary origin of chloroplasts via endosymbiosis. This type of DNA is key for the autonomous functions that chloroplasts perform, separate from nuclear DNA. Quick Tip: The structure of chloroplast DNA can be crucial for studies in plant genetics and biotechnology, especially when engineering plants for better photosynthetic efficiency.

Understanding the characteristics of each plant.

Rose has a perigynous flower, which means that the ovary is positioned in the middle, and the other floral parts (sepals, petals, stamens) are attached around the ovary. Hence, \( A \) matches with \( II \) (Perigynous flower).

Pea has marginal placentation, where the ovules are borne on the margins of the ovary. Thus, \( B \) matches with \( IV \) (Marginal placentation).

Cotton has twisted aestivation, where the petals are arranged in a spiral manner with overlapping edges. Therefore, \( C \) matches with \( I \) (Twisted aestivation).

Mango produces a drupe, which is a fleshy fruit with a single seed enclosed by a hard endocarp. Hence, \( D \) matches with \( III \) (Drupe).

Thus, the correct match is:

- A-II: Rose has a perigynous flower.

- B-IV: Pea has marginal placentation.

- C-I: Cotton has twisted aestivation.

- D-III: Mango has a drupe.


Therefore, the correct answer is option (3). Quick Tip: Knowing floral anatomy and fruit types is crucial for botany and helps in plant identification and classification.

Step 1: Understanding the Tricarboxylic Acid Cycle

The TCA cycle, also known as the Krebs cycle or citric acid cycle, is a series of chemical reactions used by all aerobic organisms to generate energy through the oxidation of acetate derived from carbohydrates, fats, and proteins into carbon dioxide and chemical energy.


Step 2: Analyzing Each Step Mentioned

Option (1) Succinyl-CoA → Succinic acid: This step involves the conversion of Succinyl-CoA to Succinic acid. It is accompanied by the cleavage of the CoA group and the conversion of GDP to GTP (or ADP to ATP in some organisms). This reaction, catalyzed by Succinyl-CoA synthetase, is a substrate-level phosphorylation rather than an oxidation reaction.

Option (2) Isocitrate \(\rightarrow\) \ch{alpha-ketoglutaric acid:
This reaction involves oxidative decarboxylation and is catalyzed by isocitrate dehydrogenase.
NAD\(^{+}\) is reduced to NADH, indicating oxidation.

Option (3) Malic acid \(\rightarrow\) \ch{Oxaloacetic acid:
This step is catalyzed by malate dehydrogenase and involves the oxidation of malate to oxaloacetate,
reducing NAD\(^{+}\) to NADH.

Option (4) Succinic acid → Malic acid: This includes two steps: Succinic acid to fumarate by succinate dehydrogenase (an oxidation where FAD is reduced to FADH2), and then fumarate to malate by fumarase (hydration).


Step 3: Conclusion
The only step listed that does not involve the oxidation of a substrate is the conversion of Succinyl-CoA to Succinic acid. While this step does involve the transformation of energy, it does not involve the direct oxidation of the molecule itself; rather, it's coupled to the synthesis of GTP/ATP, marking it as a step of substrate-level phosphorylation. Quick Tip: This step in the TCA cycle is unique as it couples the conversion of Succinyl-CoA to the synthesis of ATP or GTP, unlike other steps which involve oxidation.

Understanding somatic hybridization.
Somatic hybridization involves the fusion of protoplasts from two different plant varieties to combine their genetic material in one cell, which can then be cultivated to grow into a hybrid plant. This technique is used to bring together desirable traits from both parent varieties. Quick Tip: Somatic hybridization is a powerful tool in plant biotechnology for creating hybrids with new traits, bypassing sexual reproduction limitations.

Step 1: Analyzing the Image

The image depicts an inflorescence characterized by long, prominent stamens that are freely hanging and well exposed to the environment, which is typical for maximizing wind dispersal of pollen.


Step 2: Evaluating the Options

Option 1 (Cleistogamous flowers showing autogamy): Incorrect as cleistogamous flowers are closed, promoting self-pollination without external pollen exposure.

Option 2 (Compact inflorescence showing complete autogamy): Incorrect as the structure does not appear compact or self-contained; rather, it is open and adapted for external pollination.

Option 3 (Wind pollinated plant inflorescence): Correct, the structure and arrangement of stamens are ideal for pollen dispersal by wind, indicating an adaptation to wind pollination.

