JEE Main 2026 5th April Evening Shift Question Paper with Solutions

Speed of light ($c$): $[c] = [L T^{-1}]$

Planck's constant ($h$): From $E = h\nu \implies [h] = \frac{[M L^2 T^{-2}]}{[T^{-1}]} = [M L^2 T^{-1}]$

Gravitational constant ($G$): From $F = \frac{G M^2}{R^2} \implies [G] = \frac{[M L T^{-2}][L^2]}{[M^2]} = [M^{-1} L^3 T^{-2}]$

$\left[ \sqrt{\frac{Gh}{c^3}} \right] = [L]$

$\left[ \sqrt{\frac{Gh}{c^5}} \right] = [T]$

$\left[ \sqrt{\frac{hc}{G}} \right] = [M]$

[$\sqrt{\frac{K^2 L^2 c^3}{Gh}}$] = $[K]$

$\frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I}$

$R = \frac{V}{I} = \frac{10\text{ V}}{5\text{ A}} = 2\ \Omega$

$\frac{\Delta R}{2} = \frac{0.5}{10} + \frac{0.2}{5}$

$\frac{\Delta R}{2} = 0.05 + 0.04 = 0.09$

$\Delta R = 0.09 \times 2 = 0.18\ \Omega$

$f = mg \sin\theta$

$f = 1 \times 10 \times \sin(30^\circ) = 10 \times \frac{1}{2} = 5\text{ N}$

$s = v \times t = 4\text{ m/s} \times 2\text{ s} = 8\text{ m}$

$\alpha = 90^\circ - 30^\circ = 60^\circ$

$W_f = f \cdot s \cdot \cos\alpha$

$W_f = 5\text{ N} \times 8\text{ m} \times \cos(60^\circ)$

$W_f = 40 \times \frac{1}{2} = 20\text{ J}$

From the velocity-time graph:

• Positive triangular region from 0 to 20 s
• Negative triangular region from 20 to 40 s

Distance travelled is total area under the speed-time graph.

First triangle:

$\text{Area}_1=\frac{1}{2}\times20\times5=50\ \text{m}$

Second triangle (take magnitude):

$\text{Area}_2=\frac{1}{2}\times20\times5=50\ \text{m}$

Total distance:

$50+50=100\ \text{m}$

Net displacement:

$50-50=0$

Average velocity:

$v_{\text{avg}}=\frac{\text{displacement}}{\text{total time}}$

$v_{\text{avg}}=\frac{0}{40}=0\ \text{m/s}$

$\theta = \omega_0 t + \frac{1}{2}\alpha t^2$

$\theta = \frac{1}{2}\alpha t^2$

$\theta_1 = \frac{1}{2}\alpha (2)^2 = \frac{1}{2}\alpha (4) = 2\alpha$

$\theta_{\text{total}} = \frac{1}{2}\alpha (4)^2 = \frac{1}{2}\alpha (16) = 8\alpha$

$\theta_2 = \theta_{\text{total}} - \theta_1$

$\theta_2 = 8\alpha - 2\alpha = 6\alpha$

$\frac{\theta_2}{\theta_1} = \frac{6\alpha}{2\alpha} = 3$

$K_{\text{total}} = K_{\text{trans}} + K_{\text{rot}} = \frac{1}{2}mv_0^2 + \frac{1}{2}I\omega_0^2$

$K_{\text{total}} = mgh$ (conservation of energy)

$K_{\text{total}} = \frac{1}{2}mv_0^2 + \frac{1}{2}(kmR^2)\left(\frac{v_0}{R}\right)^2$

$K_{\text{total}} = \frac{1}{2}mv_0^2 + \frac{1}{2}kmv_0^2 = \frac{1}{2}mv_0^2(1 + k)$

$\frac{1}{2}mv_0^2(1 + k) = mgh$

$\frac{1}{2}v_0^2(1 + k) = gh \implies h = \frac{v_0^2(1 + k)}{2g}$

$\frac{v_0^2(1 + k)}{2g} = \frac{7v_0^2}{10g}$

$\frac{1 + k}{2} = \frac{7}{10}$

$k = 1.4 - 1 = 0.4 = \frac{2}{5}$

Hence, it is a solid sphere.

