JEE Main 2025 April 4 Shift 1 Physics Question Paper, Exam Analysis, and Answer Keys (Available)

1. Given:
- Mean free path, \(\lambda = 3 \times 10^{-7} \mathrm{~m}\)
- Average speed, \(v_{avg} = 600 \mathrm{~m/s}\)

2. Frequency of collisions:
\[ Frequency = \frac{v_{avg}}{\lambda} = \frac{600}{3 \times 10^{-7}} = 2 \times 10^{9} \mathrm{~s^{-1}} \]

Therefore, the correct answer is (3) \(2 \times 10^{9} / \mathrm{s}\). Quick Tip: The frequency of collisions is given by the average speed divided by the mean free path.

1. Force due to the laser beam:
\[ F = \frac{2P}{c} = \frac{2}{c} \frac{dE}{dt} \]

2. Change in momentum of the mirror:
\[ m(V - 0) = \int F dt = \frac{2}{c} \int dE = \frac{2E}{c} \]

3. Using work-energy theorem:
\[ W_g = \Delta K \]
\[ -\frac{mg l (1 - \cos \theta)}{l} = \frac{1}{2} m V^2 \]
\[ \frac{g l (1 - \cos \theta)}{l} = \frac{1}{2} \left( \frac{4E^2}{m^2 c^2} \right) \]
\[ \frac{g l \theta^2}{2} = \frac{2E^2}{m^2 c^2} \]
\[ \theta = \frac{2E}{mc \sqrt{gl}} \]

Therefore, the correct answer is (4) \(\theta=\frac{2 \mathrm{E}}{\mathrm{mc} \sqrt{\mathrm{gl}}}\). Quick Tip: Use the work-energy theorem to find the deflection angle.

1. Given:
\[ K = \frac{\cos \theta_{\mathrm{A}}}{\cos \theta_{\mathrm{B}}} \]

2. Interpretation:
- If \(K\) is negative, \(\cos \theta_{\mathrm{A}}\) and \(\cos \theta_{\mathrm{B}}\) are of opposite signs.
- This implies that one liquid has a concave meniscus and the other has a convex meniscus.

Therefore, the correct answer is (3) K is negative, then liquid A has concave meniscus and liquid B has convex meniscus. Quick Tip: The sign of \(K\) indicates the nature of the meniscus of the liquids.

1. Correct expressions for torque:
- B. \(\ddot{\tau}=\frac{\mathrm{d}}{\mathrm{dt}}(\ddot{\mathrm{r}} \times \ddot{\mathrm{p}})\)
- C. \(\ddot{\tau}=\ddot{\mathrm{r}} \times \frac{\mathrm{d} \dot{\mathrm{p}}}{\mathrm{dt}}\)
- D. \(\ddot{\tau}=\mathrm{I} \dot{\alpha}\)
- E. \(\ddot{\tau}=\ddot{\mathrm{r}} \times \ddot{\mathrm{F}}\)

Therefore, the correct answer is (3) B, C, D and E Only. Quick Tip: Torque can be expressed in terms of position vector, linear momentum, angular momentum, and force.

1. Fringe width formula:
\[ \beta = \frac{D \lambda}{d} \]

2. Percentage change of fringe width:
- If \(d\) is doubled, \(\beta\) is halved.
- Therefore, the percentage change is \(50%\).

Therefore, the correct answer is (3) \(50 %\). Quick Tip: The fringe width is inversely proportional to the slit separation.

1. RMS value of current:
\[ i_{rms} = \frac{i_0}{\sqrt{2}} = 100 \mathrm{~A} \]

2. Frequency of the current:
\[ f = \frac{\omega}{2\pi} = \frac{100\pi}{2\pi} = 50 \mathrm{~Hz} \]

Therefore, the correct answer is (3) \(100 \mathrm{~A}, 50 \mathrm{~Hz}\). Quick Tip: The RMS value of an alternating current is given by the peak value divided by \(\sqrt{2}\).