Option 4 (Water pollinated flowers with mucilaginous covering): Incorrect as the structure does not show any adaptations like mucilaginous coverings that would be associated with water pollination.


Step 3: Concluding with the Correct Answer

The correct answer is Option 3, which accurately describes the botanical characteristics observed in the image, aligning with adaptations known for wind pollination. Quick Tip: Wind-pollinated plants often have flowers that are not brightly colored or fragrant, as these features are unnecessary for attracting pollinators.

Detailed Analysis of Each Match:

\(\alpha\)-1 Antitrypsin (A): This is a protein that protects tissues from enzymes of inflammatory cells, especially elastase. Deficiency in \(\alpha\)-1 antitrypsin can lead to diseases like emphysema due to uncontrolled enzyme activity degrading lung tissue. Thus, A matches with III (Emphysema).


Cry IAb (B): Cry IAb is a gene from the bacterium Bacillus thuringiensis (Bt) that codes for a protein toxic to certain insects, including the corn borer. This gene is used in genetically modified crops to provide resistance against this pest. Therefore, B matches with IV (Corn borer).


Cry IAc (C): Similar to Cry IAb, Cry IAc is another variant of the Bt toxin protein targeting different pests, such as the cotton bollworm. This makes crops like cotton resistant to such insects. Hence, C matches with I (Cotton bollworm).


Enzyme replacement therapy (D): This is a medical treatment used to replace a deficient or absent enzyme in patients with enzyme deficiencies. A common example is the treatment of ADA (adenosine deaminase) deficiency, which can lead to severe combined immunodeficiency (SCID). Thus, D matches with II (ADA deficiency).
Quick Tip: Understanding the application of biotechnology in medicine and agriculture highlights the practical importance of genetic engineering.

Step 1: Analyzing Statement I.

Statement I is true because the hymen can be absent or stretched for reasons unrelated to sexual activity, such as sports or physical activity. Therefore, it is not a reliable indicator of virginity.

Step 2: Analyzing Statement II.

Statement II is false because the hymen may tear or stretch during various activities, not just during the first coitus. For example, tampon use or vigorous exercise can cause the hymen to tear or stretch. Quick Tip: It’s important to understand that biological markers like the hymen do not necessarily correlate with virginity, and various factors can affect its appearance.

Matching the species with their common names.

Pterophyllum (A) is commonly known as the Angel fish, hence A-III.

Myxine (B) is also known as Hag fish, hence B-I.

Pristis (C) is commonly called the Saw fish, hence C-II.

Exocoetus (D) is known as the Flying fish, hence D-IV.
Quick Tip: Familiarizing yourself with common names and scientific names helps in identifying species correctly in taxonomy.

Step 1: Analyzing Assertion A.

The assertion is incorrect because FSH (Follicle Stimulating Hormone) acts on ovarian follicles in females but stimulates Sertoli cells in males, not Leydig cells. Leydig cells are primarily stimulated by LH (Luteinizing Hormone) to produce testosterone.

Step 2: Analyzing Reason R.

The reason is correct as it accurately describes the function of ovarian follicles in females (secreting estrogen) and interstitial cells (Leydig cells) in males (secreting androgen, primarily testosterone). Quick Tip: FSH is involved in stimulating gametogenesis and hormone production, but its targets differ between males and females.

Identifying the parts of the female reproductive system.

The Fallopian tube consists of four main parts: the infundibulum, ampulla, isthmus, and fimbriae. The uterine fundus, however, refers to the upper part of the uterus, not part of the Fallopian tube. Quick Tip: The Fallopian tube is a key part of the female reproductive system, responsible for transporting eggs from the ovaries to the uterus.

Identifying non-natural contraceptive methods.

Vaults refer to barrier methods such as cervical caps, which are not natural methods but rather physical barriers used to prevent pregnancy. Quick Tip: Differentiating between natural and artificial contraceptive methods is essential for understanding reproductive health options.

Step 1: Matching diseases with their causative agents.

Typhoid (A) is caused by a bacterium, hence A-IV.

Leishmaniasis (B) is caused by a protozoan, hence B-III.

Ringworm (C) is a fungal infection, hence C-I.

Filariasis (D) is caused by a nematode, hence D-II.
Quick Tip: Knowing the causative agents of diseases helps in understanding their treatment and prevention.

Step 1: Analyzing the Assertion.