The initial potential energy of the body at the surface of the earth is:

$U_i=-\frac{GMm}{R_e}$

The body is taken to a height $h=2R_e$.

The final distance of the body from the center of the earth is:

$r=R_e+h=R_e+2R_e=3R_e$

The final potential energy of the body at this height is:

$U_f=-\frac{GMm}{3R_e}$

The increase in potential energy ($\Delta U$) is the difference between the final and initial potential energies:

$\Delta U=U_f-U_i$

$\Delta U=\left(-\frac{GMm}{3R_e}\right)-\left(-\frac{GMm}{R_e}\right)$

$\Delta U=\frac{GMm}{R_e}-\frac{GMm}{3R_e}$

$\Delta U=GMm\left(\frac{1}{R_e}-\frac{1}{3R_e}\right)$

$\Delta U=GMm\left(\frac{3-1}{3R_e}\right)$

$\Delta U=\frac{2GMm}{3R_e}$

$g=\frac{GM}{R_e^2}\implies GM=gR_e^2$

$\Delta U=\frac{2(gR_e^2)m}{3R_e}$

$\Delta U=\frac{2}{3}mgR_e$

$E = \text{Surface Tension } (S) \times \text{Surface Area } (A)$

$\Delta E = E_{\text{initial}} - E_{\text{final}} = S \cdot (A_{\text{initial}} - A_{\text{final}})$

$\text{Volume of big drop} = 8 \times \text{Volume of one small drop}$

$\frac{4}{3}\pi R^3 = 8 \times \left(\frac{4}{3}\pi r^3\right)$

$R^3 = 8r^3 \implies R = \sqrt[3]{8r^3} = 2r$

$A_{\text{initial}} = 8 \times (4\pi r^2) = 32\pi r^2$

$A_{\text{final}} = 4\pi R^2 = 4\pi (2r)^2 = 4\pi (4r^2) = 16\pi r^2$

$\Delta E = S \cdot \left(A_{\text{initial}} - A_{\text{final}}\right)$

$\Delta E = S \cdot \left(32\pi r^2 - 16\pi r^2\right)$

$\Delta E = 16\pi r^2 S$

$PT^3 = \text{constant}$

For an ideal gas, the ideal gas equation of state is:

$PV = nRT \implies P = \frac{nRT}{V}$

$\left(\frac{nRT}{V}\right) T^3 = \text{constant}$

$\frac{T^4}{V} = \frac{\text{constant}}{nR}$

Since $n$ (number of moles) and $R$ (universal gas constant) are constants, we can replace the entire right side of the equation with a new constant, let's call it $K$:

$V = KT^4$

volume expansion ($\gamma$)

$\gamma = \frac{1}{V} \frac{dV}{dT}$

$\frac{dV}{dT} = \frac{d}{dT}(KT^4)$

$\frac{dV}{dT} = 4KT^3$

$\gamma = \frac{1}{KT^4} (4KT^3)$

$\gamma = \frac{4KT^3}{KT^4}$

$\gamma = \frac{4}{T}$

A. $\sin^2\omega t$ can be rewritten as $\frac{1}{2} - \frac{1}{2}\cos(2\omega t)$. Since it consists of a single sinusoidal term, it represents SHM with a time period $T = \frac{2\pi}{2\omega} = \frac{\pi}{\omega}$ $\rightarrow \text{III}$

$d\varepsilon = B(r) \cdot v \cdot dr = B(r) \cdot (r\omega) \cdot dr$

$\varepsilon = \int_{0}^{L} B_0 e^{-\lambda r} \cdot (r\omega) \, dr$

$\varepsilon = B_0 \omega \int_{0}^{L} r \cdot e^{-\lambda r} \, dr$

$\int r e^{-\lambda r} \, dr = r \left( -\frac{1}{\lambda} e^{-\lambda r} \right) - \int \left( -\frac{1}{\lambda} e^{-\lambda r} \right) \, dr$ (Integration by parts)

$\int r e^{-\lambda r} \, dr = -\frac{r}{\lambda} e^{-\lambda r} + \frac{1}{\lambda} \int e^{-\lambda r} \, dr$

