JEE Main 2025 April 2 Shift 2 Physics Question Paper, Exam Analysis, and Answer Keys (Available)

(A): Since polar dielectrics are randomly oriented, \( \mathbf{P_{net}} = \mathbf{0} \).

(R): If \( \mathbf{E} \) is absent, polar dielectrics remain polar and are randomly oriented.
Quick Tip: When analyzing assertions and reasons in physics, remember that the reason should adequately explain the assertion. If not, mark them as correct but not the correct explanation.

The voltage sensitivity is given by: \[ \frac{\theta}{V} = \frac{NAB}{cR} \]
where \( N \) is the number of turns, \( A \) is the area, \( B \) is the magnetic field, \( c \) is the torsional constant, and \( R \) is the resistance.

Thus, the ratio of the voltage sensitivity of \( M_1 \) and \( M_2 \) is: \[ Ratio = \frac{N_1 A_1 B_1}{N_2 A_2 B_2} = \frac{15 \times 3.6 \times 10^{-3} \times 0.25}{21 \times 1.8 \times 10^{-3} \times 0.50} = \frac{1}{1} \]

Thus, the ratio of voltage sensitivity of \( M_1 \) and \( M_2 \) is 1 : 1. Quick Tip: In moving coil galvanometers, the voltage sensitivity is directly proportional to the number of turns, the area of the coil, and the magnetic field strength, and inversely proportional to the resistance.

Diameter is given as \( R \).
\[ Radius = \frac{R}{2} \]

The formula for the moment of inertia about a tangential axis is given by:
\[ I_{tangent} = \frac{3}{2} m \left( \frac{R}{2} \right)^2 = \frac{3}{8} m R^2 \]

Thus, the moment of inertia is:
\[ I_{tangent} = \frac{3}{8} M r^2 \] Quick Tip: For a circular ring, the moment of inertia about an axis tangent to the ring is derived from the parallel axis theorem by adjusting for the offset from the center.

Let the radius of each drop be \( r \) and the radius of the bigger drop be \( R \).

The volume of the two smaller drops:
\[ V_{small} = 2 \times \frac{4}{3} \pi r^3 \]

The volume of the larger drop:
\[ V_{big} = \frac{4}{3} \pi R^3 \]

Equating the volumes:
\[ 2 \times \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3 \Rightarrow r = R^{2/3} \]

The surface energy for the smaller drops is:
\[ U_i = 2 \times 4 \pi r^2 T = 8 \pi r^2 T \]

The surface energy for the bigger drop is:
\[ U_f = 4 \pi R^2 T = 4 \pi R^{4/3} T \]

The heat lost in the process is:
\[ Heat lost = U_i - U_f = 8 \pi r^2 T - 4 \pi R^{4/3} T = 4 \pi r^2 T \left[ 2^2 - 2^3 \right] \]

Thus, the energy released is:
\[ Energy released = 4\pi r^2 T \left[ 2^2 - 2^3 \right] \] Quick Tip: To solve problems involving surface tension, always use the principle of volume conservation and the fact that energy is related to the surface area of the drop.

Magnetic field does not work on the speed of the electron because magnetic forces only act perpendicular to the velocity.

Thus, the speed of the electron will not change, and consequently, its de-Broglie wavelength will remain the same.

The de-Broglie wavelength is given by:
\[ \lambda = \frac{h}{mv} \]

Since the speed \( v \) does not change due to the magnetic field, the de-Broglie wavelength remains constant at \( \lambda_0 \).

Thus, the de-Broglie wavelength at time \( t \) is the same as at \( t = 0 \):
\[ \lambda(t) = \lambda_0 \] Quick Tip: When dealing with an electron in a magnetic field, remember that the magnetic force only changes the direction of motion, not the speed. Thus, the de-Broglie wavelength remains unaffected.