Breast-feeding is indeed recommended for the health of newborns because it provides essential nutrients and antibodies during the critical early stages of growth.

Step 2: Analyzing the Reason.

Colostrum, the first milk produced by the mother after delivery, contains high levels of antibodies that help protect the newborn from infections. This makes the Reason R correct and an explanation for Assertion A. Quick Tip: Colostrum is a vital first milk that helps in building immunity in the newborn, which is why early breastfeeding is so important.

Understanding the Hardy-Weinberg equilibrium.

The Hardy-Weinberg equilibrium assumes no evolution is occurring, meaning factors like gene migration, genetic recombination, and genetic drift can all affect allele frequencies and cause evolution. A constant gene pool, however, implies no changes in allele frequencies, which is the condition for Hardy-Weinberg equilibrium to hold true. Quick Tip: The Hardy-Weinberg equilibrium principle is used to study allele frequencies in a population, assuming no evolutionary forces act on it.

Understanding the role of bio-reactors.
Bio-reactors are typically used for large-scale cultivation of microorganisms, including bacteria, fungi, or cells for the production of various products. They are designed to optimize growth conditions, not typically for small-scale cultures, which would be produced in flasks or smaller vessels. Quick Tip: Bio-reactors are critical for industrial-scale fermentation processes, providing controlled environments for optimal microbial growth.

Analysis of Muscle Types and Locations:

(a) Skeletal Muscle: Characterized by striations and voluntary control. Common examples include muscles such as the biceps and triceps which are used for movement of bones.

Correct identification: Triceps, a large muscle on the back of the upper limb of many vertebrates. It is responsible for extension of the elbow joint.


(b) Smooth Muscle: Lacks striations and is under involuntary control. It is found in walls of hollow organs like intestines and stomach.

Correct identification: Stomach, which extensively uses smooth muscle for the process of digestion through peristalsis.

(c) Cardiac Muscle: Striated like skeletal muscle but operates under involuntary control, found only in the heart.

Correct identification: Heart, the primary location of cardiac muscle which is specialized for continuous contractions to pump blood throughout the body.\

Conclusion:

Each muscle type is correctly identified in option (4) with its location, matching the unique structural and functional characteristics necessary for their respective roles in the human body. Quick Tip: Distinguishing muscle types is key in anatomy and helps in understanding the physiological roles of different muscle tissues in the body.

Identifying the correct plant or drug source.

Cocaine comes from the plant Erythroxylum, hence A-III.

Heroin is derived from the opium poppy, Papaver somniferum, so B-IV.

Morphine is also derived from Papaver somniferum, hence C-I.

Marijuana comes from Cannabis sativa, making D-II.
Quick Tip: Being familiar with the sources of narcotics and their medicinal uses can help differentiate them in pharmacology-related questions.

Step 1: Analyzing each brain structure's function.

Pons (A) connects different regions of the brain, facilitating communication, hence A-III.

Hypothalamus (B) contains neurosecretory cells and is involved in regulating endocrine functions, hence B-IV.

Medulla (C) controls basic life functions, such as respiration and gastric secretions, hence C-II.

Cerebellum (D) is responsible for maintaining posture and balance, hence D-I. Quick Tip: Understanding the function of different brain regions helps in recognizing their role in physiological processes.

Analyzing the disorders.

Myasthenia gravis (A) is an autoimmune disorder in which the body's immune system attacks the neuromuscular junction, leading to muscle weakness.

Rheumatoid arthritis (B) is an autoimmune disorder where the immune system attacks the joints, leading to inflammation and damage.

Gout (C) is not an autoimmune disorder; it is a metabolic condition caused by the accumulation of uric acid crystals in joints.

Muscular dystrophy (D) is a group of genetic disorders affecting muscle strength but is not autoimmune.

Systemic Lupus Erythematosus (SLE) (E) is a classic autoimmune disorder where the immune system attacks various tissues in the body. Quick Tip: Autoimmune diseases are conditions where the immune system mistakenly attacks the body's own tissues, and recognizing them is important for diagnostic purposes.

Understanding the role of the Ti plasmid.

The Ti plasmid (tumor-inducing plasmid) of Agrobacterium tumefaciens is responsible for causing crown gall disease in plants by transferring a part of the plasmid into the plant's genome, which induces tumor formation. Quick Tip: The Ti plasmid is a critical tool in genetic engineering as it is used to transfer genes into plant cells.

Analyzing the characteristics of each sub-phase.

