JEE Main 2025 April 2 Shift 1 Physics Question Paper, Exam Analysis, and Answer Keys (Available)

The direction of propagation of light is perpendicular to the wave front and is symmetric about the \( x \), \( y \), and \( z \) axes. The angle made by the direction of wave propagation with the \( x \)-axis is the same as that with the \( y \)-axis and the \( z \)-axis. Thus, the equation can be written as:
\[ \cos \theta = \cos \beta = \cos \gamma \quad (where \alpha, \beta, \gamma are the angles made by light with the x, y, z axes respectively) \]

Also, we know that \( \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \). Since the angles are equal, we have:
\[ \cos^2 \alpha + \cos^2 \alpha + \cos^2 \alpha = 1 \quad \Rightarrow \quad 3 \cos^2 \alpha = 1 \quad \Rightarrow \quad \cos \alpha = \frac{1}{\sqrt{3}} \]

Thus, the angle is \( \cos^{-1} \left( \frac{1}{\sqrt{3}} \right) \). Quick Tip: The symmetry of the wave propagation allows us to use the property that the angle made with the \( x \), \( y \), and \( z \) axes is the same.

From the given equation \( \left( P + \frac{a}{V^2} \right)(V - b) = RT \), we have the following dimensions for each variable: \[ [a] = \left[ P \right] \left[ V \right]^2 = ML^{-1}T^{-2}L^2 = M L T^{-2} \] \[ [b] = [V] = L^3 \]
Now, \( [ab] = (M L T^{-2})(L^3) = M L^4 T^{-2} \).

Thus, the dimensions of \( ab \) correspond to the dimension of **compressibility**. Quick Tip: To solve for dimensional analysis problems, break down each term into its basic dimensions and multiply accordingly.

The work done \( W_f \) by the force \( F = 20 \, N \) is given by: \[ W_f = F \cdot d = 20 \times 1 = 20 \, J \]

This is the change in kinetic energy of the wheel: \[ KE = \frac{1}{2} I \omega^2 \]

Using \( I = MR^2 \) where \( M = 10 \, kg \) and \( R = 0.1 \, m \): \[ I = 10 \times (0.1)^2 = 0.1 \, kg m^2 \]

Now equating the work done to the change in kinetic energy: \[ 20 = \frac{1}{2} \times 0.1 \times \omega^2 \quad \Rightarrow \quad \omega = 20 \, rad/s \]


Quick Tip: When a steady force is applied and the object rotates, the work done on the object is equal to its kinetic energy.

The location of the image of A can be found using the lens formula: \[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \]
where \( f = 20 \, cm \), \( u = -30 \, cm \), and \( v = 60 \, cm \).

Using the magnification formula: \[ m = \frac{v}{u} = \frac{60}{-30} = -2 \]

Since the object size is small with respect to the location, we can calculate the small change \( dv \) in the image: \[ dv = m^2 du = 4 \times 1 = 4 \, cm \]
This gives us the size of the image at \( P \) as \( h_i = m h_o = 2 \times 2 = 4 \, cm \).

The angle made by the image with the principal axis is \( -45^\circ \), which corresponds to the correct answer.


Quick Tip: When the object is slanted and small compared to the location, use magnification and lens formulas to find the angle made by the image.

Let the charge distribution on the two plates be \( \sigma \) and \( -\sigma \), with the point charge \( q \) placed at the midpoint between the plates.

The electric field due to each plate at the midpoint is as follows:





For Plate 1, the electric field is \( \frac{\sigma}{2 \epsilon_0} \) directed away from the plate, and for Plate 2, the electric field is \( \frac{\sigma}{2 \epsilon_0} \) directed towards the plate.

Thus, the net electric field experienced by the charge \( q \) is:
\[ E_{net} = \frac{3\sigma}{2 \epsilon_0} \]

Now, the force on the charge \( q \) is given by:
\[ F = qE = q \times \frac{3\sigma}{2 \epsilon_0} = \frac{3q\sigma}{2 \epsilon_0} \]


Quick Tip: When calculating the electric field due to uniformly charged infinite plates, use the formula \( E = \frac{\sigma}{2\epsilon_0} \) for each plate and then sum the fields considering the direction.