1. Speed of sound formula:
\[ v_{sound} = \sqrt{\frac{\gamma P}{\rho}} \]

2. Ratio of specific heats (\(\gamma\)):
- \(\gamma_{\mathrm{He}} = \frac{5}{3}\)
- \(\gamma_{\mathrm{CH}_{4}} \approx \frac{4}{3}\)
- \(\gamma_{\mathrm{CO}_{2}} \approx \frac{4}{3}\)

3. Ratio of speeds of sound:
\[ \sqrt{\frac{5}{3}}: \sqrt{\frac{4}{3}}: \sqrt{\frac{4}{3}} \]

Therefore, the correct answer is (3) \(\sqrt{\frac{5}{3}}: \sqrt{\frac{4}{3}}: \sqrt{\frac{4}{3}}\). Quick Tip: The speed of sound in a gas depends on the ratio of specific heats (\(\gamma\)).

1. Dimensions of electric flux (\(\phi_E\)) and magnetic flux (\(\phi_M\)):
\[ \phi_E = EA \quad and \quad \phi_M = BA \]

2. Ratio of electric flux to magnetic flux:
\[ \frac{\phi_E}{\phi_M} = \frac{E}{B} \]

3. Dimensions of \(\frac{E}{B}\):
\[ \left[ \frac{E}{B} \right] = \frac{[E]}{[B]} = \frac{MLT^{-2}A^{-1}}{MT^{-2}A^{-1}} = LT^{-1} \]

4. Dimensions of \(\frac{E}{B}\):
\[ \left[ \frac{E}{B} \right] = LT^{-1} \]

Therefore, the correct answer is (4) \((1,-1)\). Quick Tip: The ratio of electric flux to magnetic flux has dimensions of \(LT^{-1}\).

1. Given:
- Object distance, \(u = -40 \mathrm{~cm}\)
- Magnification, \(m = \frac{1}{2}\)

2. Magnification formula:
\[ m = \frac{f}{f - u} \]
\[ \frac{1}{2} = \frac{f}{f - (-40)} \]
\[ f + 40 = 2f \implies f = 40 \mathrm{~cm} \]

3. For magnification \(\frac{1}{3}\):
\[ \frac{1}{3} = \frac{40}{40 - u} \]
\[ 40 - u = 120 \implies u = -80 \mathrm{~cm} \]

Therefore, the correct answer is (1) 40 cm away from the mirror. Quick Tip: Use the magnification formula to find the object distance.

1. Assertion A:
- The stopping potential does not depend on the intensity of light.
- Therefore, Assertion A is false.

2. Reason R:
- Increasing the intensity of light increases the rate of photoelectrons emitted.
- Therefore, Reason R is true.

Therefore, the correct answer is (2) \(\mathbf{A}\) is false but \(\mathbf{R}\) is true. Quick Tip: The stopping potential in the photoelectric effect does not depend on the intensity of light.

1. Position vector:
\[ \overrightarrow{\mathrm{r}} = \mathrm{a}(\hat{\mathrm{i}} \cos \omega \mathrm{t} + \hat{\mathrm{j}} \sin \omega \mathrm{t}) \]

2. Acceleration:
\[ \overrightarrow{\mathrm{a}} = -\omega^2 \overrightarrow{\mathrm{r}} \]

3. Force:
\[ \overrightarrow{\mathrm{F}} = m \overrightarrow{\mathrm{a}} = -m \omega^2 \overrightarrow{\mathrm{r}} \]

Therefore, the direction of force is opposite to the direction of \(\overrightarrow{\mathrm{r}}\). Quick Tip: The force is proportional to the acceleration, which is opposite to the position vector in this case.

