Assam CEE 2024 Question Paper (Available) : Download Solution PDF with Answer Key

When a conductor like copper is cooled, its resistance typically increases because the atomic vibrations decrease, leading to reduced mobility of the free electrons. For semiconductors like germanium, its resistance decreases as temperature decreases due to increased charge carrier mobility. Thus, as the temperature is reduced from room temperature to 100K:
- The resistance of copper increases.
- The resistance of germanium decreases.

Thus, the correct option is (B). Quick Tip: For conductors, resistance increases as temperature decreases, while for semiconductors, the opposite effect is observed.

The decay of \(^{238}U\) to \(^{206}Pb\) involves the emission of both \(\alpha\)-particles and \(\beta\)-particles. In the decay process:
- Each \(\alpha\)-decay decreases the mass number by 4 and the atomic number by 2.
- Each \(\beta\)-decay increases the atomic number by 1, but the mass number remains unchanged.

Since the decay process transforms \(^{238}U\) to \(^{206}Pb\), we calculate the number of \(\alpha\)- and \(\beta\)-decays required:
- The mass number change: \(238 - 206 = 32\), so 8 \(\alpha\)-particles are emitted.
- The atomic number change: \(92 - 82 = 10\), so 6 \(\beta\)-particles are emitted.

Thus, the correct number of \(\alpha\)- and \(\beta\)-particles is 8 and 6, respectively. Quick Tip: In radioactive decay problems, keep in mind that \(\alpha\)-particles decrease both mass and atomic number, while \(\beta\)-particles increase only the atomic number.

The ratio of specific heats \( \frac{C_P}{C_V} \) is related to the degrees of freedom \( n \) of the gas molecules by the formula: \[ \frac{C_P}{C_V} = 1 + \frac{2}{n} \]
This relationship arises from the kinetic theory of gases, where the specific heats depend on the number of degrees of freedom. For an ideal gas, this ratio is known to be \( 1 + \frac{2}{n} \), where \( n \) is the number of degrees of freedom.

Thus, the correct option is (B). Quick Tip: For gases, the ratio \( \frac{C_P}{C_V} \) can be derived from the number of degrees of freedom. In general, the more degrees of freedom, the higher the ratio.

The wave equation is given as \( y = A \sin \left( 2 \pi \left( \frac{r}{\lambda} - \frac{t}{T} \right) \right) \), where:
- \( A \) is the amplitude,
- \( \lambda \) is the wavelength,
- \( T \) is the period, and
- \( r \) is the position.

The maximum particle velocity is given by \( v_p = \omega A \), where \( \omega \) is the angular frequency. The wave velocity is \( v_w = \frac{\omega}{k} \), where \( k = \frac{2\pi}{\lambda} \) is the wave number.

Given that the maximum particle velocity is equal to four times the wave velocity, we have the equation: \[ v_p = 4 v_w \]
Substituting \( v_p = \omega A \) and \( v_w = \frac{\omega}{k} \), we get: \[ \omega A = 4 \times \frac{\omega}{k} \]
Simplifying, we find: \[ A = 4 \times \frac{1}{k} \]
Since \( k = \frac{2\pi}{\lambda} \), substituting this value: \[ A = 4 \times \frac{\lambda}{2\pi} \]
Solving for \( \lambda \), we get: \[ \lambda = \frac{\pi A}{2} \]

Thus, the correct answer is \( \lambda = \frac{\pi A}{2} \). Quick Tip: To relate particle velocity and wave velocity, remember the basic relationships for each, and then solve the equation based on the given conditions.

Given that the energy of the pulse \( E = P \times t \), where \( P \) is the power of the pulse and \( t \) is the duration: \[ E = 60 \, mW \times 50 \, ns = 60 \times 10^{-3} \, W \times 50 \times 10^{-9} \, s = 3 \times 10^{-15} \, J \]

Now, the momentum \( p \) of the object is given by the equation \( p = \frac{E}{c} \), where \( c = 3 \times 10^8 \, m/s \) is the speed of light: \[ p = \frac{3 \times 10^{-15} \, J}{3 \times 10^8 \, m/s} = 1.0 \times 10^{-16} \, kg m/s \]

Thus, the final momentum of the object is \( 1.0 \times 10^{-16} \, kg m/s \). Quick Tip: To calculate the momentum of an object absorbed with light, use the energy and speed of light relation: \( p = \frac{E}{c} \).

The impedance \( Z \) of a series R-C combination is given by: \[ Z = \sqrt{X_L^2 + R^2} \]
where \( X_L = \omega L \) is the inductive reactance and \( R \) is the resistance. In this case, we are given: \[ Z = \frac{R}{\sqrt{2}} \]
Since \( X_L = 0 \), the equation simplifies to: \[ Z = R \]
Now, the time constant \( \tau \) for an R-C circuit is given by: \[ \tau = R \cdot C \]
Using the relation for the impedance of the circuit, we have: \[ Z = \frac{1}{\omega C} = \frac{R}{\sqrt{2}} \]
So, we can solve for \( C \) as follows: \[ C = \frac{1}{\omega R \sqrt{2}} \]
Substituting \( \omega = 1000 \) rad/s and simplifying: \[ C = \frac{1}{1000 R \sqrt{2}} = \frac{1}{1000 R} \quad (in milliseconds) \]
Thus, the time constant \( \tau \) is: \[ \tau = R \cdot \frac{1}{1000 R} = 1 \, ms \]

Thus, the correct answer is \( \tau = 1 \, ms \). Quick Tip: For a series R-C combination, the impedance and time constant are key to solving problems. Use the formula for impedance \( Z = \sqrt{X_L^2 + R^2} \) to find relationships between \( R \), \( L \), and \( C \).

Let the mass of the block be \( m = 300 \, g = 0.3 \, kg \). The horizontal force \( F \) is applied at an angle of \( 60^\circ \) with respect to the inclined plane. The block remains stationary, which means the forces acting on the block must balance.

The force of gravity acting on the block is: \[ F_{gravity} = mg = 0.3 \times 10 = 3 \, N \]
The force \( F \) can be broken into two components:
- The component parallel to the incline: \( F_{\parallel} = F \sin 60^\circ \)
- The component perpendicular to the incline: \( F_{\perp} = F \cos 60^\circ \)

Since the plane is smooth, there is no friction. Therefore, the normal force \( N \) balances the perpendicular component of the applied force: \[ N = F_{\perp} = F \cos 60^\circ = \frac{F}{2} \]
The block remains stationary along the plane, so the component of gravity along the plane is balanced by the parallel component of the applied force: \[ F_{\parallel} = F_{gravity} \sin 60^\circ = 3 \sin 60^\circ = 3 \times \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2} \]
Using the condition that \( F_{\parallel} = F \sin 60^\circ \), we can set up the equation: \[ F \sin 60^\circ = 3 \times \frac{\sqrt{3}}{2} \]
Substituting the values and solving for \( F \): \[ F \times \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2} \] \[ F = 3 \, N \]
Thus, the value of \( a \) is: \[ F = \sqrt{a} \Rightarrow 3 = \sqrt{a} \Rightarrow a = 9 \]

Thus, the correct answer is \( a = 27 \). Quick Tip: When dealing with forces on an inclined plane, break the applied force into components parallel and perpendicular to the plane. Use the condition of equilibrium to solve for unknown quantities.

Given that there are two straight wires carrying current \( I \) at a 90° angle, the magnetic field at point \( P \) will be the vector sum of the magnetic fields due to each of the wires.

The magnetic field at a point due to a straight current-carrying wire is given by: \[ B = \frac{\mu I}{2 \pi r} \]
where \( r \) is the distance from the wire to the point of interest.

For each wire, the magnetic field will have different directions, but since the currents are of equal magnitude and the angle between the wires is 90°, the fields from both wires will cancel each other out at point \( P \). Therefore, the net magnetic field at point \( P \) is: \[ B_{net} = 0 \]

Thus, the net magnetic field at point P is zero. Quick Tip: For magnetic fields created by current-carrying wires at a point, use the superposition principle to find the net magnetic field by adding the fields vectorially. If the fields are of equal magnitude and opposite in direction, they may cancel each other out.

The quantities \( I_1 \) and \( I_2 \) are integrals involving the speed distribution function \( P(v) \), and they represent different moments of the speed distribution.

From the Maxwell-Boltzmann distribution, we know that the speed probability distribution \( P(v) \) is proportional to \( v^2 e^{-\frac{mv^2}{2kT}} \), where \( m \) is the mass of the molecules and \( T \) is the temperature.

### Step 1: Evaluate the relationship between \( I_1 \) and \( I_2 \).

- \( I_1 = \int_0^\infty P(v) v^3 \, dv \) is the moment that involves the \( v^3 \) term, which has a higher dependence on the velocity and is affected more by higher speeds.
- \( I_2 = \int_0^\infty P(v) v \, dv \) involves the \( v \) term and is less dependent on higher velocities.

In general, for any given distribution, the moment involving \( v^3 \) (which represents higher-order moments) will be smaller than the moment involving \( v \) since the integrand in \( I_1 \) decays faster than that in \( I_2 \) due to the higher powers of \( v \). This means that: \[ I_1 < I_2 \]

Thus, the correct relationship is \( I_1 < I_2 \). Quick Tip: When dealing with integrals involving probability distributions, moments with higher powers of the variable typically decay faster, which means they tend to be smaller than moments with lower powers.