$\int r e^{-\lambda r} \, dr = -\frac{r}{\lambda} e^{-\lambda r} - \frac{1}{\lambda^2} e^{-\lambda r}$

$\int r e^{-\lambda r} \, dr = -e^{-\lambda r} \left( \frac{r}{\lambda} + \frac{1}{\lambda^2} \right)$

$\varepsilon = B_0 \omega \left[ -e^{-\lambda r} \left( \frac{r}{\lambda} + \frac{1}{\lambda^2} \right) \right]_{0}^{L}$

$\varepsilon = B_0 \omega \left[ -e^{-\lambda L} \left( \frac{1}{\lambda^2} + \frac{L}{\lambda} \right) - \left( -\frac{1}{\lambda^2} \right) \right]$

$\varepsilon = B_0 \omega \left[ \frac{1}{\lambda^2} - e^{-\lambda L} \left( \frac{1}{\lambda^2} + \frac{L}{\lambda} \right) \right]$

Under steady-state DC conditions, the capacitor behaves as an open circuit.

Therefore, no current flows through the middle branch containing the capacitor and the $4\,\Omega$ resistor.

Hence, the current flows only through the outer loop containing $2\,\Omega$ and $6\,\Omega$ resistors and the $2\,V$ battery.

Let the current in the outer loop be $I$.

Applying Kirchhoff's voltage law,

$2I+6I=2$

$8I=2$

$I=0.25\,A$

The potential difference between the left and right nodes is the potential drop across the $2\,\Omega$ resistor:  $V=IR$

$V=0.25\times2$

$V=0.5\,V$

Since no current flows through the $4\,\Omega$ resistor, there is no potential drop across it. Therefore, the potential difference across the capacitor is the same as the potential difference between the two nodes.

Thus,  $V_C=0.5\,V$

Hence, the correct option is A.

$\Delta K = W_{\text{total}} = W_{\text{electric}} + W_{\text{magnetic}}$

The magnetic force acting on a moving charge is given by the Lorentz force expression $\vec{F}_B = q(\vec{v} \times \vec{B})$. Since $\vec{F}_B$ is always perpendicular to the velocity vector $\vec{v}$ (and consequently perpendicular to the instantaneous displacement vector $d\vec{r}$), it does zero work on the particle:

$W_{\text{magnetic}} = \int \vec{F}_B \cdot d\vec{r} = 0$

$W_{\text{electric}} = \int \vec{F}_E \cdot d\vec{r}$

$\Delta K = W_{\text{electric}} = \int_{x=0}^{x=L} (q\vec{E}) \cdot d\vec{r}$

$\Delta K = \int_{0}^{L} \left( q E_0 e^{-\lambda x} \hat{i} \right) \cdot \left( dx\hat{i} + dy\hat{j} + dz\hat{k} \right)$

$\Delta K = \int_{0}^{L} q E_0 e^{-\lambda x} \, dx$

$\Delta K = q E_0 \left[ \frac{e^{-\lambda x}}{-\lambda} \right]_{0}^{L}$

$\Delta K = -\frac{q E_0}{\lambda} \left( e^{-\lambda L} - 1 \right)$

$\Delta K = \frac{q E_0}{\lambda} \left( 1 - e^{-\lambda L} \right)$

Assertion: EM waves carry energy ($U$) and momentum ($p$). For a completely absorbing surface, the momentum delivered is $p = \frac{U}{c}$. The rate of change of this momentum produces a net force. Therefore, EM waves do exert a physical pressure (radiation pressure) on any surface they strike. Hence, Assertion is true.

Reason: EM fields are configurations of energy, and their constituent particles (photons) possess zero rest mass. Hence, Reason is true.

EM waves exert pressure precisely because they carry linear momentum and energy, which can be transferred to a surface during a collision or absorption. The fact that they have no rest mass does not cause this pressure; rather, it highlights the unique relativistic property that a particle or wave can carry linear momentum ($p = \frac{E}{c}$) even when its rest mass is zero.

Since both statements are independently true but the Reason does not provide the causal explanation for the Assertion, the correct option is B.