Diakinesis is the final sub-phase of prophase I, where the chiasmata (points of crossing over) are fully terminalized. Hence, A-II.

Pachytene is characterized by the appearance of recombination nodules, which are important for the process of genetic recombination. Hence, B-IV.

Zygotene is the phase where the synaptonemal complex, which helps in pairing homologous chromosomes, begins to form. Hence, C-I.

Leptotene is the earliest sub-phase, where chromosomes first become visible as thin threads. Hence, D-III.
Quick Tip: Familiarizing yourself with the stages of meiosis and the events specific to each phase will help in answering questions related to genetic processes.

Understanding each contraceptive method.

Non-medicated IUD, such as the Lippes Loop, does not release hormones or metals.

Copper releasing IUD, like Multiload 375, uses copper as a spermicidal agent.

Hormone releasing IUD, exemplified by LNG-20, releases levonorgestrel, a hormone used to prevent pregnancy.

Implants, such as those releasing Progestogens, are placed under the skin to release hormones over a long period.
Quick Tip: Familiarize yourself with the types of contraceptives and their mechanisms to better understand their applications in medical practice.

Analyzing the coelom status of the organisms.

Annelids (A) are true coelomates because they have a well-developed coelom (body cavity) that is completely lined with mesoderm.

Poriferans (B) are not pseudocoelomates; they lack a true coelom or pseudocoelom and are classified as acoelomates.

Aschelminthes (C) are pseudocoelomates, not acoelomates, as they have a fluid-filled body cavity not entirely lined by mesoderm.

Platyhelminthes (D) are also acoelomates and do not have a pseudocoelom. Quick Tip: Always remember that true coelomates have a mesoderm-lined cavity, while pseudocoelomates have a body cavity not fully surrounded by mesoderm.

Understanding the Factors Influencing Oxyhaemoglobin Formation:

The formation of oxyhaemoglobin, the complex of oxygen and hemoglobin, is influenced by several physiological factors. This process is essential for transporting oxygen from the lungs to the tissues.


Analysis of Each Factor:

Partial Pressure of Oxygen pO\(_2\): Higher pO\(_2\) increases the affinity of hemoglobin for oxygen, facilitating the formation of oxyhaemoglobin. In the alveoli, where oxygen concentration is high due to fresh air intake, the pO\(_2\) is naturally higher, promoting this binding.

Partial Pressure of Carbon Dioxide pCO\(_2\): Lower pCO\(_2\) reduces the competition between carbon dioxide and oxygen for binding sites on hemoglobin, further enhancing oxygen binding.


Hydrogen Ion Concentration H\(^+\): Lower H\(^+\) concentration, or a higher pH, shifts the oxygen-hemoglobin dissociation curve to the left (known as the Bohr effect). This shift increases hemoglobin's affinity for oxygen, facilitating the formation of oxyhaemoglobin.

Detailed Analysis of Options:
Option (1) Low pCO\(_2\) and High H\(^+\) concentration is not favorable because high H\(^+\) concentration (or low pH) would decrease hemoglobin's oxygen affinity.

Option (2) Low pCO\(_2\) and High temperature: While lower pCO\(_2\) is favorable, higher temperatures actually decrease the affinity of hemoglobin for oxygen, promoting oxygen release rather than uptake in tissues.

Option (3) High pO\(_2\) and High pCO\(_2\): While high pO\(_2\) is favorable, high pCO\(_2\) would lower hemoglobin's oxygen affinity due to competitive inhibition and a shift in pH.

Option (4) High pO\(_2\) and Lesser H\(^+\) concentration: This is the most favorable condition in the alveoli for oxyhaemoglobin formation. High pO\(_2\) enhances oxygen loading, and lesser H\(^+\) concentration increases hemoglobin's affinity for oxygen, facilitating efficient oxygen transport. Quick Tip: The Bohr effect describes how pH levels and CO\(_2\) concentrations influence hemoglobin's oxygen-binding capacity, crucial for understanding respiratory physiology.

Understanding joint classifications.

Fibrous joints (A) are found in the skull where there is no movement, hence A-III.

Cartilaginous joints (B) are found in adjacent vertebrae, allowing limited movement, hence B-I.

Hinge joints (C) are found in the knee and allow bending, hence C-IV.

Ball and socket joints (D) are found in the shoulder (humerus and pectoral girdle) and allow rotational movement, hence D-II. Quick Tip: Joints are classified based on their structure and the movement they allow, so knowing the anatomical features will help in their identification.