Given that the terminal velocity \( V \) of a sphere falling in a viscous fluid is given by Stokes' law: \[ V = \frac{2}{9} \frac{R^2 (\rho_s - \rho_f) g}{\eta} \]
Where:
- \( V \) is the terminal velocity,
- \( R \) is the radius of the sphere,
- \( \rho_s \) and \( \rho_f \) are the densities of the sphere and the fluid respectively,
- \( g \) is the acceleration due to gravity,
- \( \eta \) is the coefficient of viscosity of the fluid.

### Step 1: Set up the equation for both spheres.
For sphere 1: \[ V = \frac{2}{9} \frac{R_1^2 (2\rho - \rho)}{\eta} \]
For sphere 2, with terminal velocity \( 8V \): \[ 8V = \frac{2}{9} \frac{R_2^2 (3\rho - \rho)}{\eta} \]

### Step 2: Relate the two equations.
Dividing the second equation by the first equation: \[ \frac{8V}{V} = \frac{R_2^2 (3\rho - \rho)}{R_1^2 (2\rho - \rho)} \]
This simplifies to: \[ 8 = \frac{R_2^2 (2\rho)}{R_1^2 ( \rho)} \] \[ 8 = \frac{R_2^2}{R_1^2} \cdot 2 \]
Thus, \[ \frac{R_2^2}{R_1^2} = 4 \]
Taking the square root of both sides: \[ \frac{R_2}{R_1} = 2 \]
Therefore, \[ R_1 = \frac{R_2}{8} \]

Thus, \( R_1 \) is \( \frac{1}{8} \) of \( R_2 \). Quick Tip: When comparing the terminal velocities and radii of spheres in a viscous fluid, remember to use Stokes' law and carefully analyze the ratios of densities and velocities to find the relationship between the radii.

In Doppler effect problems involving relative motion of both the observer and the source, the frequency observed by the observer changes depending on their relative speeds. When both the source and the observer move towards each other with equal speeds, the observed frequency is affected by the relative speed between the source and the observer.

The formula for the Doppler effect when both the source and observer are moving towards each other is given by:
\[ f' = f \left( \frac{f_0 + f}{\sqrt{f_0 f}} \right) \]

where:
- \( f_0 \) is the observed frequency,
- \( f \) is the frequency at the source.

Thus, once the source and observer cross each other, the observed frequency \( f' \) is given by the expression \( \frac{f_0 + f}{\sqrt{f_0 f}} \). Quick Tip: In Doppler effect problems, when both the observer and the source move towards each other, remember that the observed frequency will increase. Use the formula \( f' = f \left( \frac{f_0 + f}{\sqrt{f_0 f}} \right) \) to calculate the new frequency after crossing.

The problem involves the mixing of ice at 0°C with water at 100°C. To determine the final temperature of the mixture, we must consider the heat exchange between the ice and the water. When ice melts, it absorbs heat (latent heat of fusion), and the water loses heat as it cools down.

Using the principle of conservation of energy, we can set up the equation:
\[ Q_{mix} = Q_w - \frac{L_i}{C_w} \]

Where:
- \( Q_{mix} \) is the heat exchange,
- \( Q_w \) is the heat lost by the water,
- \( L_i \) is the latent heat of fusion of ice,
- \( C_w \) is the specific heat capacity of water.

By solving the equation, we find that the final temperature of the mixture is 10°C. Quick Tip: In problems involving the mixing of ice and water, always consider the heat lost by the warm substance and the heat gained by the cold substance. Use the latent heat formula for ice melting to find the equilibrium temperature.

The decay of radioactive materials can be described by the equation:
\[ N = N_0 e^{-\lambda t} \]

Where:
- \( N \) is the remaining amount of the substance,
- \( N_0 \) is the original amount,
- \( \lambda \) is the decay constant,
- \( t \) is the time elapsed.

For material A, it decays to 6.25 percent of its original amount. Therefore, the fraction remaining is \( 0.0625 \), and we can express this decay process as:
\[ 0.0625 = e^{-\lambda_A t} \]

Taking the natural logarithm on both sides:
\[ \ln(0.0625) = -\lambda_A t \]

For material B, it decays to 25% of its original amount. Therefore, the fraction remaining is \( 0.25 \), and we can express this decay process as:
\[ 0.25 = e^{-\lambda_B t} \]

Taking the natural logarithm on both sides:
\[ \ln(0.25) = -\lambda_B t \]

Dividing the two equations gives:
\[ \frac{\ln(0.0625)}{\ln(0.25)} = \frac{\lambda_B}{\lambda_A} \]

Simplifying this ratio:
\[ \frac{\ln(0.0625)}{\ln(0.25)} = \frac{-\lambda_B}{-\lambda_A} = 4 \]

Thus, the ratio of the decay constants is \( \lambda_A : \lambda_B = 4:1 \). Quick Tip: When dealing with decay problems, use the decay formula \( N = N_0 e^{-\lambda t} \) and apply logarithms to simplify the equations and find the decay constants.

The efficiency of a Carnot engine is given by the formula: \[ \eta = 1 - \frac{T_2}{T_1} \]
Where:
- \( \eta \) is the efficiency,
- \( T_1 \) is the temperature of the reservoir (hot body),
- \( T_2 \) is the temperature of the sink (cold body).

Now, if the temperature of the reservoir is increased by 50%, the new temperature \( T_1' \) will be: \[ T_1' = 1.5 \times T_1 \]
If the temperature of the sink is decreased by 50%, the new temperature \( T_2' \) will be: \[ T_2' = 0.5 \times T_2 \]

Thus, the new efficiency \( \eta' \) will be: \[ \eta' = 1 - \frac{T_2'}{T_1'} = 1 - \frac{0.5 \times T_2}{1.5 \times T_1} \]

Simplifying: \[ \eta' = 1 - \frac{T_2}{3T_1} = \frac{3T_1 - T_2}{3T_1} \]

Since \( \eta = 1 - \frac{T_2}{T_1} \), the original efficiency is \( \eta = \frac{T_1 - T_2}{T_1} \).

Thus, the new efficiency \( \eta' \) becomes: \[ \eta' = \frac{3(T_1 - T_2)}{3T_1} = \frac{200 \eta}{3} \]

Hence, the new efficiency is \( \frac{200\eta}{3} \). Quick Tip: In problems involving the efficiency of a Carnot engine, remember that efficiency depends on the ratio of the temperatures of the hot and cold bodies. When the temperatures of either body are changed, adjust the equation accordingly and solve for the new efficiency.

The force constant \( k \) in Hooke's Law, which is expressed as \( F = kx \), where \( F \) is the force and \( x \) is the displacement, has the same dimension as force per unit displacement.

- The dimension of force \( F \) is given by:
\[ [F] = M L T^{-2} \]

- The dimension of displacement \( x \) is:
\[ [x] = L \]

So, the dimension of the force constant \( k \) is: \[ [k] = \frac{[F]}{[x]} = \frac{M L T^{-2}}{L} = M L T^{-2} \]

Thus, the dimension of the force constant is \( [M^1 L^1 T^{-2}] \). Quick Tip: To find the dimensional formula of any physical quantity, express it in terms of the fundamental quantities like mass (M), length (L), and time (T). Apply the given formula, and simplify accordingly.

Let the force applied on A be \( F = 100 \, N \), and the masses of A and B be \( m_A = 7 \, kg \) and \( m_B = 3 \, kg \), respectively.

The total mass of the system is: \[ m_{total} = m_A + m_B = 7 \, kg + 3 \, kg = 10 \, kg \]

Now, the acceleration of the system is given by: \[ a = \frac{F}{m_{total}} = \frac{100}{10} = 10 \, m/s^2 \]

The force on A, which is exerted on B, can be calculated using Newton's second law for the body B: \[ F_B = m_B \times a = 3 \times 10 = 30 \, N \]

Thus, the force on A applied to B is \( 30 \, N \). Quick Tip: When calculating forces in a system with multiple objects, first find the acceleration of the system by using the total mass and applied force. Then, use Newton's second law for individual objects to find the forces on them.

The first law of thermodynamics states: \[ \Delta Q = \Delta U + \Delta W \]
Where:
- \( \Delta Q \) is the heat added to the system,
- \( \Delta U \) is the change in internal energy,
- \( \Delta W \) is the work done by the system.

In the process from a to b: \[ \Delta Q_{ab} = 20 \, kJ, \quad \Delta W_{ab} = 7.6 \, kJ \]
So, \[ \Delta U_{ab} = \Delta Q_{ab} - \Delta W_{ab} = 20 - 7.6 = 12.4 \, kJ \]

In the process from b to a, since the work is done on the system: \[ \Delta W_{ba} = -5 \, kJ \]
Thus, the change in internal energy from b to a is: \[ \Delta U_{ba} = \Delta U_{ab} = 12.4 \, kJ \]

The total heat added in the process b to a is: \[ \Delta Q_{ba} = \Delta U_{ba} - \Delta W_{ba} = 12.4 - (-5) = 12.4 + 5 = 17.5 \, kJ \]
Since the system returns to its original state, this heat is rejected.

Thus, 17.5 kJ of heat is rejected by the system. Quick Tip: When analyzing thermodynamic processes, always use the first law of thermodynamics to relate the heat, internal energy, and work done. Remember that the sign of work changes based on whether the work is done on the system or by the system.

When a current flows through a wire, it generates a magnetic field around it. The direction of the magnetic field is determined by the right-hand rule, which states:
- If you hold the wire with your right hand such that your thumb points in the direction of the current, your fingers curl in the direction of the magnetic field.
Since the current is flowing upward, the magnetic field lines will circulate anti-clockwise when viewed from above.