1. Given:
\[ \mathrm{T}_{1} \sin \theta_{1} + \mathrm{T}_{2} \sin \theta_{2} = \mathrm{mg} \]
\[ \mathrm{T}_{1} = \sqrt{3} \mathrm{~T}_{2} \]

2. Substitute \(\mathrm{T}_{1}\):
\[ \sqrt{3} \mathrm{~T}_{2} \sin \theta_{1} + \mathrm{T}_{2} \sin \theta_{2} = \mathrm{mg} \]
\[ \mathrm{T}_{2} (\sqrt{3} \sin \theta_{1} + \sin \theta_{2}) = \mathrm{mg} \]

3. For \(\theta_{1} = 60^{\circ}\) and \(\theta_{2} = 30^{\circ}\):
\[ \mathrm{T}_{2} (\sqrt{3} \cdot \frac{\sqrt{3}}{2} + \frac{1}{2}) = \mathrm{mg} \]
\[ \mathrm{T}_{2} = \frac{\mathrm{mg}}{2} \]

Therefore, the correct answer is (2) \(\theta_{1}=60^{\circ} \theta_{2}=30^{\circ}\) with \(\mathrm{T}_{2}=\frac{\mathrm{mg}}{2}\). Quick Tip: Use the equilibrium condition to find the tensions and angles.

1. Charge calculation:
\[ q = \int I(t) dt \]
\[ q = \int_{1}^{2} (0.02t + 0.01) dt \]
\[ q = \left[ 0.02 \frac{t^2}{2} + 0.01t \right]_{1}^{2} \]
\[ q = \left[ 0.01(4) + 0.01(2) \right] - \left[ 0.01(1) + 0.01(1) \right] \]
\[ q = 0.04 \mathrm{~C} \]

Therefore, the correct answer is (4) 0.04 C. Quick Tip: Integrate the current function to find the charge.

1. Assertion A:
- The kinetic energy needed to project a body of mass \(m\) from earth surface to infinity is \(\frac{1}{2} \mathrm{mgR}\).
- This is incorrect.

2. Reason R:
- The maximum potential energy of a body is zero when it is projected to infinity from earth surface.
- This is correct.

Therefore, the correct answer is (1) A False but \(\mathbf{R}\) is true. Quick Tip: The kinetic energy needed to project a body to infinity is equal to the potential energy at the earth's surface.

1. Boolean expression:
\[ Y = A \overline{B} C + \overline{AC} \]

2. Gate configurations:
- A. One 3-input AND gate, 3 NOT gates and one 2-input OR gate, One 2-input AND gate
- B. One 3-input AND gate, 1 NOT gate, One 2-input NOR gate and one 2-input OR gate

Therefore, the correct answer is (2) A,B Only. Quick Tip: Use the Boolean expression to determine the correct gate configurations.

1. Initial frequency:
\[ f_1 = \frac{v}{4l} \]

2. Frequency with water:
\[ f_2 = \frac{v}{4(4l/5)} = \frac{5v}{16l} \]

3. Percentage change:
\[ \frac{\Delta f}{f} = \frac{\frac{5v}{16l} - \frac{v}{4l}}{\frac{v}{4l}} \times 100 = 25% \]

Therefore, the correct answer is (1) \(25 %\). Quick Tip: The frequency of the fundamental note changes with the length of the pipe.

1. Angular acceleration:
\[ \alpha = -\omega^2 \theta \]
\[ \omega = \sqrt{\frac{g}{l}} \]

2. Equating angular accelerations:
\[ \frac{g}{l_1} \theta_1 = \frac{g}{l_2} \theta_2 \]
\[ \theta_1 l_2 = \theta_2 l_1 \]

Therefore, the correct answer is (4) \(\theta_{1} l_{2}=\theta_{2} l_{1}\). Quick Tip: The angular acceleration of a pendulum depends on its length and angular displacement.

1. Electric field at points A and B:
- \(\vec{E}_{A} > \vec{E}_{B}\)

2. Electric field at points C and D:
- \(\vec{E}_{C} \neq \vec{E}_{D}\)

Therefore, the correct answer is (3) \(\vec{E}_{C} \neq \vec{E}_{D} ; \vec{E}_{A}>\vec{E}_{B}\). Quick Tip: The electric field depends on the arrangement and charge density of the sheets and spherical body.