The speed of the boat relative to the river is \( 27 \, km/hr \), and the boat crosses the river at an angle of \( 150^\circ \) to the direction of the river flow.

Using the formula for the effective speed component of the boat in the direction perpendicular to the flow of the river:
\[ V_{L} = 27 \, km/hr \times \cos 60^\circ = \frac{27}{2} = 13.5 \, km/hr \]

The time taken to cross the river is \( 30 \, seconds \) or \( \frac{1}{2} \) minute. Using the formula for distance:
\[ S = V_t \times t = 13.5 \, km/hr \times \frac{30}{60} \, hr = 13.5 \times \frac{1}{2} = 112.5 \, m \]

Thus, the width of the river is \( 112.5 \, m \).


Quick Tip: When solving river crossing problems, decompose the boat's velocity into components: one parallel to the river flow and one perpendicular to it. The perpendicular component gives the speed for crossing the river.

Let the electric field at point \( P \) in between the charges be zero. Let the position of \( P \) be at a distance \( x \) from the origin, where the electric field due to both charges cancels each other.





The electric field due to a point charge is given by: \[ E = \frac{kq}{r^2} \]
For the electric field to be zero at point \( P \), the fields due to both charges must be equal and opposite. So: \[ \frac{kq}{x^2} = \frac{k(9q)}{(d - x)^2} \]
Simplifying: \[ \frac{1}{x^2} = \frac{9}{(d - x)^2} \]
Solving for \( x \): \[ d - x = 3x \quad \Rightarrow \quad d = 4x \quad \Rightarrow \quad x = \frac{d}{4} \]

Thus, the coordinate of point \( P \) is \( \left(\frac{d}{4}, 0, 0\right) \). Quick Tip: In problems involving multiple charges, use the principle of superposition for electric fields and set the total field equal to zero to find the point of cancellation.

Given the voltage \( V = 4.2 \) volts and the battery capacity \( 5800 \) mAh, we can calculate the energy stored in the battery using the formula: \[ Energy supplied by battery = Vq \]
where \( q \) is the charge in coulombs. Converting \( 5800 \) mAh to coulombs: \[ q = 5800 \times 3600 \times 10^{-3} \, C = 5800 \times 3.6 \, C = 20880 \, C \]
Thus, the energy supplied by the battery is: \[ Energy = 4.2 \times 5800 \times 3600 \times 10^{-3} = 87.696 \, kJ \]
Therefore, the energy stored in the battery when fully charged is approximately \( 87.7 \, kJ \). Quick Tip: To calculate energy stored in a battery, use the formula \( E = Vq \), where \( V \) is the voltage and \( q \) is the charge in coulombs.

Given:
\( x_1 = \sqrt{7} \sin 5t \), \quad \( x_2 = 2 \sqrt{7} \sin \left( 5t + \frac{\pi}{3} \right) \)


From phasor, the displacement is represented as:

\[ \sqrt{7} \quad and \quad 2 \sqrt{7} \quad with angle \, 60^\circ \] \[ Amplitude of resultant SHM = 7 \]
\[ \phi = \tan^{-1} \left( \frac{2 \sqrt{7} \times \frac{\sqrt{3}}{2}}{\sqrt{7} + 2 \sqrt{7} \times \frac{1}{2}} \right) = \tan^{-1} \left( \frac{\sqrt{3}}{2} \right) = \tan^{-1} \left( \sqrt{3} \right) \] \[ X_R = 7 \sin \left( 5t + \phi \right) \]
\[ a_R = 7 \times 25 \sin \left( 5t + \phi \right) \]
\[ a_{max} = 175 \, cm/sec = 175 \times 10^{-2} \, m/sec \]


Quick Tip: In problems involving the superposition of simple harmonic motions, phasor addition simplifies the calculation of resultant amplitude and phase.