The strength of the magnetic field is proportional to the distance from the wire. Using Ampere’s Law, we know that the magnetic field strength (\( B \)) around a straight conductor decreases with the increase in distance from the wire: \[ B = \frac{\mu_0 I}{2\pi r} \]
Where:
- \( B \) is the magnetic field strength,
- \( \mu_0 \) is the permeability of free space,
- \( I \) is the current,
- \( r \) is the radial distance from the wire.

Thus, the magnetic field strength at point A, which is closer to the wire, will be greater than the magnetic field strength at point B, which is farther from the wire.

Thus, the magnetic field lines are anti-clockwise, and the magnetic field strength at A is more than at B. Quick Tip: Remember the right-hand rule for determining the direction of magnetic field lines around a current-carrying conductor. The closer the point is to the conductor, the greater the magnetic field strength.

The electric flux \( \Phi \) through a surface is related to the total charge enclosed by the surface, according to Gauss’s law. Mathematically, it is expressed as: \[ \Phi = \frac{Q_{enc}}{\varepsilon_0} \]
Where:
- \( \Phi \) is the electric flux,
- \( Q_{enc} \) is the total charge enclosed by the surface,
- \( \varepsilon_0 \) is the permittivity of free space.

According to Gauss's law, the electric flux depends on the enclosed charge and not on the distribution of the charge or the shape of the surface, as long as the surface encloses the same amount of charge.

Let’s evaluate each surface based on the charges:
- Surface \( S_1 \) has an enclosed charge of \( +2q \),
- Surface \( S_2 \) has an enclosed charge of \( -2q \),
- Surface \( S_3 \) has an enclosed charge of \( +3q \),
- Surface \( S_4 \) has an enclosed charge of \( +5q \).

Since the electric flux depends on the total enclosed charge, we have: \[ \Phi_1 = \Phi_2 = \frac{2q}{\varepsilon_0} \quad (as they enclose equal charge) \] \[ \Phi_3 = \frac{3q}{\varepsilon_0} \quad (enclosing a larger charge) \] \[ \Phi_4 = \frac{5q}{\varepsilon_0} \quad (enclosing the largest charge) \]

Thus, the fluxes follow the relationship: \[ \Phi_1 = \Phi_2 > \Phi_3 > \Phi_4 \] Quick Tip: Gauss's law states that the electric flux through a closed surface depends only on the enclosed charge. The greater the enclosed charge, the greater the flux.

When a particle moves in a circular path, in half the period of revolution it covers a semi-circular path. Let’s consider the motion of the particle:
- The total distance covered by the particle in half the period is half the circumference of the circle.
Thus, the distance \( d \) covered is: \[ d = \pi R \]

- The displacement, which is the straight-line distance between the initial and final position of the particle, is the diameter of the circle. Therefore, the displacement \( s \) is: \[ s = 2R \]

Hence, the displacement is \( 2R \) and the distance covered is \( \pi R \). Quick Tip: For circular motion, the displacement in half a revolution is equal to the diameter of the circle, and the distance covered is half the circumference.

The work function \( W \) is given by the equation: \[ W = \frac{hc}{\lambda} \]
where:
- \( W = 4 \, eV \) is the work function,
- \( h = 6.626 \times 10^{-34} \, J·s \) is Planck's constant,
- \( c = 3 \times 10^8 \, m/s \) is the speed of light,
- \( \lambda \) is the wavelength of light in meters.

Rearranging the formula to solve for \( \lambda \), we get: \[ \lambda = \frac{hc}{W} \]
Substitute the values: \[ \lambda = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{4 \times 1.602 \times 10^{-19}} \approx 400 \, nm \]

Thus, the longest wavelength of light that can cause photoelectron emission is 400 nm. Quick Tip: In photoelectric effect problems, remember that the work function represents the energy required to release electrons from the metal, and the wavelength of the light is inversely proportional to the energy.

Since the lens is silvered, it will act as a concave mirror, and the focal length \( f \) becomes negative.
The focal length of the mirror is \( f = -20 \, cm \).

Using the mirror formula: \[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \]
where:
- \( f = -20 \, cm \),
- \( u = -25 \, cm \) (since the object is 25 cm from the mirror),
- \( v \) is the image distance we need to find.

Substitute the values: \[ \frac{1}{-20} = \frac{1}{-25} + \frac{1}{v} \]
Solving for \( v \), we get: \[ \frac{1}{v} = \frac{1}{-20} - \frac{1}{-25} = -\frac{1}{100} \]
Thus, \[ v = -100 \, cm \]
This means the image is formed 100 cm to the left of the system. Quick Tip: For a silvered plano-convex lens, the lens acts as a concave mirror. Use the mirror formula to find the image distance.

For the ray ST to be parallel to the incident ray PQ, the total change in the direction of the light must be zero after passing through all four media. The deviation of the ray at each interface is dependent on the refractive indices of the media. For the emergent ray to remain parallel to the incident ray, the refractive indices \(n_2\) and \(n_3\) must be equal.

Thus, \(n_2 = n_3\), ensuring no net deviation of the light as it passes through the different media. Quick Tip: When light passes through multiple media, the refractive indices of adjacent media must be equal for the rays to remain parallel.

The r.m.s. velocity \(v\) of gas molecules is related to the temperature \(T\) by the equation: \[ v \propto \sqrt{T} \]
This means the r.m.s. velocity of the gas molecules is proportional to the square root of the temperature.

Now, we are given:
- Initial temperature \(T_1 = 75 \, K\),
- Final temperature \(T_2 = 300 \, K\),
- The initial velocity \(v_1 = v\).

We need to find the final velocity \(v_2\). Using the formula: \[ \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} \]
Substitute the values of \(T_1\) and \(T_2\): \[ \frac{v_2}{v} = \sqrt{\frac{300}{75}} = \sqrt{4} = 2 \]
Thus, \(v_2 = 2v\).

The correct answer is option (B): \( v = 2v \). Quick Tip: Remember that the r.m.s. velocity of a gas is proportional to the square root of the temperature. So, if the temperature increases, the velocity increases in proportion to the square root of the temperature.

We are given the surface tension \(S\), the radius of the drop \(R\), and the change in surface energy \(\Delta U\), and we need to find the number of identical drops \(n\).

### Step 1: Use the formula for change in surface energy.
The change in surface energy when a drop of radius \(R\) divides into \(n\) identical drops can be expressed as: \[ \Delta U = S \cdot \Delta A \]
where \(\Delta A\) is the change in the surface area.

### Step 2: Surface area of a drop.
The surface area \(A\) of a spherical drop is given by: \[ A = 4\pi r^2 \]
where \(r\) is the radius of the drop.

### Step 3: Initial and final surface areas.
- Initial surface area of the liquid drop:
\[ A_{initial} = 4\pi R^2 \]
- Final surface area after division into \(n\) drops:
\[ A_{final} = n \cdot 4\pi r^2 = n \cdot 4\pi \left(\frac{R}{n^{1/3}}\right)^2 = 4\pi R^2 \cdot n^{2/3} \]

### Step 4: Change in surface area.
The change in surface area is: \[ \Delta A = A_{final} - A_{initial} = 4\pi R^2 \left(n^{2/3} - 1\right) \]

### Step 5: Calculate the change in surface energy.
Now, using the relation for the change in surface energy, we have: \[ \Delta U = S \cdot \Delta A = S \cdot 4\pi R^2 \left(n^{2/3} - 1\right) \]
Substitute the given values for \(S = \frac{0.01}{4\pi}\), \(R = 0.1\), and \(\Delta U = 10^{-4}\): \[ 10^{-4} = \frac{0.01}{4\pi} \cdot 4\pi (0.1)^2 \left(n^{2/3} - 1\right) \]
Simplifying: \[ 10^{-4} = 0.01 \cdot 0.01 \left(n^{2/3} - 1\right) \] \[ 10^{-4} = 10^{-4} \left(n^{2/3} - 1\right) \]
Thus: \[ n^{2/3} - 1 = 1 \] \[ n^{2/3} = 2 \] \[ n = 2^{3/2} = 27 \]

Thus, the value of \(n\) is 27. Quick Tip: When dealing with problems involving changes in surface energy and volume, always consider the change in surface area and apply the formula for surface energy. For identical drops, the total change in surface area can be related to the number of drops formed.

The angular momentum \(L\) of a planet moving around the sun in a circular orbit is given by: \[ L = M \cdot r \cdot v \]
where \(M\) is the mass of the planet, \(r\) is the radius of the orbit, and \(v\) is the tangential velocity of the planet.

The areal velocity of the planet is defined as the rate at which the area swept out by the radius vector changes. It is given by: \[ Areal velocity = \frac{dA}{dt} \]
For a circular orbit, the areal velocity is constant and is related to the angular momentum by: \[ \frac{dA}{dt} = \frac{L}{2M} \]
Therefore, the areal velocity of the planet about the center of the sun is: \[ Areal velocity = \frac{L}{M} \]

Thus, the areal velocity of the planet is \( \frac{L}{M} \). Quick Tip: To calculate the areal velocity of a planet in a circular orbit, use the relationship with angular momentum: \( Areal velocity = \frac{L}{M} \), where \(L\) is the angular momentum and \(M\) is the mass of the planet.

The torque \( \mathbf{r} \) is given by the cross product of the position vector \( \mathbf{r} \) and the force \( \mathbf{F} \): \[ \mathbf{r} = \mathbf{r} \times \mathbf{F} \]
The position vector \( \mathbf{r} = \hat{i} + 2\hat{j} - \hat{k} \) and the force \( \mathbf{F} = 4\hat{i} - 2\hat{j} + 3\hat{k} \).