Understanding the bond types.

Lipase (A) breaks down ester bonds, which are found in lipids, hence A-II.

Nuclease (B) breaks down phosphodiester bonds in nucleic acids, hence B-IV.

Protease (C) breaks peptide bonds between amino acids, hence C-I.

Amylase (D) breaks down glycosidic bonds in carbohydrates, hence D-III.
Quick Tip: Knowing the types of bonds broken by different enzymes can help identify their function in digestion and metabolism.

Identifying the correct chromosomal abnormality for each condition.

Down’s syndrome (A) is caused by trisomy of the 21st chromosome, so A-III.

\(\alpha\)-Thalassemia (B) is related to a mutation in the 16th chromosome, so B-IV.

\(\beta\)-Thalassemia (C) involves mutations on chromosome 11, so C-I.

Klinefelter’s syndrome (D) is caused by an extra X chromosome, so D-II.
Quick Tip: Chromosomal abnormalities are often linked to specific syndromes and conditions, so understanding the genetics behind these can help in diagnosis.

Understanding the evolutionary sequence.

Homo habilis (A) is one of the earliest members of the genus Homo, appearing around 2.4 million years ago.

Homo erectus (D) evolved next and is considered the first species to have similar body proportions to modern humans.

Homo neanderthalensis (C) appeared after Homo erectus and is known to have coexisted with early Homo sapiens.

Homo sapiens (B) represents modern humans and appeared most recently. Quick Tip: Understanding the timeline of human evolution helps in identifying the major stages of human ancestry and development.

Analyzing each organism and characteristic.

Pleurobrachia is a type of comb jelly, which belongs to the phylum Ctenophora. Hence, \( A \) matches with \( II \) (Ctenophora).


Radula is a characteristic feature of mollusks and is used for feeding. Therefore, \( B \) matches with \( I \) (Mollusca).


Stomochord is a structure found in Hemichordata, which is a phylum closely related to chordates. Thus, \( C \) matches with \( IV \) (Hemichordata).


Air bladder is found in Osteichthyes (bony fish) and helps in buoyancy. Hence, \( D \) matches with \( III \) (Osteichthyes).

Thus, the correct match is:

- A-II: Pleurobrachia belongs to Ctenophora.

- B-I: Radula is found in Mollusca.

- C-IV: Stomochord is found in Hemichordata.

- D-III: Air bladder is found in Osteichthyes.


Therefore, the correct answer is option (4). Quick Tip: Familiarizing yourself with common names and scientific names helps in identifying species correctly in taxonomy.

Understanding the concept of convergent evolution.

Convergent evolution occurs when unrelated species evolve similar traits independently, usually due to similar environmental pressures. Penguins (birds) and dolphins (mammals) both developed flippers for swimming, despite their different evolutionary lineages, which is an example of convergent evolution. Quick Tip: Convergent evolution highlights how different species adapt in similar ways to similar environmental challenges.

Understanding the function of plasmid genes.
The X gene is typically responsible for regulating the copy number of the plasmid in the bacterial cell. It ensures that there is an appropriate number of plasmid copies during cell division.

The Y gene encodes for a protein that is involved in the replication of the plasmid, which is essential for the plasmid's propagation in the host cell.
Quick Tip: In plasmids like pBR322, the replication control and antibiotic resistance genes are crucial for maintaining plasmid stability and selecting bacteria that contain the plasmid.

Understanding the sequence of the cell cycle.

The cell cycle begins with Gap 1 phase (E), where the cell grows and prepares for DNA synthesis.

During the Synthesis phase (C), DNA is replicated.

Gap 2 phase (A) follows, where the cell prepares for mitosis.

Karyokinesis (D) is the process of nuclear division.

Finally, Cytokinesis (B) is the division of the cytoplasm, completing cell division.
Quick Tip: Remember the sequence of the cell cycle: G1 → S → G2 → Mitosis (karyokinesis) → Cytokinesis.

Matching each term with the correct disease or cause.

Common cold is caused by rhinoviruses, so A-III.

Haemozoin is a waste product of the malaria parasite, Plasmodium, hence B-I.

Widal test is used for the diagnosis of typhoid, so C-II.

Allergy can be caused by dust mites, which are common allergens, making D-IV.
Quick Tip: Understanding the causes of common diseases and their diagnostic tests can greatly help in answering medical science questions.