We have: \[ \mu_r = (1 + \chi) \quad so \quad \chi = (\mu_r - 1) \]

Also, \[ \mu = \mu_0 \mu_r \quad \Rightarrow \quad \mu_r = \frac{\mu}{\mu_0} \]

Thus, \[ \chi = \frac{\mu}{\mu_0} - 1 \] Quick Tip: In problems involving magnetic susceptibility, remember that the relative permeability \( \mu_r \) is directly related to \( \chi \), and permeability \( \mu \) is proportional to \( \mu_0 \times \mu_r \).

From the circuit diagram, we have the following: \[ i = \frac{20}{400} \, A = 10 \, mA \quad (Load current \( I_L \)) \] \[ V_L = 5V \quad (Zener voltage) \]
Also, \[ R_L = \frac{V_L}{i} = \frac{5}{10 \times 10^{-3}} = 500 \, \Omega \]


Quick Tip: In zener diode circuits, the load current and load resistance can be found by using Ohm’s law and the given zener voltage.

For an adiabatic process, \( dQ = 0 \).


Thus, the molar heat capacity is zero: \[ dQ = 0 \Rightarrow dU = -dW \]

Also, \[ dU = \frac{f}{2} nR dT \]

Thus, the correct option is:


Only option (3) is correct.
Quick Tip: In an adiabatic process, the system is thermally isolated, so there is no heat transfer. This implies that the change in internal energy is equal to the work done by the system.

Since the lamina is in equilibrium, the net force and net torque must be zero. Thus: \[ F_{net} = 0 \quad and \quad T_{net} = 0 \]

The torque due to force \( F \) at point \( O \) is given by the equation: \[ T = 10 \cdot 10 - F \cdot \ell = 0 \]
Thus, \( F = 10 \) N.


Quick Tip: In problems involving torque, the point of rotation is essential. The sum of torques about any point in equilibrium is zero.

The magnetic field at the center of a circular coil is given by: \[ B_1 = \frac{\mu_0 I}{2R} \]

The magnetic field at an axial distance \( x \) from the center is given by: \[ B_2 = B_1 \sin \theta = \frac{B_1 \cdot \left( \frac{R}{x} \right)^2}{5} \]

Substituting \( x : R = 3 : 4 \), we get: \[ \frac{B_2}{B_1} = \frac{64}{125} \] Quick Tip: When dealing with the magnetic field produced by a circular coil, remember that the magnetic field is strongest at the center and weakens with distance along the axis.

The energy of an electron in a specific orbit is given by: \[ E \propto \frac{Z}{n^2} \]
For hydrogen atom, \( Z_H = 1 \), for He\(^+\), \( Z_{He^+} = 2 \), and for Li\(^2+\), \( Z_{Li^{2+}} = 3 \).

1st excited state \( n = 2 \) and 2nd excited state \( n = 3 \).

From the given statements, only (A) and (B) are correct.


Quick Tip: In Bohr's atomic model, the energy levels depend on the atomic number \( Z \) and the principal quantum number \( n \). For ions like He\(^+\) and Li\(^2+\), the energy will differ due to their increased nuclear charge.

Let the moment of inertia of the rod about the axis passing through its center and normal to its length be \( \alpha = \frac{ML^2}{12} \), where \( M \) is the mass and \( L \) is the length.







Now, the rod is cut into two equal parts, each having mass \( \frac{M}{2} \) and length \( \frac{L}{2} \). Each part has a moment of inertia \( \alpha' \).

For the cross shape, the total moment of inertia will be the sum of the moments of inertia of the two parts, considering the distance from the center of the rod. After using the parallel axis theorem, we get:
\[ \alpha' = 2 \times \frac{M L^2}{48} = \frac{\alpha}{4} \]

Thus, the correct option is \( \frac{\alpha}{4} \).


Quick Tip: To calculate the moment of inertia of composite shapes, use the parallel axis theorem and add the individual moments of inertia.