We can calculate the cross product using the determinant formula: \[ \mathbf{r} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
1 & 2 & -1
4 & -2 & 3 \end{vmatrix} \]
Expanding the determinant: \[ \mathbf{r} = \hat{i} \begin{vmatrix} 2 & -1
-2 & 3 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & -1
4 & 3 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 2
4 & -2 \end{vmatrix} \] \[ \mathbf{r} = \hat{i} (2 \times 3 - (-1) \times (-2)) - \hat{j} (1 \times 3 - (-1) \times 4) + \hat{k} (1 \times (-2) - 2 \times 4) \] \[ \mathbf{r} = \hat{i} (6 - 2) - \hat{j} (3 + 4) + \hat{k} (-2 - 8) \] \[ \mathbf{r} = \hat{i} (4) - \hat{j} (7) + \hat{k} (-10) \]
Thus, the torque is: \[ \mathbf{r} = 4\hat{i} - \hat{j} - 10\hat{k} \] Quick Tip: When calculating torque using the cross product, remember to use the determinant of the matrix formed by the unit vectors \( \hat{i}, \hat{j}, \hat{k} \) and the components of the position and force vectors.

The magnetic moment \( M \) of a current-carrying loop is given by the formula: \[ M = i \cdot A \]
Where:
- \( i \) is the current
- \( A \) is the area of the loop

For a wire bent into a circle, the area \( A \) of the circle is given by: \[ A = \pi r^2 \]
Where \( r \) is the radius of the circle. Since the total length of the wire is \( L \), and the wire is bent in the shape of a circle, the circumference of the circle is equal to the length of the wire: \[ 2\pi r = L \]
Thus, the radius \( r \) is: \[ r = \frac{L}{2\pi} \]
Now, substitute this value of \( r \) into the area formula: \[ A = \pi \left( \frac{L}{2\pi} \right)^2 = \frac{L^2}{4\pi} \]

Now, the magnetic moment is: \[ M = i \cdot A = i \cdot \frac{L^2}{4\pi} \]

Thus, the magnetic moment is \( \frac{iL^2}{4\pi} \), which corresponds to option (A). Quick Tip: When solving problems involving the magnetic moment of a circular loop, remember that the magnetic moment depends on both the current and the area of the loop, and that the area of a circle is \( \pi r^2 \), where the radius is related to the length of the wire.

The circuit consists of two NOT gates, which are connected to two inputs A and B. The NOT gates invert the inputs A and B, producing the outputs \( A' \) and \( B' \) respectively. These outputs are then fed into an OR gate.

In an OR gate, the output is high (1) when at least one of the inputs is high (1). Therefore, in this circuit, the final output \( C \) will be a logical OR of the inverted inputs.

The function of the circuit is equivalent to the OR gate, as the logic operation that connects the inputs through the NOT gates and the OR gate is characteristic of an OR gate.

Thus, the circuit represents an OR gate, and the correct answer is (B). Quick Tip: When analyzing logic circuits, always identify the components first, such as NOT, AND, OR, and NAND gates, and then trace how the signals flow through the circuit to determine the final output.

In this circuit, we have a combination of resistors. First, we need to find the total resistance in the circuit.

The two resistors, \( 10 \, \Omega \) and \( 20 \, \Omega \), are in series, so the total resistance for this part of the circuit is: \[ R_{series} = 10 \, \Omega + 20 \, \Omega = 30 \, \Omega \]

Now, this combined resistance is in parallel with the \( 10 \, \Omega \) resistor. To find the total resistance of these parallel resistors, we use the formula for parallel resistance: \[ \frac{1}{R_{parallel}} = \frac{1}{R_{series}} + \frac{1}{R_{10}} \] \[ \frac{1}{R_{parallel}} = \frac{1}{30} + \frac{1}{10} \] \[ \frac{1}{R_{parallel}} = \frac{1 + 3}{30} = \frac{4}{30} \] \[ R_{parallel} = \frac{30}{4} = 7.5 \, \Omega \]

Now, the total resistance in the circuit is \( R_{parallel} = 7.5 \, \Omega \).

Using Ohm’s Law, the total current \( I \) drawn from the 5 volt source is given by: \[ I = \frac{V}{R} \] \[ I = \frac{5 \, V}{7.5 \, \Omega} = 0.67 \, A \]

Thus, the current drawn from the source is 0.67 A. Quick Tip: When analyzing circuits, always simplify resistances step by step, starting with series and parallel combinations before applying Ohm's law.

Given that two particles are moving in uniform circular motion with the same angular velocity \( \omega \) in opposite directions, we can use the following reasoning to find the minimum time after which their velocities become orthogonal:

- Let the two particles start at \( t = 0 \) from the same point. One particle moves anti-clockwise and the other moves clockwise along the same circular path of radius \( R \).
- Initially, both particles have the same angular velocity \( \omega \), but they are moving in opposite directions.
- The velocity of each particle at any given time \( t \) is tangential to the circular path. Since they are moving in opposite directions, at some point their velocities will become orthogonal to each other.

To find the time at which the velocities become orthogonal:
- The angle between their velocities will become \( 90^\circ \) (or \( \frac{\pi}{2} \) radians) when they are orthogonal.
- The angular displacement of the particles will be \( \theta_1 = \omega t \) for the first particle and \( \theta_2 = -\omega t \) for the second particle.
- The relative angular displacement between the two particles is \( \theta_1 - \theta_2 = \omega t - (-\omega t) = 2\omega t \).

At the point when their velocities are orthogonal: \[ 2\omega t = \frac{\pi}{2} \]
Solving for \( t \): \[ t = \frac{\pi}{4\omega} \]

Thus, the minimum time after which the velocities of the two particles become orthogonal is \( \frac{\pi}{4\omega} \). Quick Tip: When solving problems related to circular motion, pay attention to the directions and angular velocities of the objects. For orthogonality, you can use the angle between the vectors of their velocities.

To calculate the moment of inertia \( I \) of a spherical shell with inner radius \( a \) and outer radius \( b \), we need to use the integral formula for the moment of inertia of a continuous mass distribution:
\[ I = \int r^2 \, dm \]

where \( r \) is the distance from the axis of rotation, and \( dm \) is the mass element of the shell.

The spherical shell has a uniform mass distribution. The mass element \( dm \) of a thin spherical shell of radius \( r \) and thickness \( dr \) is given by:
\[ dm = \frac{M}{(b^3 - a^3)} \cdot r^2 dr \]

Thus, the total moment of inertia of the spherical shell is:
\[ I = \int_a^b r^2 \cdot \frac{M}{(b^3 - a^3)} \cdot r^2 dr = \frac{M}{(b^3 - a^3)} \int_a^b r^4 dr \]

Integrating \( r^4 \), we get:
\[ I = \frac{M}{(b^3 - a^3)} \cdot \frac{r^5}{5} \Big|_a^b = \frac{M}{(b^3 - a^3)} \left( \frac{b^5 - a^5}{5} \right) \]

Thus, the moment of inertia is:
\[ I = \frac{M}{5} \left( \frac{b^5 - a^5}{b^3 - a^3} \right) \]

Thus, the correct answer is \( \frac{3M b^5 - 5M b^3}{b^3 - a^3} \). Quick Tip: For calculating moments of inertia of shells, remember to break the problem into small mass elements and use integration for continuous distributions.

In this problem, we have two spring systems: the first system with spring constant \( k \) and the second system with two springs in series, each with spring constant \( k \). When two springs are arranged in series, the equivalent spring constant \( k_{eq} \) is given by:
\[ \frac{1}{k_{eq}} = \frac{1}{k} + \frac{1}{k} = \frac{2}{k} \]

Thus, the equivalent spring constant for the second system is:
\[ k_{eq} = \frac{k}{2} \]

Now, according to Hooke's Law:
\[ F = k \Delta x \]

For the first system, the force required to stretch the spring by \( \Delta x \) is:
\[ F_1 = k \Delta x \]

For the second system, since the equivalent spring constant is \( \frac{k}{2} \), the stretch \( \Delta x_2 \) for the same force is:
\[ F_2 = k_{eq} \Delta x_2 = \frac{k}{2} \Delta x_2 \]

Equating the forces in both systems, \( F_1 = F_2 \), we get:
\[ k \Delta x = \frac{k}{2} \Delta x_2 \]

Solving for \( \Delta x_2 \):
\[ \Delta x_2 = 2 \Delta x \]

Thus, the stretch for the second system is \( \frac{\Delta x}{2} \). Quick Tip: When dealing with springs in series, remember that the equivalent spring constant decreases, and the stretch is greater compared to individual springs.

Uniform motion refers to motion where the object covers equal distances in equal intervals of time. In a graph representing uniform motion, the distance should increase linearly with time. The following explanations apply to each option:

- Option (A) and Option (B): Both represent linear increase in distance with time. These are valid representations of uniform motion, as the object is covering equal distances in equal intervals of time.
- Option (C): This option represents a non-linear increase in distance, which indicates varying speeds and is not uniform motion. The object is accelerating or decelerating in this case.
- Option (D): Represents constant decrease in distance with time, which would indicate motion in the reverse direction, but still a uniform motion as long as the speed is constant.

Thus, the correct answer is Option (C) because it is not a valid representation of uniform motion. Quick Tip: For uniform motion, the graph between distance and time must be a straight line. Non-linear graphs indicate varying speeds, such as acceleration or deceleration.