Analyzing the anatomy of cockroaches.

In cockroaches, anal cerci are located at the posterior end of the body, typically on the 10th segment. These cerci are sensory structures that help the cockroach detect environmental changes, particularly air currents. Quick Tip: Understanding insect anatomy, especially sensory organs like cerci, is useful in identifying their roles in behavior and physiology.

Understanding the structure and functions.

Axoneme (A) is a structure found in cilia and flagella that supports their movement, hence A-II.

Cartwheel pattern (B) is a characteristic arrangement of microtubules found in centrioles, hence B-I.

Crista (C) refers to the folds inside the mitochondria, where ATP is synthesized, hence C-IV.

Satellite (D) is associated with the chromosomes, particularly the small bodies found around them, hence D-III.
Quick Tip: Understanding the cellular structures and their functions will help in identifying their respective components during biological analysis.

Step 1: Identifying steroid hormones.

Steroid hormones are derived from cholesterol and include hormones like progesterone, cortisol, and testosterone.


Step 2: Identifying the exception.

Glucagon is a peptide hormone, not a steroid hormone, as it is derived from a protein and not cholesterol.
Quick Tip: Steroid hormones are lipid-soluble and derived from cholesterol, while peptide hormones like glucagon are water-soluble.

Step 1: Analyzing Statement I.

Statement I says that the descending limb of the loop of Henle is impermeable to water and permeable to electrolytes. This is false because the descending limb is permeable to water but impermeable to electrolytes. The primary function of the descending limb is to allow water to be reabsorbed into the bloodstream.

Step 2: Analyzing Statement II.

Statement II mentions that the proximal convoluted tubule (PCT) is lined by simple columnar brush border epithelium and increases the surface area for reabsorption. This statement is also false. The proximal convoluted tubule is lined by simple cuboidal epithelium with a brush border of microvilli that increase the surface area for reabsorption. However, it is not columnar epithelium.

Thus, both statements are false.

Therefore, the correct answer is option (4). Quick Tip: - The descending limb of the loop of Henle is permeable to water and impermeable to electrolytes, while the ascending limb is impermeable to water but permeable to electrolytes. - The proximal convoluted tubule is lined by simple cuboidal epithelium with a brush border to increase surface area for reabsorption.

Step 1: Understanding the transcription process.

DNA-dependent RNA polymerase synthesizes RNA using a DNA template. The RNA sequence is complementary to the DNA strand, except that uracil (U) replaces thymine (T).

Step 2: Determining the complementary RNA sequence.

The given DNA template is:

3’TACATGGCAAATATCCATTCA5’

The complementary RNA sequence will be:

5’AUGUACCGUUUAUAGGUAAGU3’

This matches option (3). Quick Tip: In transcription, remember that thymine (T) in DNA pairs with adenine (A) in RNA, and cytosine (C) pairs with guanine (G), while uracil (U) replaces thymine in RNA.

Step 1: Understanding the respiratory volumes and capacities.

Expiratory capacity (A) is the total volume of air that can be exhaled after a normal inhalation. This includes tidal volume and expiratory reserve volume, so A-II.

Functional residual capacity (B) is the volume of air remaining in the lungs after normal exhalation. It is the sum of expiratory reserve volume and residual volume, hence B-IV.

Vital capacity (C) is the total volume of air that can be exhaled after a maximum inhalation, which is the sum of tidal volume, inspiratory reserve volume, and expiratory reserve volume, so C-I.

Inspiratory capacity (D) is the maximum amount of air that can be inhaled after a normal exhalation, which is the sum of tidal volume and inspiratory reserve volume, hence D-III.
Quick Tip: Familiarize yourself with the different lung volumes and capacities, as they are crucial for understanding respiratory mechanics.

Understanding the conduction pathway in the heart.

The action potential starts at the SA node (E), which initiates the electrical impulse.

The impulse then travels to the AV node (C), where it is briefly delayed.

From there, it passes to the AV bundle (A) and moves down to the bundle branches (D).

Finally, it reaches the Purkinje fibres (B), which spread the impulse to the ventricles. Quick Tip: Remember the flow of the heart’s electrical impulse: SA node → AV node → AV bundle → bundle branches → Purkinje fibres.

Matching the epithelium types with their examples.

Unicellular glandular epithelium (A) refers to individual secretory cells such as goblet cells, hence A-III.