Using the lens formula: \[ \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \]

Substitute the values:
\[ \frac{1.5}{v} - \frac{1}{-0.2} = \frac{1.5 - 1}{0.4} \]

Simplifying: \[ \frac{1.5}{v} + 5 = \frac{0.5}{0.4} = 1.25 \]

Solving for \( v \): \[ \frac{1.5}{v} = 1.25 - 5 = -3.75 \]
\[ v = -0.4 \, m \]

Hence, the image is located 0.24 m left to the spherical surface.


Quick Tip: When solving for the position of an image in spherical surfaces, always use the appropriate sign convention for distances and refractive indices.

\begin{tabular{l l
% Option
(A) Coefficient of viscosity & \([n] = [ML^{-1}T^{-1}]\)

% Option
(B) Intensity & \([I] = [ML^{1}T^{-3}]\)

% Option
(C) Pressure gradient & \([K] = [ML^{-1}T^{-2}]\)

% Option
(D) Compressibility & \([K] = [ML^{-1}T^{-2}]\)

\end{tabular Quick Tip: In dimensional analysis, the dimensions of physical quantities are crucial for understanding their relationships. Pay attention to how exponents are used to represent various physical properties.

From the diagram in the solution, we have the force acting on the charge due to the electric field of the sheet:





The force is given by: \[ F_e = qE = mg \]
where \( q \) is the charge and \( E \) is the electric field due to the sheet. The electric field is related to the charge density \( \sigma \) as: \[ E = \frac{\sigma}{2 \epsilon_0} \]
Thus, the equation becomes: \[ q \left( \frac{\sigma}{2 \epsilon_0} \right) = mg \]
Rearranging to solve for \( \sigma \): \[ \sigma = \frac{2g m}{q} \]
Substitute the known values: \[ \sigma = \frac{2 \times 8.85 \times 10^{-12} \times 100 \times 10^{-6} \times 10}{10 \times 10^{-6}} \] \[ \sigma = 17.7 \times 10^{-10} C/m^2 \] \[ \sigma = 1.77 nC/cm^2 \]

Thus, the charge density of the sheet is \( 1.77 \) nC/cm\(^2\). Quick Tip: When calculating the charge density from the angle, consider the forces acting on the object and how the electric field interacts with the charge. Use the equilibrium condition to set up the necessary equations.

The maximum kinetic energy \( K_E \) of the electron is given by: \[ K_E = \frac{hc}{\lambda} - \phi \]
where \( p \) is the momentum of the electron, and the relation for momentum is: \[ p = \sqrt{2m K_E} = \sqrt{2m \left( \frac{hc}{\lambda} - \phi \right)} \]
Since the motion is in a magnetic field, the radius of the circular path is: \[ d_{A-B} = 2R = \frac{p}{qB} \]
Thus, the distance between A and B becomes: \[ d_{A-B} = \frac{2}{eB} \sqrt{2m \left( \frac{hc}{\lambda} - \phi \right)} = \frac{\sqrt{8m \left( \frac{hc}{\lambda} - \phi \right)}}{eB} \] Quick Tip: In problems involving magnetic fields, the radius of the electron’s path is related to its momentum and the magnetic field. The formula for the distance is derived by equating the magnetic force to the centripetal force.

The force \( F_{ext} \) required to prevent the door from opening is given by: \[ F_{ext} + F_w = F_t \]
where \( F_w \) is the force due to water and \( F_t \) is the total force on the window.

In equilibrium: \[ F_{ext} = F_t - F_w \]

Now, \( F_t \) is the total force on the window, which is: \[ F_t = (\rho_1 + \rho_2) g h A \]
and \[ F_w = (\rho_1 + \rho_2) g h A \]

Thus, the force needed: \[ F_{ext} = (1500 - 1000) \times 10 \times 10^{-4} \times 150 \] \[ = 150 \, N \]


Quick Tip: For problems involving fluid pressure and forces, the total force on an object can be computed by integrating the pressure over the area. Pay attention to the different fluid densities and heights.

Given: \[ \ell = 2 \, m, \quad Y = 2 \times 10^{11} \, N/m^2 \]

The elastic potential energy density \( \mu \) is given by: \[ \mu = \frac{\Delta \varepsilon}{\ell} = \frac{Y \Delta r}{r} \]
where \( \Delta r \) is the elongation.