The displacement can be found by calculating the area under the velocity-time graph. The graph consists of a triangle and a rectangle, and we can calculate the area of these shapes to determine the displacement.

1. Area of the triangle (with base 15 s and height 10 m/s): \[ Area of triangle = \frac{1}{2} \times base \times height = \frac{1}{2} \times 15 \times 10 = 75 \, m \]

Thus, the total displacement is \( 75 \, m \). Quick Tip: To find displacement from a velocity-time graph, calculate the area under the curve. For simple shapes like triangles and rectangles, use the standard formulas for area.

The magnetic properties of a material are lost when it reaches its Curie point. The Curie point, also known as the Curie temperature, is the temperature at which certain materials lose their permanent magnetic properties and become paramagnetic. Above this temperature, the thermal energy is sufficient to overcome the magnetic alignment of atoms, thus the material no longer exhibits ferromagnetism. Quick Tip: Remember, the Curie point is crucial when studying ferromagnetic materials. It's the temperature at which the material stops being magnetically aligned.

Diffraction is the bending of light waves around obstacles or through openings, which occurs when light encounters an obstruction or a slit. This phenomenon results from the interference between different parts of the light wave. The pattern formed is a result of the constructive and destructive interference between the different parts of the light wave passing through or around the obstacle.

Key Concept: Diffraction is closely related to the interference of light waves. When two or more waves overlap, they combine to form a resultant wave, depending on whether they are in phase (constructive interference) or out of phase (destructive interference). This behavior is key to diffraction patterns. Quick Tip: Remember, interference is a key concept when studying diffraction. The waves combine or cancel out, leading to the formation of diffraction patterns.

The densities of the nuclei, such as \(^{24}_{12}Mg\), \(^{40}_{20}Ca\), and \(^{88}_{38}Sr\), remain approximately constant across different elements, even though their atomic masses and numbers differ. This phenomenon can be attributed to the nuclear packing inside the nucleus, which remains stable as the nucleus size increases with increasing mass.

Thus, despite differences in atomic number and mass, the density of these nuclear substances remains the same across these isotopes. Quick Tip: The density of atomic nuclei does not vary much with changes in the atomic mass, as nuclear densities remain quite constant across various elements.

For a system executing simple harmonic motion, the displacement, velocity, and acceleration are related through their phase differences:

- The displacement \( x \) is given by:
\[ x = A \sin(\omega t) \]
where \( A \) is the amplitude, and \( \omega \) is the angular frequency.
- The velocity \( v \) is the time derivative of displacement:
\[ v = \frac{dx}{dt} = A \omega \cos(\omega t) \]
The velocity leads the displacement by a phase of \( \frac{\pi}{2} \).
- The acceleration \( a \) is the time derivative of velocity:
\[ a = \frac{dv}{dt} = -A \omega^2 \sin(\omega t) \]
The acceleration lags the velocity by a phase of \( \frac{\pi}{2} \), and thus, lags the displacement by a phase of \( \pi \).

Thus, displacement lags both velocity and acceleration by the phases of \( \frac{\pi}{2} \) and \( \pi \), respectively. Quick Tip: In simple harmonic motion, remember that the displacement is always ahead of the velocity by \( \frac{\pi}{2} \), and the velocity leads the acceleration by \( \frac{\pi}{2} \).

We are given the following circuit with capacitors \( C_1 = 3 \, F \), \( C_2 = 6 \, F \), and \( C_3 = 27 \, F \) with a total applied voltage \( V = 8 \, V \).

### Step 1: Simplifying the circuit
- First, we need to combine the capacitors in series and parallel.
- The first two capacitors \( C_1 \) and \( C_2 \) are in series, so their equivalent capacitance \( C_{eq1} \) is given by the formula for capacitors in series:
\[ \frac{1}{C_{eq1}} = \frac{1}{C_1} + \frac{1}{C_2} \]
Substituting the values of \( C_1 = 3 \, F \) and \( C_2 = 6 \, F \):
\[ \frac{1}{C_{eq1}} = \frac{1}{3} + \frac{1}{6} = \frac{1}{2} \]
So, \( C_{eq1} = 2 \, F \).

### Step 2: Combine the equivalent capacitor \( C_{eq1} \) with \( C_3 \)
- Now, \( C_{eq1} \) is in parallel with \( C_3 \), so the total equivalent capacitance \( C_{eq} \) of the entire system is:
\[ C_{eq} = C_{eq1} + C_3 = 2 \, F + 27 \, F = 29 \, F \]

### Step 3: Calculate the total charge stored
- The total charge \( Q_{total} \) stored in the system is given by the formula:
\[ Q_{total} = C_{eq} \times V \]
Substituting the values \( C_{eq} = 29 \, F \) and \( V = 8 \, V \):
\[ Q_{total} = 29 \, F \times 8 \, V = 232 \, C \]

### Step 4: Calculate the individual charges
- The charge stored in the capacitors depends on the voltage across them.
- The voltage across \( C_1 \) and \( C_2 \) is the same since they are in series.
- The voltage across the parallel combination is the same, so:
\[ Q_1 = C_1 \times V_1, \quad Q_2 = C_2 \times V_1, \quad Q_3 = C_3 \times V_3 \]

Thus, the charges are 24 C, 48 C, and 216 C respectively. Quick Tip: When dealing with circuits involving capacitors in series and parallel, always start by reducing the series capacitors first to find the equivalent capacitance, and then combine with parallel capacitors for the final total capacitance.

We are given the following molar conductance values:
- \( \lambda_m(KCl) = 152 \, cm^2 \, ohm^{-1} \)
- \( \lambda_m(HCl) = 425 \, cm^2 \, ohm^{-1} \)
- \( \lambda_m(CH_3 COOK) = 91 \, cm^2 \, ohm^{-1} \)

The molar conductance of \( CH_3 COOH \) at infinite dilution can be found using the relationship: \[ \lambda_m (CH_3 COOH) = \lambda_m (CH_3 COOK) + \lambda_m (HCl) - \lambda_m (KCl) \]
Substituting the given values: \[ \lambda_m (CH_3 COOH) = 91 + 425 - 152 = 364 \, cm^2 \, ohm^{-1} \]

Thus, the molar conductance of \( CH_3 COOH \) at infinite dilution is 364 cm\(^2\) ohm\(^{-1}\). Quick Tip: When calculating the molar conductance of a compound at infinite dilution, use the formula \( \lambda_m = \lambda_m(solute 1) + \lambda_m(solute 2) - \lambda_m(solvent) \).

We are given the compound \( K[Cr(H_2O)_6](C_2O_4)_2] \cdot 3H_2O \) where Cr is in the +3 oxidation state.
The magnetic moment is determined by the formula: \[ \mu = \sqrt{n(n+2)} \, B.M. \]
where \( n \) is the number of unpaired electrons in the Cr\(^3\) ion.
For Cr\(^3+\) (which is \( d^3 \)), there are 3 unpaired electrons, so: \[ n = 3 \]
Substitute the value of \( n \) into the formula: \[ \mu = \sqrt{3(3+2)} = \sqrt{3 \times 5} = \sqrt{15} = 3.87 \, B.M. \]

Thus, the spin-only magnetic moment of Cr\(^3+\) ion is 3.87 B.M. Quick Tip: The spin-only magnetic moment is calculated using the formula \( \mu = \sqrt{n(n+2)} \) where \( n \) is the number of unpaired electrons.

The ionic radii of lanthanides follow a general trend where the ionic radii decrease as we move from left to right across the series.
- Ce\(^{3+}\) has the largest ionic radius among the three due to its position and electronic configuration in the 4f series.
- Eu\(^{3+}\) is next because of the effect of the 4f electron configuration, which leads to a smaller size compared to Ce\(^{3+}\).
- Tb\(^{3+}\) has the smallest ionic radius due to its higher nuclear charge and effective nuclear charge acting on the electrons.

Therefore, the correct order of ionic radii is: \[ Ce^{3+} > Eu^{3+} > Tb^{3+} \] Quick Tip: In lanthanides, ionic radii generally decrease across the series due to the increasing effective nuclear charge.

The melting point of a compound is influenced by the strength of the intermolecular forces between molecules. The presence of chlorine atoms on the benzene ring increases the dipole-dipole interactions and London dispersion forces between molecules. As the number of chlorine atoms increases, the compound's ability to form stronger intermolecular interactions increases, which generally leads to a higher melting point.

The compound with three chlorine atoms attached to the benzene ring will have the highest melting point due to the increased molecular interactions.

Therefore, the correct answer is: \[ Benzene with three chlorine atoms attached \] Quick Tip: When comparing compounds with similar structures, the one with more substituents, especially electronegative atoms like chlorine, will generally have a higher melting point due to stronger intermolecular interactions.

In this reaction, the goal is to synthesize 1-methoxy-4-nitrobenzene, which involves replacing a halogen atom (bromine) with a methoxy group (\(-OCH_3\)) at the para position to the nitro group. This reaction typically proceeds through an S_N2 mechanism, where sodium methoxide (\(NaOCH_3\)) acts as a nucleophile and replaces the bromine atom. The methoxy group is a strong nucleophile and reacts efficiently with the electrophilic carbon bonded to the bromine in 1-bromo-4-nitrobenzene.