1. Electric field due to the charge sheet:
\[ E = \frac{\sigma}{2 \epsilon_0} \]

2. Torque acting on the rod:
\[ \tau = PE \sin \theta \]
\[ \tau = \left[ 2 \times 10^{-6} \times \frac{2}{10} \right] \left[ \frac{100 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}} \right] \frac{1}{2} \]
\[ \tau = \frac{10}{8.85} = 1.12 \mathrm{~Nm} \]

Therefore, the correct answer is (2) 1.12 Nm. Quick Tip: The torque acting on a dipole in an electric field is given by the product of the dipole moment and the electric field.

1. Radius of the \(5^{th}\) orbit for \(\mathrm{Li}^{2+}\):
\[ r_{5} = \frac{5^2}{3} a_0 \]

2. Radius of the \(5^{th}\) orbit for \(\mathrm{He}^{+}\):
\[ r_{5} = \frac{5^2}{2} a_0 \]

3. Ratio of the radii:
\[ \frac{r_{\mathrm{Li}^{2+}}}{r_{\mathrm{He}^{+}}} = \frac{2}{3} \]

Therefore, the correct answer is (4) \(\frac{2}{3}\). Quick Tip: The radius of an orbit in the Bohr model depends on the principal quantum number and the atomic number.

1. Mechanical energy conservation:
\[ K_i + U_i = K_f + U_f \]
\[ 0 + Mgh = \frac{1}{2} mv^2 \left(1 + \frac{k^2}{R^2}\right) + 0 \]
\[ v = \sqrt{\frac{2gh}{1 + \frac{k^2}{R^2}}} \]

2. Ratio of velocities:
\[ \frac{v_{ring}}{v_{solid sphere}} = \sqrt{\frac{1 + \frac{2}{5}}{1 + 1}} = \sqrt{\frac{7}{10}} \]
\[ x = 3.5 \approx 4 \]

Therefore, the correct answer is (4) 4. Quick Tip: Use mechanical energy conservation to find the velocities of the ring and the solid sphere.

1. Deformation angle:
\[ 2 \theta_{1} = \theta_{2} \]
\[ 2 \frac{\sigma_{1}}{\eta_{1}} = \frac{\sigma_{2}}{\eta_{2}} \]
\[ 2 \left( \frac{F}{l d_{1} \eta_{1}} \right) = \frac{F}{l d_{2} \eta_{2}} \]
\[ \eta_{2} = \frac{\eta_{1}}{4} = 1 \times 10^{9} \]
\[ x = 1 \]

Therefore, the correct answer is (1) 1. Quick Tip: Use the deformation angle to find the shear modulus of the material.

1. Given:
\[ m = -\frac{1}{3} \]
\[ -\frac{v}{u} = \frac{1}{3} \implies v = \frac{u}{3} \]

2. Distance between object and image:
\[ u - v = 30 \mathrm{~cm} \]
\[ u - \frac{u}{3} = 30 \implies u = 45 \mathrm{~cm}, \quad v = 15 \mathrm{~cm} \]

3. Focal length:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{15} + \frac{1}{45} = \frac{4}{60} \]
\[ f = \frac{60}{4} = 15 \mathrm{~cm} \]
\[ x = 45 \]

Therefore, the correct answer is (45). Quick Tip: Use the magnification formula to find the object and image distances, then calculate the focal length.

1. Redraw the circuit:
- The capacitors are connected in parallel.

2. Equivalent capacitance:
\[ C_{eq} = 4C = 4 \times 16 \mu \mathrm{F} = 64 \mu \mathrm{F} \]

Therefore, the correct answer is (64) \(64 \mu \mathrm{F}\). Quick Tip: Use the parallel connection of capacitors to find the equivalent capacitance.

1. Induced emf:
\[ \varepsilon = Bv \ell_{\mathrm{AB}} \]
\[ \varepsilon = \frac{1}{\sqrt{2}} \times \frac{10 \mathrm{~cm}}{s} \times 2 \left( 10 \sin 45^{\circ} \right) \mathrm{cm} \]
\[ \varepsilon = 10 \mathrm{~mV} \]

Therefore, the correct answer is (10) 10 mV. Quick Tip: Use Faraday's law to calculate the induced emf.