Compound epithelium (B) is found on surfaces that need protection from wear and tear, such as the buccal cavity, hence B-IV.

Multicellular glandular epithelium (C) can be found in structures like salivary glands, hence C-I.

Endocrine glandular epithelium (D) is found in glands like the pancreas that have internal secretions, hence D-II. Quick Tip: Each type of epithelium has specific characteristics and functions; knowing these can help identify where they are found in the body.

Analyzing hormone functions and cell types involved in spermatogenesis.

FSH (Follicle Stimulating Hormone) directly stimulates the Sertoli cells, which are crucial for nurturing the sperm cells during spermatogenesis.

LH (Luteinizing Hormone) acts primarily on Leydig cells, which are responsible for producing androgens (testosterone) necessary for the final stages of spermatogenesis.


Step 2: Understanding the terms spermatogenesis and spermiogenesis.

Spermatogenesis refers to the entire process of sperm production from spermatogonial stem cells.

Spermiogenesis refers specifically to the final stage of spermatogenesis, where spermatids are transformed into mature spermatozoa.


Step 3: Mapping the correct flow of hormone action and cell response.

Given the description and the process diagram, the sequence should correctly associate FSH with Leydig cells, and LH's influence through androgens on Sertoli cells, leading to spermiogenesis, making option (3) the correct sequence.
Quick Tip: Understanding the hormonal control and stages of spermatogenesis is crucial for comprehending male reproductive physiology.

Ordering the enzymatic reaction steps.

The enzyme first binds the substrate at the active site (E).

This binding leads to the formation of a substrate-enzyme complex (A).

Chemical bonds in the substrate are then altered or broken (D).

This leads to the release of products from the enzyme (C).

Finally, the enzyme is freed up to bind to another substrate (B).
Quick Tip: Understanding the enzyme catalytic cycle is crucial for studying biochemical reactions and how enzymes facilitate these processes.

Understanding the electrocardiogram (ECG) components.

The P wave (A) represents atrial depolarization, leading to atrial contraction, hence A-III.

The QRS complex (B) represents the rapid depolarization of the ventricles, crucial for ventricular contraction, hence B-II.

The T wave (C) represents the repolarization of the ventricles, resetting the ventricular cells to their resting state, hence C-IV.

The T-P gap (D) represents the phase where the heart muscles are electrically silent between heartbeats, hence D-I.
Quick Tip: Each wave and segment of an ECG represents a specific electrical activity in the heart, crucial for understanding heart function and diagnosing conditions.

Step 1: Verifying Statement I.

Statement I is correct as both mitochondria and chloroplasts are bound by double membranes.

Step 2: Verifying Statement II.

Statement II is incorrect; it is the inner membrane of chloroplasts that is less specialized in ion transport compared to the highly impermeable inner membrane of mitochondria, which is involved in the electron transport chain and ATP synthesis. Quick Tip: Understanding the structure and function of cell organelles helps in comprehending cellular energy production mechanisms.

Matching the molecular biology terms with their functions.
RNA polymerase III (A) is responsible for transcribing snRNAs and tRNA, hence A-IV.

Termination of transcription (B) often involves the Rho factor, which helps in stopping the transcription process at specific sites, hence B-III.

Splicing of Exons (C) involves snRNPs, which are part of the spliceosome complex that removes introns from pre-mRNA, hence C-I.

TATA box (D) is a crucial component of the promoter region of genes, helping in the initiation of transcription, hence D-II.
Quick Tip: Understanding the roles of various components in transcription and post-transcriptional modifications is essential for grasping gene expression regulation.

Step 1: Analyzing Statement I.

Statement I is correct; the corpus callosum is indeed the nerve tract that connects the two cerebral hemispheres, facilitating interhemispheric communication.

Step 2: Analyzing Statement II.

Statement II is incorrect because the brain stem consists of the medulla oblongata, pons, and midbrain, not the cerebrum. Quick Tip: Understanding the structural organization of the brain helps in accurately identifying its components and their functions.

Step 1: Analyzing Statement I.

Statement I is correct as bone marrow is indeed a primary lymphoid organ responsible for the production of all types of blood cells, including lymphocytes.

Step 2: Analyzing Statement II.

Statement II is correct. The bone marrow is the site where all T-lymphocytes are originally produced, and the thymus is where T-lymphocytes mature and differentiate, thus both providing critical environments for T-cell development. Quick Tip: The bone marrow and thymus are key to the lymphatic and immune systems, playing crucial roles in the production and maturation of lymphocytes.