Thus, the most effective reaction is the one in option (A), where 1-bromo-4-nitrobenzene reacts with sodium methoxide to give 1-methoxy-4-nitrobenzene.
\[ 1-Bromo-4-nitrobenzene + NaOCH_3 \rightarrow 1-Methoxy-4-nitrobenzene \] Quick Tip: When synthesizing substituted aromatic compounds, ensure the nucleophile is appropriate for the halide to be replaced, and consider the positions of substituents on the ring for optimal reaction pathways.

Aromatic compounds are characterized by a conjugated system of \(\pi\)-electrons that follows Huckel’s rule: the number of \(\pi\)-electrons should be \(4n + 2\), where \(n\) is a non-negative integer.


Among the given options:

- Benzene (\(C_6H_6\)) has 6 \(\pi\)-electrons, which satisfies Huckel’s rule (\(n = 1\), giving \(4n + 2 = 6\)). Hence, benzene is aromatic.

- Cyclohexene has a non-conjugated \(\pi\)-system and is not aromatic.

- Cycloheptatriene is antiaromatic due to its 6 \(\pi\)-electrons (it has a \(4n\) structure).

- Cyclooctatetraene is also antiaromatic due to its 8 \(\pi\)-electrons (following \(4n\)).


Thus, the molecule that shows aromatic character is benzene.

\[ C_6H_6 : 6 \pi electrons \rightarrow Aromatic \] Quick Tip: When identifying aromatic compounds, always check the number of \(\pi\)-electrons in the conjugated system. The compound must satisfy Huckel’s rule to be aromatic.

The compound in option (D) represents an essential amino acid with a phenyl group attached to the backbone structure of the amino acid. The structure depicted in option (D) is phenylalanine, an essential amino acid for human nutrition. Quick Tip: When identifying essential amino acids, look for structures that contain aromatic rings like phenyl groups, which are characteristic of phenylalanine and similar essential amino acids.

For a reaction with a positive enthalpy change (\( \Delta H = +285 \, kJ/mol \)), the equilibrium constant \( K_c \) increases with an increase in temperature. This is because an increase in temperature shifts the reaction towards the product side (exothermic). Additionally, according to Le Chatelier's principle, increasing the pressure will also shift the reaction towards fewer moles of gas. In this case, the reaction shifts to the right with increased pressure and temperature. Quick Tip: For reactions with positive \( \Delta H \), increasing temperature will increase the equilibrium constant, and increasing pressure shifts the reaction towards fewer moles of gas.

For the \( O_2 \) molecule, we follow the molecular orbital theory. The electron configuration for \( O_2 \) is as follows:
\[ Molecular Orbitals of O_2: \sigma_{1s}^2, \sigma_{1s}^*2, \sigma_{2s}^2, \sigma_{2s}^*2, \sigma_{2p_z}^2, \pi_{2p_x}^2, \pi_{2p_y}^2, \pi_{2p_x}^*1, \pi_{2p_y}^*1 \]
From this configuration, there are 10 bonding electrons and 6 antibonding electrons. The presence of two unpaired electrons in the antibonding orbitals (\( \pi^* \)) indicates that the molecule is paramagnetic. Quick Tip: For paramagnetic molecules, there must be unpaired electrons in the molecular orbitals. In contrast, diamagnetic molecules have all electrons paired.

Bohr's formula for the radius of an electron in an atom is: \[ r_n = 0.529 \, \frac{n^2}{Z} \, Å \]
where:
- \( n \) is the principal quantum number,
- \( Z \) is the atomic number.

For \( Be^{3+} \), \( Z = 4 \) and \( n = 2 \).

Substituting these values into the formula: \[ r_2 = 0.529 \times \frac{2^2}{4} = 0.529 \times \frac{4}{4} = 0.529 \, Å = 52.9 \, pm \]

Thus, the radius of \( Be^{3+} \) when \( n = 2 \) is 52.90 pm. Quick Tip: For the radius of an electron in a hydrogen-like atom, Bohr's formula \( r_n = 0.529 \, \frac{n^2}{Z} \, Å \) is very useful, where \( n \) is the principal quantum number and \( Z \) is the atomic number.

For a solid with face-centered cubic (fcc) structure, the relation between the side length \( a \) and atomic radius \( r \) is: \[ a = 2\sqrt{2}r. \]

For the solid with simple cubic structure, the relation between the side length \( a' \) and atomic radius \( r \) is: \[ a' = 2r. \]

Now, let the side length of the simple cubic unit cell be \( a' \). Given the relation for the fcc unit cell side length \( a = 2A \), we can solve for the side length \( a' \) of the simple cubic unit cell:
\[ 2\sqrt{2}r = 4r \quad \Rightarrow \quad a' = \sqrt{2} \, A. \]

Thus, the side length of the unit cell for the simple cubic lattice will be \( \sqrt{2} \, A \). Quick Tip: For different lattice types, use the known relations between atomic radius and side length. For fcc: \( a = 2\sqrt{2}r \), and for simple cubic: \( a = 2r \).

We are given the initial rate data and we need to determine the rate law. The rate law is in the form: \[ r = k[A]^m[B]^n. \]

- Comparing Experiments 1 and 2: \[ \frac{Rate_2}{Rate_1} = \frac{k[A_2]^m[B_2]^n}{k[A_1]^m[B_1]^n} = \frac{1.6}{0.8} = 2. \]
This gives: \[ \frac{[A_2]^m[B_2]^n}{[A_1]^m[B_1]^n} = 2 \quad \Rightarrow \quad \frac{(0.4)^m(0.2)^n}{(0.2)^m(0.4)^n} = 2. \]
Solving this gives \( m = 1 \) and \( n = 2 \).

Thus, the rate law is \( r = k[A]^1[B]^2 \). Quick Tip: To determine the rate law, compare the rate data from different experiments. The rate law can be determined by analyzing how the change in the concentration of reactants affects the rate of the reaction.

We are given:
- \( K_b = 0.52 \, K kg mol^{-1} \) (Boiling point elevation constant),
- \( i = 3 \) (The solute dissociates into 3 moles of particles),
- \( m = 2 \, moles in 1 kg of water \).

The formula for boiling point elevation is: \[ \Delta T_b = i \times K_b \times m \]

Substituting the given values: \[ \Delta T_b = 3 \times 0.52 \times 2 = 3.12 \, K \]

Thus, the elevation of boiling point is \( \Delta T_b = 3.18 \, K \). Quick Tip: To calculate the elevation in boiling point, use the formula \( \Delta T_b = i \times K_b \times m \), where \( i \) is the van't Hoff factor (number of particles produced per formula unit), \( K_b \) is the ebullioscopic constant, and \( m \) is the molality of the solution.

We are given:
- Boiling point of pure solvent = 353.3 K
- Melting point of pure solvent = 278.6 K
When a solute is dissolved, the following changes occur:
- The boiling point increases (boiling point elevation).
- The melting point decreases (freezing point depression).

Since the solute is non-volatile and non-electrolyte, the change in both boiling and melting points can be calculated using the colligative properties formulas:
- \( \Delta T_b = i \times K_b \times m \) (boiling point elevation),
- \( \Delta T_f = i \times K_f \times m \) (freezing point depression).

Here, \( i \) is the van’t Hoff factor, which is 1 for non-electrolytes, and \( m \) is the molality.

After applying the equations, the corrected values for boiling and melting points are:
- Boiling point = 354.11 K
- Melting point = 279.2 K

Thus, the answer is \( 354.11 \, K \) for boiling point and \( 279.2 \, K \) for melting point. Quick Tip: To solve problems involving colligative properties, remember that the change in boiling and freezing points depends on the number of particles in the solution, the solvent's boiling point elevation constant (\( K_b \)) and freezing point depression constant (\( K_f \)).

In this reaction, the compound undergoes an electrophilic addition reaction with HBr. The reaction involves a vinyl group (C=CH2), which can undergo Markovnikov's rule for the addition of HBr. According to Markovnikov’s rule, the hydrogen will add to the carbon atom of the double bond that has the greater number of hydrogen atoms, while the bromine will add to the carbon atom that has fewer hydrogen atoms.

Thus, the addition will proceed as follows:
- The proton (H+) will add to the carbon of the double bond that has more hydrogen atoms (CH2 group).
- The bromine (Br) will add to the other carbon (CH) of the double bond.

This leads to the formation of 2-Bromobutane as the major product. Quick Tip: When solving addition reactions, remember Markovnikov's rule, which tells you how the H+ and halogen (like Br) will add across the double bond. The halogen attaches to the carbon with fewer hydrogen atoms.

In the compound \( CH_3(CH_2)_{16}COO^{-}Na^{+} \), the hydrophilic part refers to the part that is attracted to water, typically involving a polar or ionic group.

- The \( COO^{-} \) group is a carboxylate anion that is polar and hydrophilic, meaning it is water-soluble. This part of the molecule is attracted to water molecules due to the negative charge and its polarity.
- The \( Na^{+} \) cation is an inorganic ion that does not participate in hydrophilic interactions but in ionic bonding.
- The \( CH_3(CH_2)_{16} \) alkyl group is hydrophobic due to its nonpolar nature, making it repelled by water.

Thus, the hydrophilic part of this molecule is the \( COO^{-} \) group, as it is the most polar and interacts effectively with water molecules. Quick Tip: When identifying the hydrophilic portion of a molecule, look for polar or charged groups, such as carboxylate (\( COO^{-} \)) or hydroxyl groups, which can form hydrogen bonds with water.

Ba(OH)_2 dissociates completely into \( Ba^{2+} \) and \( 2OH^{-} \). The concentration of hydroxide ions, \( [OH^{-}] \), is double the concentration of Ba(OH)_2 because each formula unit of Ba(OH)_2 provides two hydroxide ions.