The velocity \( v \) of the wave is given by:
\[ v = \frac{distance}{time} = \frac{12 \, cm}{0.3 \, s} = 4 \, cm/s \]

Next, the wave number \( k \) and angular frequency \( \omega \) are related to the wavelength \( \lambda \) and frequency \( f \) as:
\[ k = \frac{2 \pi}{\lambda} = \frac{2 \pi}{7.5} = 0.83 \, cm^{-1} \]
\[ \omega = v k = 4 \times 0.83 = 3.35 \, rad/s \]

Thus, the wave equation is:
\[ y = A \cos(kx - \omega t) = A \cos(0.83x - 3.35t) \]

Given that the amplitude \( A \) is 2 cm (maximum displacement), the equation becomes:
\[ y = 2 \cos(0.83x - 3.35t) \, cm \] Quick Tip: Remember, the general form of the wave equation is \( y = A \cos(kx - \omega t) \), where \( A \) is the amplitude, \( k \) is the wave number, and \( \omega \) is the angular frequency. Use these relationships to derive the wave equation.

We are given:
\[ Q = AT \] \[ I = A \] \[ \mu_0 = ML^3 T^{-2} A^{-2} \]
Now, we need to find the dimensions of the product \( P = Q \mu_0 I \).

The dimension of \( P \) is calculated as follows:
\[ P = Q \mu_0 I = [AT] [ML^3 T^{-2} A^{-2}] [A] \]

This simplifies to:
\[ P = [M^1 L^1 T^{-2} A^1] \]

Now, we check the dimensions of momentum:
\[ Momentum = M \cdot L \cdot T^{-1} \]

We find that the dimensions of \( P \) are the same as that of momentum. Therefore, the correct answer is:
\[ q \mu_0 I \] Quick Tip: To match the dimensions of momentum, use the fact that momentum has the dimensions \( M \cdot L \cdot T^{-1} \) and check for the correct expression.

We are given the energy stored in a solenoid \( U \), and the relation for magnetic energy density \( U/V \) is:
\[ U/V = \frac{B^2}{2\mu_0} \]

This implies that:
\[ U = \frac{B^2}{2\mu_0} \times V \]

Where \( V = A \ell \), the volume of the solenoid. Substituting:
\[ U = \frac{B^2}{2\mu_0} \times A \ell \]

Thus, the magnetic energy stored in the solenoid is:
\[ U = \frac{B^2 A \ell}{4 \mu_0} \]

Hence, the correct answer is:
\[ \frac{B^2 A \ell}{4 \mu_0} \] Quick Tip: For problems involving energy stored in a magnetic field, use the formula for the energy density in the magnetic field \( U/V = \frac{B^2}{2 \mu_0} \), and apply the volume of the solenoid to find the total energy.

We are given the potential difference between two plates as \( V \), and the separation between the plates is 10 cm. The distance between points A and B is 3 cm and 4 cm, respectively, with the total distance between the plates being 10 cm.

Using \( \Delta V = E \Delta d \), where \( E \) is the electric field and \( \Delta d \) is the distance:
\[ V = E \times 10 \, cm \]

From the diagram, we know that \( E = \frac{V}{10} \). The potential difference between points A and B is:
\[ V_{AB} = E \times 4 \, cm = \frac{V}{10} \times 4 = \frac{2V}{5} \]

Thus, the potential difference between points A and B is \( \frac{2}{5} V \). Quick Tip: To calculate potential difference in a parallel plate setup, use the relationship \( V_{AB} = \frac{E \times d}{V} \), where \( d \) is the distance between the points of interest.

For an adiabatic process, the heat exchanged \( Q = 0 \), hence, the change in internal energy is equal to the work done on or by the system:
\[ Q = \Delta U + W = 0 \implies \Delta U = - W \]

For an adiabatic process, the work done \( W \) is proportional to the change in temperature. Also, we know that:
\[ W = -n C_V \Delta T \quad or \quad |W| = n C_V \Delta T \propto T_2 - T_1 \]

Thus, both (B) and (E) are correct, making option (3) the correct answer. Quick Tip: In an adiabatic process, internal energy change is equal to the work done, and the work is proportional to the change in temperature.

The radius for a hydrogen-like ion is given by the formula:
\[ r = r_0 \frac{n^2}{z} \]

where \( r_0 \) is the radius for hydrogen, \( n \) is the principal quantum number, and \( z \) is the atomic number. For \( Li^{2+} \), we have \( n = 1 \) and \( z = 3 \), so the radius is:
\[ r = r_0 \frac{1^2}{3} = \frac{r_0}{3} \]

Thus, \( X = 3 \). Quick Tip: For hydrogen-like ions, the radius decreases as the atomic number increases, following the formula \( r = r_0 \frac{n^2}{z} \).