Assign the correct function to each organ based on its anatomical and physiological role.

A. The Crop in cockroaches is used for the storage of food, so A matches with IV.

B. Gastric Caeca are the ring of 6-8 blind tubules at the junction of the foregut and midgut, serving to increase the surface area for enzyme secretion and absorption, so B matches with II.

C. Malpighian tubules, a ring of 100-150 yellow coloured thin filaments, are involved in excretion and osmoregulation at the junction of the midgut and hindgut, so C matches with III.

D. The Gizzard is equipped with chitinous teeth and helps in grinding the food, so D matches with I.
Quick Tip: Understanding the anatomical structure of the digestive system in insects like cockroaches can provide insights into their efficient nutrient absorption and waste management strategies, which are crucial for their survival in diverse environments.

Understanding Gause's competitive exclusion principle.

Statement I is incorrect because Gause's principle states that two species competing for the same limited resources cannot coexist at constant population values, not different resources.

Statement II correctly summarizes the competitive exclusion principle where one species will outcompete the other when resources are limited. Quick Tip: Gause's principle is a foundational concept in ecology that explains how competition influences species diversity and population dynamics.

Understanding nephron structure.

The juxta medullary nephrons have their Loop of Henle extending deep into the renal medulla, which is crucial for concentrating urine. Quick Tip: Juxta medullary nephrons are essential for producing concentrated urine, a key adaptation in mammals for water conservation.

Analyzing non-chordate characteristics.

Notochord is absent (B) is a defining trait of non-chordates.

Heart is dorsal if present (D) and Post anal tail is absent (E) are also characteristics typical of many non-chordates, unlike chordates that typically have a ventral heart and a post-anal tail. Quick Tip: Remember, non-chordates lack a notochord at any stage of their life, which is a fundamental difference from chordates.

Understanding geological eras and their dominant life forms.
Mesozoic Era (A) is known as the age of reptiles, including dinosaurs, hence A-III.

Proterozoic Era (B) predates most complex life and is characterized by lower forms of life, such as single-celled organisms, hence B-I.

Cenozoic Era (C) is known as the age of mammals, hence C-IV.

Paleozoic Era (D) saw the rise of fish and early amphibians, hence D-II.
Quick Tip: Linking geological eras with their characteristic life forms can aid in understanding evolutionary history.

Matching diseases with their symptoms.

Exophthalmic goiter (A) is related to hyperthyroidism and features like protruding eyeballs, hence A-III.

Acromegaly (B) is due to excessive growth hormone, leading to enlarged features, hence B-IV.

Cushing’s syndrome (C) results from excess cortisol, associated with symptoms like moon face and hyperglycemia, hence C-I.

Cretinism (D) involves hypo-secretion of thyroid hormones, causing stunted growth and developmental delays, hence D-II. Quick Tip: Identifying endocrine disorders requires an understanding of hormone function and the physical manifestations of hormone imbalances.

Step 1: Understand the inheritance of the ABO blood group.

The ABO blood group system is based on the presence of antigens and antibodies. The alleles \(I^A\) and \(I^B\) are dominant over \(i\), which does not produce any antigen. Therefore, the presence of \(i i\) genotype results in type O blood, which lacks A and B antigens.

Step 2: Determine the potential genotypes of the parents.

For a child to be of type O blood group (\(ii\)), each parent must contribute an \(i\) allele. This means:

The father with blood group B+ can either be \(I^B I^B\) or \(I^B i\). However, the child having \(ii\) mandates the father's genotype must include \(i\), thus \(I^B i\).

The mother with blood group A+ can either be \(I^A I^A\) or \(I^A i\). Similar to the father, for the child to be \(ii\), the mother must also be heterozygous, thus \(I^A i\).


Step 3: Review the child’s genotype and match the options.

The child's genotype is \(ii\), clearly indicating both parents have contributed an \(i\) allele.


Step 4: Assess each option against the genotypes required.

Option A (\(I^B i/I^A i/ii\)) fits perfectly as it allows both parents to carry and pass the \(i\) allele to the child, resulting in blood group O+.
Options B, C, D, and E do not consistently allow the transmission of an \(i\) allele from each parent to result in \(ii\).
Quick Tip: When solving genetics problems involving blood types, always start by determining the possible genotypes that could result from the alleles provided by the parents.