- The concentration of hydroxide ions \( [OH^{-}] = 2 \times 0.005 \, M = 0.010 \, M \).

Now, we calculate the pOH using the formula: \[ pOH = -\log[OH^{-}] \]
Substitute the value of \( [OH^{-}] \): \[ pOH = -\log(0.010) = 2 \]

Since \( pH + pOH = 14 \), we can calculate the pH: \[ pH = 14 - pOH = 14 - 2 = 12 \]

Thus, the pH value of the solution is 12. Quick Tip: When calculating pH or pOH, remember that the sum of pH and pOH always equals 14 in aqueous solutions. Also, for a base like Ba(OH)_2, the concentration of \( OH^{-} \) ions is directly related to the concentration of the base.

The wavelength of an electron can be calculated using the de Broglie relation: \[ \lambda = \frac{h}{mv} \]
Where:
- \( h \) is Planck's constant, \( 6.626 \times 10^{-34} \, J \cdot s \),
- \( m \) is the mass of the electron, \( 9.1 \times 10^{-31} \, kg \),
- \( v \) is the velocity of the electron, \( 2.2 \times 10^7 \, ms^{-1} \).

Substitute the values into the formula: \[ \lambda = \frac{6.626 \times 10^{-34}}{(9.1 \times 10^{-31})(2.2 \times 10^7)} \approx 3.3 \times 10^{-11} \, m. \]

Thus, the wavelength of the electron is \( 3.3 \times 10^{-11} \, m \). Quick Tip: To calculate the de Broglie wavelength of a particle, use the formula \( \lambda = \frac{h}{mv} \), where the mass \( m \) and velocity \( v \) are crucial in determining the wavelength.

Quantum numbers describe the state of an electron in an atom. They must satisfy the following restrictions:
1. \( n \) (the principal quantum number) must be a positive integer (\( n = 1, 2, 3, \dots \)).
2. \( l \) (the angular momentum quantum number) can take integer values from \( 0 \) to \( n-1 \), i.e., \( l = 0, 1, 2, \dots, n-1 \).
3. \( m_l \) (the magnetic quantum number) can take integer values from \( -l \) to \( +l \), i.e., \( m_l = -l, -l+1, \dots, l-1, l \).
4. \( m_s \) (the spin quantum number) can take values \( +\frac{1}{2} \) or \( -\frac{1}{2} \).

Now, checking the given combinations:
- (A) is correct because \( n = 1, l = 0, m_l = 0, m_s = -\frac{1}{2} \) is allowed.
- (B) is incorrect because for \( n = 1 \), \( l \) can only be 0, but here \( l = 1 \). Hence, this combination is not allowed.
- (C) is correct because \( n = 2, l = 1, m_l = +1, m_s = -\frac{1}{2} \) is allowed.
- (D) is correct because \( n = 3, l = 1, m_l = 0, m_s = +\frac{1}{2} \) is allowed.

Thus, the incorrect combination is option (B). Quick Tip: In quantum mechanics, pay close attention to the allowed values of quantum numbers. For \( n = 1 \), \( l = 0 \) is the only possible value, which makes (B) an incorrect combination.

N/A Quick Tip: For first-order reactions, remember the half-life formula \( t_{1/2} = \frac{0.693}{k} \). This is crucial when calculating half-life from the rate constant.

The oxidation states of iodine in the given compounds are as follows:
- In \( HI \), iodine has an oxidation state of \(-1\).
- In \( ICl \), iodine has an oxidation state of \(-1\).
- In \( HIO_3 \), iodine has an oxidation state of \(+5\).
- In \( HIO_2 \), iodine has an oxidation state of \(+3\).
- In \( KIO_3 \), iodine has an oxidation state of \(+5\).

Thus, the correct order of increasing oxidation states of iodine is: \[ HI < ICl < HIO_2 < KIO_3 \] Quick Tip: Remember, in compounds like \( HI, ICl, HIO_3, KIO_3 \), the oxidation number of iodine increases as it moves from \( -1 \) to \( +5 \). The more oxygen atoms present, the higher the oxidation state.

N/A Quick Tip: In general, alkylamines (e.g., ethylamine) are stronger bases than arylamines (e.g., aniline) due to the electron-donating effect of alkyl groups.

In this reduction reaction, the nitro group (\( NO_2 \)) needs to be reduced to an amine group (\( NH_2 \)) or similar. Sodium borohydride (\( NaBH_4 \)) is a common reducing agent used for reducing nitro groups to amines or hydroxyl groups.

- \( NaBH_4 \) is a mild and selective reducing agent that can reduce nitro groups to amines without affecting other functional groups like alcohols.
- Lithium aluminium hydride (\( LiAlH_4 \)) is a stronger reducing agent but is generally used for reducing esters, aldehydes, and ketones.
- \( B_2H_6 \) (diborane) is typically used for the reduction of unsaturated compounds.
- Calcium oxide (\( CaO \)) is not a reducing agent in this context.

Thus, the suitable reducing agent is \( NaBH_4 \). Quick Tip: Sodium borohydride (\( NaBH_4 \)) is a selective and milder reducing agent suitable for the reduction of nitro groups to amines.

The presence of electron-withdrawing groups (such as nitro groups) in aromatic compounds increases the acidity of hydroxyl groups attached to the ring. Among the given options, the nitro group (\( NO_2 \)) at the para position exerts a strong electron-withdrawing effect, making the hydroxyl group more acidic.

- In \( p-O_2N-C_6H_4OH \), the nitro group at the para position is most effective in withdrawing electron density, thus making this compound the most acidic.
- \( m-O_2N-C_6H_4OH \) also has a nitro group, but its effect is weaker because it is in the meta position.
- \( o-O_2N-C_6H_4OH \) has the nitro group in the ortho position, but the effect is still less than the para position due to steric hindrance.
- \( C_6H_5OH \) has no electron-withdrawing group and, therefore, is the least acidic.

Thus, the most acidic compound is \( p-O_2N-C_6H_4OH \). Quick Tip: Electron-withdrawing groups, especially when located at the para position, increase the acidity of phenolic compounds.

The reaction involves an alkene (\( CH_2 = CH_2 \)) undergoing hydration in the presence of sulfuric acid. This is a typical electrophilic addition reaction, where the alkene undergoes protonation, followed by nucleophilic attack by water to form the alcohol.

1. The alkene, \( CH_2 = CH_2 \), reacts with \( H_2SO_4 \) (sulfuric acid), leading to the formation of an alkyl hydrogen sulfate intermediate.
2. The intermediate then undergoes hydrolysis, leading to the formation of ethanol, \( CH_3CH_2OH \).

Thus, the major product ‘A’ is \( CH_2 = CH_2 \), and the major product ‘B’ is \( CH_3CH_2OH \). Quick Tip: In reactions with alkene and sulfuric acid, the hydration typically follows Markovnikov’s rule, where the proton adds to the carbon with the greater number of hydrogen atoms.

The spontaneity of a process is determined by the Gibbs free energy change, \( \Delta G \), which is given by the equation: \[ \Delta G = \Delta H - T \Delta S \]
For a process to be spontaneous, \( \Delta G \) must be negative.

In this case:
- \( \Delta H \) is positive, which means the system absorbs heat.
- \( \Delta S \) is negative, indicating a decrease in entropy.

Thus, \( \Delta G = \Delta H - T \Delta S \) will always be positive (since both \( \Delta H > 0 \) and \( \Delta S < 0 \)). Therefore, the process will always be non-spontaneous at all temperatures. Quick Tip: To determine the spontaneity of a process, check the signs of \( \Delta H \) and \( \Delta S \). If \( \Delta H \) is positive and \( \Delta S \) is negative, the process will always be non-spontaneous.

For an irreversible expansion, the work done \( W \) is given by: \[ W = -P_{ext} \Delta V \]
For a reversible expansion, the work done is given by: \[ W = -nRT \ln \left( \frac{V_f}{V_i} \right) \]
For the same isothermal expansion, the relationship between irreversible and reversible work done is: \[ W_{reversible} = W_{irreversible} \times 2.303 \]
Thus: \[ W_{reversible} = -2.303 \times W_{irreversible} = -2.303 \times (-1 \cdot 2RT) = -1.386 RT \] Quick Tip: When dealing with isothermal processes, remember that the work done in reversible and irreversible processes are related by a factor of \( 2.303 \) in the case of isothermal expansion.

To determine the molecule with a non-zero dipole moment, we need to consider the symmetry of the molecule and the electronegativity difference between the atoms.
- In BeF\(_2\), BF\(_3\), and CO\(_2\), the molecules are linear or symmetrical, and the individual dipoles cancel each other out, resulting in zero dipole moment.
- In NF\(_3\), the molecule has a trigonal pyramidal shape with nitrogen as the central atom and fluorine atoms at the corners. The individual dipoles do not cancel, leading to a non-zero resultant dipole moment.

Thus, NF\(_3\) has a non-zero dipole moment. Quick Tip: To determine if a molecule has a non-zero dipole moment, check if it has a symmetrical geometry. If the geometry is asymmetrical, like in NF\(_3\), it likely has a non-zero dipole moment.

To determine which oxide is the most acidic, we look at the oxides of metals and non-metals. Oxides of non-metals (like Cl\(_2\)O\(_7\)) are generally acidic, while oxides of metals tend to be basic.
- Na\(_2\)O and CaO are basic oxides of metals.
- Al\(_2\)O\(_3\) is amphoteric, meaning it can act as either an acid or a base depending on the conditions.
- Cl\(_2\)O\(_7\) is an acidic oxide and is the strongest acid among the options. It reacts vigorously with water to form acids.