Given:
- Binding energy per nucleon of \( H_2^1 \) = 1.1 MeV
- Binding energy per nucleon of \( He_4^2 \) = 7.0 MeV

The energy released \( Q \) is the difference between the binding energy of the reactants and products:
\[ E_B = BE_{reactant} - BE_{product} \]
\[ E_B = 1.1 \times 2 + 1.1 \times 2 - 7 \times 4 = 23.6 \, MeV \]

Thus, the energy released is:
\[ Q = 23.6 \, MeV \] Quick Tip: The energy released in a nuclear fusion reaction can be calculated using the binding energy per nucleon of the reactants and products.

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For the given digital circuit, follow the logic gates step by step. Using the inputs \( P = 0 \) and \( Q = 1 \), we compute the outputs as follows:

- The AND gate gives \( 0 \)
- The NOT gate inverts the inputs appropriately.

Hence, the correct output for the given circuit is:
\[ P = 0, Q = 0 \] Quick Tip: When analyzing logic circuits, work step by step through each gate (AND, OR, NOT) to determine the final outputs.

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Using the magnification formula for mirrors:
\[ m = \frac{f}{u-f} \]

For the concave mirror, the object distance is \( u = -18 \, cm \), and the focal length is \( f = \frac{R}{2} = 6 \, cm \), where \( R = 12 \, cm \):
\[ m_1 = \frac{6}{18 - 6} = \frac{1}{2} \]

For the convex mirror, the object distance is the same, and the focal length is positive:
\[ m_2 = \frac{6}{18 + 6} = \frac{1}{4} \]

Hence, the ratio of the sizes of the images formed by the convex mirror and the concave mirror is:
\[ \frac{m_2}{m_1} = \frac{1/4}{1/2} = \frac{1}{2} \]

Thus, the correct answer is:
\[ \boxed{\frac{1}{2}} \] Quick Tip: Remember, the magnification formula for mirrors relates the image size to the object distance and the focal length. For concave mirrors, the object distance is negative, while for convex mirrors it is positive.

Displacement is the straight-line distance from the initial point to the final point. Since the sportsman runs around the circular track and ends up at the same position (A), the displacement is the straight-line distance through the circle’s center. Therefore:
\[ Displacement = 2r \]

The distance travelled is the total path length covered by the sportsman, which consists of two complete laps around the circular track. Thus, the total distance is:
\[ Distance = 2\pi r + \pi r = 3\pi r \]

Thus, the correct answer is:
\[ \boxed{3\pi r, 2r} \] Quick Tip: When dealing with circular motion, remember that the displacement is the straight-line distance from start to end, and the distance is the total path length travelled.

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Given that the body is in equilibrium, we can resolve the forces in the vertical and horizontal directions. The weight of the body is \( mg = 1 \times 9.8 = 9.8 \, N \).

For the vertical direction: \[ T_1 \sin 30^\circ + T_2 \sin 30^\circ = mg \]
For the horizontal direction: \[ T_1 \cos 30^\circ = T_2 \cos 30^\circ \]

Thus: \[ T_1 = T_2 \]
Now, solving for the tensions using the vertical direction equation: \[ T_1 \sin 30^\circ + T_1 \sin 30^\circ = 9.8 \, N \] \[ 2T_1 \sin 30^\circ = 9.8 \] \[ 2T_1 \times \frac{1}{2} = 9.8 \] \[ T_1 = 5 \, N, \, T_2 = 5\sqrt{3} \, N \]

Thus, the correct answer is: \[ \boxed{T_1 = 5 \, N, \, T_2 = 5\sqrt{3} \, N} \] Quick Tip: When dealing with problems of forces in equilibrium, remember to resolve the forces into vertical and horizontal components and apply the equilibrium conditions.

This will happen when \[ \frac{1}{f_1} = \frac{1}{f_2} \] \[ (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = (\mu - 1)\left(\frac{2}{R}\right) \] \[ \frac{1}{R_1} + \frac{1}{R_2} = \frac{2}{R} \]

Thus, the possible combination for \( R_1 \) and \( R_2 \) is \( \frac{1}{5} \, cm \) and \( \frac{1}{7} \, cm \). Quick Tip: When dealing with lens formulas, always ensure that the radii of curvature for both sides of the lens match the conditions for the required lens power.