Now, for transverse strain \( u \), we use the formula: \[ u = \frac{1}{2} \times Poisson's ratio \times \left(\frac{\Delta \varepsilon}{\ell}\right) \]

Substitute the values to get the energy density: \[ \mu = \frac{5 \times 10^{-3}}{2} \times 2 \times 10^{11} \times \left[ 5 \times 10^{-3} \right]^2 = 25 \]

Thus, the elastic potential energy density is \( 25 \times 10^6 \, N/m^2 \). Quick Tip: In problems involving Young's modulus and strain, the potential energy density can be calculated using the relationship between the strain and the applied stress, considering Poisson's ratio and the transverse strain.

The angular separation for the minima in a single-slit diffraction is given by:
\[ \theta_1 = \sin^{-1}\left( \frac{2\lambda}{a} \right), \quad \theta_2 = \sin^{-1}\left( \frac{3\lambda}{a} \right) \]
where \( \lambda = 628 \, nm \) is the wavelength and \( a \) is the slit width. Also, we know: \[ \theta_1 + \theta_2 = 30^\circ \] \[ \Rightarrow \sin^{-1}\left( \frac{2\lambda}{a} \right) + \sin^{-1}\left( \frac{3\lambda}{a} \right) = \frac{\pi}{6} \]

Solving this, we find: \[ a = 6.07 \, \mu m \]

Thus, the width of the slit is \( a = 6 \, \mu m \). Quick Tip: In single-slit diffraction, the angular separation between adjacent minima is used to calculate the width of the slit. Ensure that you convert the wavelength to meters and carefully solve for the slit width.

\[ \frac{\gamma_A}{\gamma_B} = \frac{f_A + 2}{f_A} : for monoatomic gas A \] \[ \frac{\gamma_B}{\gamma_B} = \frac{f_B + 2}{f_B} : for polyatomic gas B \]
For monoatomic gas A: \[ f_A = 3 \quad (translational degrees of freedom) \]
For polyatomic gas B: \[ f_B = 3 + 3 + 1 = 7 \quad (translational, rotational, and vibrational modes) \]
Substituting these values into the formula: \[ \frac{\gamma_A}{\gamma_B} = \frac{3+2}{3} : \frac{7+2}{7} = \frac{5}{3} : \frac{9}{7} \] \[ \frac{5}{3} : \frac{9}{7} = \left( 1 + \frac{1}{n} \right) \] \[ \frac{5}{3} \cdot \frac{7}{9} = 1 + \frac{1}{n} \] \[ \frac{35}{27} = 1 + \frac{1}{n} \] \[ \frac{35}{27} - 1 = \frac{1}{n} \] \[ \frac{8}{27} = \frac{1}{n} \Rightarrow n = 3 \] Quick Tip: For problems involving specific heat ratios, break down the equation and substitute values for the translational, rotational, and vibrational degrees of freedom separately. This will help you arrive at the correct value of \(n\).

Given: \[ v_{avg} = \frac{x_1 + x_2}{t_1 + t_2} \]
Where \( x_1 = x \), \( x_2 = \frac{3x}{2} \), \( v_1 = 5 \, m/s \), and \( v_2 \) is the unknown velocity. Substituting the values: \[ v_{avg} = \frac{50}{7} \, m/s \] \[ \Rightarrow \frac{50}{7} = \frac{x + \frac{3x}{2}}{\frac{x}{v_1} + \frac{3x}{2v_2}} \] \[ \Rightarrow \frac{50}{7} = \frac{\frac{5x}{2}}{\frac{x}{5} + \frac{3x}{2v_2}} \]
Simplifying the equation: \[ \Rightarrow \frac{50}{7} = \frac{5x}{2} \times \frac{5}{x} \quad (by cross-multiplying) \] \[ \Rightarrow v_2 = 10 \, m/s \] Quick Tip: When calculating average velocity in non-uniform motion, break the total distance and time into separate parts. Apply the formula \( v_{avg} = \frac{total distance}{total time} \) to each segment and solve for the unknown variable.