Thus, Cl\(_2\)O\(_7\) is the most acidic oxide. Quick Tip: In general, oxides of non-metals are acidic, while oxides of metals are basic or amphoteric. For non-metals, the acidity increases as we move from left to right across the periodic table.

The second ionisation enthalpy refers to the energy required to remove a second electron from an ion. This value is highest for ions where the removal of the second electron would break a stable, fully-filled or half-filled electron configuration.
- For Na, after the first ionisation, the ion Na\(^+\) has a stable noble gas configuration (Ne), and it becomes very difficult to remove a second electron, resulting in a very high second ionisation enthalpy.
- For F, the second ionisation would remove an electron from a highly electronegative ion (F\(^-\)), but it’s not as high as Na.
- For Ne, it already has a stable noble gas configuration and does not easily form a positive ion with a second ionisation.
- Mg also has a relatively high second ionisation enthalpy, but it’s not as high as that of Na, as Mg’s second ionisation takes it to a stable configuration.

Thus, the second ionisation enthalpy is highest for Na. Quick Tip: When considering ionisation enthalpy, remember that it is highest when an electron is removed from a stable or noble gas configuration. This typically leads to a large increase in ionisation energy.

The electronic configuration of Chromium (Cr) is \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^5\). Chromium has an atomic number of 24, which means it has 24 electrons. According to the Aufbau principle and Hund’s rule, the 4s orbital is filled first before the 3d orbitals. However, Cr has an exceptional configuration where one electron from the 4s orbital is promoted to the 3d orbital, leading to a half-filled \(3d^5\) configuration, which is more stable than the expected \(3d^4 4s^2\) configuration.

Thus, the correct electronic configuration is \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^5\). Quick Tip: For transition elements like Cr, the stability of half-filled or fully filled \(d\)-orbitals can lead to electron promotion from the \(s\)-orbital to the \(d\)-orbital. Always remember that stability of the \(d\)-block elements plays a role in their electronic configuration.

The correct order of orbital energy for a multielectron atom is determined by the general rule of orbital filling in the \(n + l\) scheme. This rule states that the lower the sum of \(n + l\), the lower the energy of the orbital. In cases where two orbitals have the same value of \(n + l\), the orbital with the lower \(n\) value will have lower energy.

- \(6s\) has \(n = 6, l = 0\), so its energy is relatively high.
- \(5p\) has \(n = 5, l = 1\), and it is higher in energy compared to \(6s\).
- \(4f\) has \(n = 4, l = 3\), and the \(f\)-orbitals are lower in energy than both \(6s\) and \(5p\), which makes \(4f > 6s > 5p > 4s\) the correct order.

Thus, the correct order of the orbital energies is \(4f > 6s > 5p > 4s\). Quick Tip: Remember the \(n + l\) rule when determining orbital energies for multielectron atoms: Lower \(n + l\) values correspond to lower energy orbitals.

We are given that the cell potential \( E_{cell} = 2 \, V \), and we need to find the standard Gibbs free energy change (\(\Delta G^\circ\)).
The relation between the standard Gibbs free energy change and the cell potential is: \[ \Delta G^\circ = -nFE_{cell} \]
where:
- \(n\) is the number of electrons transferred in the reaction (which is 6 electrons based on the balanced equation),
- \(F\) is the Faraday constant (\(F = 96,485 \, C/mol\)),
- \(E_{cell}\) is the cell potential in volts.

Substituting the values: \[ \Delta G^\circ = -6 \times 96,485 \times 2 = -1,158,000 \, J/mol = -12 \, F \]

Thus, the standard Gibbs free energy change is \(-12 \, F\). Quick Tip: Use the formula \(\Delta G^\circ = -nFE_{cell}\) to relate Gibbs free energy change to the cell potential.

The standard EMF (\(E_{cell}^\circ\)) of a Galvanic cell is given by the equation: \[ E_{cell}^\circ = E^\circ_{cathode} - E^\circ_{anode}. \]
In this case, the cathode will be the electrode with the more positive electrode potential, and the anode will be the electrode with the more negative electrode potential.

Given:
- \(E^\circ_{A} = -0.25 \, V\),
- \(E^\circ_{B} = -0.76 \, V\).

Thus, the anode will be electrode-B (more negative), and the cathode will be electrode-A. \[ E_{cell}^\circ = (-0.25) - (-0.76) = 0.51 \, V. \]
Therefore, the standard EMF of the cell is \(0.51 \, V\), and electrode-A will act as the anode. Quick Tip: Remember, the cathode is the electrode with the higher (more positive) potential, while the anode is the electrode with the lower (more negative) potential.

In formaldehyde (\( HCHO \)), the carbon atom is double-bonded to oxygen and single-bonded to a hydrogen atom and a methyl group (\( -CH_2 \)). This requires three regions of electron density around the carbon atom. Hence, the carbon atom undergoes \( sp^2 \) hybridization to form three sp² hybrid orbitals, which results in a trigonal planar shape.

The shape of the formaldehyde molecule is trigonal planar due to the \( sp^2 \) hybridization of the central carbon atom. Quick Tip: In molecules with \( sp^2 \) hybridization, the central atom has three regions of electron density, leading to a trigonal planar geometry.

In the given compound, we need to follow the IUPAC naming rules for substituted benzene rings. First, we number the benzene ring starting from the substituent with the highest priority (in this case, Nitro \( NO_2 \)). After the nitro group, the fluoro group (\( F \)) is positioned in the 4th position and the ethyl group (\( C_2H_5 \)) is in the 1st position.

Thus, the IUPAC name for this compound is 1-Ethyl-4-Fluoro-3-nitro benzene. Quick Tip: When naming substituted benzene compounds, number the ring so that the substituents receive the lowest possible numbers. The priority of substituents generally follows: nitro > fluoro > ethyl.

Given that the molecular weight (MW) of the dibasic acid is 118, the mass of the acid is 35.4 g, and the density of the solution is 1.00 g/cm\(^3\), we can calculate the normality and molality.

Step 1: Calculating Normality (N)
Normality is given by the formula: \[ N = \frac{Gram equivalent weight}{Volume of solution in liters} \]
The equivalent weight of the acid is the molecular weight divided by 2, as it is dibasic. So, \[ Equivalent weight = \frac{118}{2} = 59 \, g/equivalent \]
The normality is: \[ N = \frac{35.4}{59 \times 1} = 0.6 \, N \]

Step 2: Calculating Molality (m)
Molality is calculated using: \[ m = \frac{moles of solute}{kg of solvent} \]
First, we calculate the moles of solute: \[ Moles of acid = \frac{35.4}{118} = 0.3 \, mol \]
Since the density of the solution is 1.00 g/cm\(^3\), 1 liter of solution weighs 1000 g. Therefore, the mass of the solvent is: \[ Mass of solvent = 1000 \, g - 35.4 \, g = 964.6 \, g = 0.9646 \, kg \]
Now, the molality is: \[ m = \frac{0.3}{0.9646} = 0.3 \, mol Kg^{-1} \]

Thus, the normality and molality of the solution are 0.6 N and 0.3 mol Kg\(^-1\), respectively. Quick Tip: For dibasic acids, remember to divide the molecular weight by 2 to find the equivalent weight when calculating normality. Molality is calculated using the mass of the solvent, not the solution.

In the \( S_N1 \) reaction, the rate-determining step involves the formation of a carbocation intermediate. The reactivity of the compound is largely dependent on the stability of the carbocation formed. The more stable the carbocation, the more reactive the compound will be in an \( S_N1 \) reaction.

The general order of reactivity for \( S_N1 \) reactions is: \[ 3^\circ > 2^\circ > 1^\circ \]
Where \( 3^\circ \) represents a tertiary carbon, \( 2^\circ \) represents a secondary carbon, and \( 1^\circ \) represents a primary carbon.

- Compound A (\( C_6H_5 -CH_2Br \)): This is a primary alkyl halide, which forms a less stable primary carbocation, making it less reactive in \( S_N1 \).
- Compound B (\( C_6H_5 -CH_2Br \)): This is a primary alkyl halide, similar to compound A.
- Compound C (\( C_6H_5 -CH_2Br \)): Again a primary halide, yet it would still undergo \( S_N1 \) in the presence of stabilizing effects from a group like \( C_6H_5 \), which increases its reactivity.
- Compound D (\( C_6H_5 -CH_2-C_6H_5 \)): This is a tertiary halide, and tertiary carbocations are more stable, so compound D is most reactive in \( S_N1 \) reactions. However, compound C is more reactive.

Thus, the compound in option \( C \) is most reactive in an \( S_N1 \) reaction. Quick Tip: In \( S_N1 \) reactions, the stability of the carbocation intermediate is key. Tertiary carbocations are the most stable, followed by secondary, then primary carbocations.

Lactose is a disaccharide composed of two monosaccharides: glucose and galactose. Specifically, it is formed by the condensation of one molecule of \( \alpha - D \)-Galactose and \( \beta - D \)-Glucose. In lactose, the glucose unit is in the \( \beta \)-configuration, while the galactose is in the \( \alpha \)-configuration, which is characteristic of lactose.

Thus, the correct combination is \( \alpha - D \)-Galactose and \( \beta - D \)-Glucose. Quick Tip: In disaccharides like lactose, pay attention to the anomeric configurations of the constituent monosaccharides, which play a crucial role in their structure and properties.