Using the formula, \[ C = \frac{1}{\sqrt{\mu_0 \epsilon_0}} \Rightarrow 1 = C^2 = L T^{-2} \]

Thus, the dimension of \( \frac{1}{\mu_0 \epsilon_0} \) is \( L^2 T^2 \). Quick Tip: The dimensions of permeability and permittivity help determine the speed of light in a vacuum, and their relationship is key in electromagnetic theory.

\[ C' = \frac{\Delta Q}{\Delta T} = JK^{-1} \]
\[ S = \frac{\Delta Q}{m \Delta T} = Jkg^{-1}K^{-1} \]
\[ L = \frac{\Delta Q}{m} = Jkg^{-1} \]
\[ \Delta Q = \frac{K A \Delta T}{L} \Rightarrow K = \frac{\Delta Q}{A \Delta T} = J m^{-1} k^{-1} s^{-1} \] Quick Tip: The formulae for heat capacity, specific heat, and thermal conductivity can be used to relate physical properties of substances.

\[ E = \frac{KQr}{(x^2 + R^2)^{3/2}} \]
\[ \frac{dE}{dx} = 0 \]
\[ \therefore x = \frac{R}{\sqrt{2}} = \sqrt{\frac{2a}{\sqrt{2}}} = a \]

Thus, the value of \( x \) is \( a \), which corresponds to option (3). Quick Tip: For maximum electric field along the axis of a charged circular loop, set the derivative of the electric field with respect to \( x \) to zero.

\[ F R = I \alpha \]
\[ \Rightarrow I = \frac{F R}{\alpha} = \frac{10 \times 0.2}{2} = 1 \, kg \, m^2 \]

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\end{figure Quick Tip: In problems involving rotational motion, torque and moment of inertia are related by the equation \( Torque = I \alpha \), where \( I \) is the moment of inertia and \( \alpha \) is the angular acceleration.

To find the internal energy of gas in the room.
\[ U = n C_v T = \frac{5}{2} R T \]
\[ = \frac{5}{2} \times P V = \frac{5}{2} \times 10^5 \times 48 = 12 \times 10^6 \, J \] Quick Tip: For a diatomic molecule, the internal energy is calculated using the formula \( U = \frac{5}{2} P V \), where \( P \) is pressure and \( V \) is volume.

\[ \mu = \frac{\sin \left( \frac{A + \delta_m}{2} \right)}{\sin \left( \frac{A}{2} \right)}, \, since \, A = 60^\circ \] \[ \therefore \delta_m = 30^\circ \] \[ \delta_m = 2i - A \, [as \, i = e] \] \[ \therefore i = 45^\circ \] Quick Tip: For minimum deviation, the angle of incidence is equal to the angle of emergence. The relationship between the refractive index and the minimum deviation is given by the formula \( \mu = \frac{\sin \left( \frac{A + \delta_m}{2} \right)}{\sin \left( \frac{A}{2} \right)} \).

\[ T = K(\ell - \ell_0) \] \[ \Rightarrow 5 = K(1.4 - \ell_0) \] \[ \Rightarrow 7 = K(1.56 - \ell_0) \] \[ \Rightarrow \frac{5}{1.4 - \ell_0} = \frac{7}{1.56 - \ell_0} \] \[ \therefore \ell_0 = 1 \, m \] Quick Tip: In problems related to the elongation of strings due to tension, the relationship between tension and elongation is often linear, and the constant of proportionality (spring constant) can be used to find the original length.

\[ KE = \frac{1}{2} mv^2 = \frac{GMm}{2r} \]
\[ KE = \frac{GMm}{2(r_e + h)} \]
\[ KE = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 1000 \times 6.4 \times 10^6}{2(6.4 \times 10^6 + 2.7 \times 10^5)} \]
\[ KE = 3 \times 10^{10} \, J \] Quick Tip: The formula for the kinetic energy of a satellite in orbit can be derived from the gravitational potential energy, where the radius includes both the radius of the Earth and the height of the satellite above the surface.