$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$

Convex lens has a positive focal length ($f_{\text{convex}} = +f_1$)

Concave lens has a negative focal length ($f_{\text{concave}} = -f_2$)

$\frac{1}{F} = \frac{1}{|f_{\text{convex}}|} - \frac{1}{|f_{\text{concave}}|}$

$\frac{1}{F} = \frac{|f_{\text{concave}}| - |f_{\text{convex}}|}{|f_{\text{convex}}| \cdot |f_{\text{concave}}|}$

$F = \frac{|f_{\text{convex}}| \cdot |f_{\text{concave}}|}{|f_{\text{concave}}| - |f_{\text{convex}}|}$

For the combination to act as a net concave lens, its equivalent focal length must be negative ($F < 0$). This occurs when the denominator is less than zero:

$|f_{\text{concave}}| - |f_{\text{convex}}| < 0 \implies |f_{\text{convex}}| > |f_{\text{concave}}|$

For the combination to act as a net convex lens, its equivalent focal length must be positive ($F > 0$). This occurs when the denominator is greater than zero:

$|f_{\text{concave}}| - |f_{\text{convex}}| > 0 \implies |f_{\text{convex}}| < |f_{\text{concave}}|$

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$m_{\text{total}} = m_1 \times m_2 = \left(\frac{v_1}{u_1}\right) \times \left(\frac{v_2}{u_2}\right)$

$\frac{1}{v_1} - \frac{1}{-15} = \frac{1}{10}$

$\frac{1}{v_1} + \frac{1}{15} = \frac{1}{10} \implies \frac{1}{v_1} = \frac{1}{10} - \frac{1}{15}$

$\frac{1}{v_1} = \frac{3 - 2}{30} = \frac{1}{30} \implies v_1 = +30\text{ cm}$

Lens $P$ forms a real image $30\text{ cm}$ to its right.

$m_1 = \frac{v_1}{u_1} = \frac{30}{-15} = -2$

Lens $Q$ is located $15\text{ cm}$ to the right of lens $P$.

Because the image from lens $P$ forms at $30\text{ cm}$ to the right of $P$, it ends up further down the path, exactly:

$\text{Distance from Q} = 30\text{ cm} - 15\text{ cm} = 15\text{ cm} \text{ (to the right of } Q)$

Since the light rays converge toward this position behind lens $Q$, it serves as a virtual object for lens $Q$:

Object distance for $Q$, $u_2 = +15\text{ cm}$

Focal length of $Q$, $f_2 = +15\text{ cm}$

$\frac{1}{v_2} - \frac{1}{+15} = \frac{1}{15}$

$\frac{1}{v_2} = \frac{1}{15} + \frac{1}{15} = \frac{2}{15} \implies v_2 = \frac{15}{2} = +7.5\text{ cm}$

$m_2 = \frac{v_2}{u_2} = \frac{7.5}{15} = +\frac{1}{2}$

$m_{\text{total}} = m_1 \times m_2 = (-2) \times \left(+\frac{1}{2}\right) = -1$

An absolute total magnification value of $|m_{\text{total}}| = 1$ means the final image has the exact same size as the original object $AB$ (the negative sign simply means it is inverted).

The point is described as being "exactly opposite to one of the slits." Since the two slits are separated by a distance $d$ symmetrically about the central axis, the distance from the central maximum to a point directly opposite one of the slits is: $y = \frac{d}{2}$

$\Delta x = \frac{y \cdot d}{D} = \frac{\left(\frac{d}{2}\right) \cdot d}{10d} = \frac{d^2}{20d} = \frac{d}{20}$

$\Delta x = \frac{5\lambda}{20} = \frac{\lambda}{4}$

$\phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{\lambda} \left(\frac{\lambda}{4}\right) = \frac{\pi}{2}$

$I = I_0 \cos^2\left(\frac{\pi / 2}{2}\right) = I_0 \cos^2\left(\frac{\pi}{4}\right)$

$I = I_0 \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{I_0}{2}$

$\lambda = \frac{h}{p} = \frac{h}{mv}$

$\lambda \propto \frac{1}{v}$

$v_m = v - 20\% \text{ of } v = v - 0.20v = 0.80v = \frac{4}{5}v$

$\frac{\lambda_m}{\lambda_0} = \frac{v_0}{v_m}$

$\frac{\lambda_m}{\lambda_0} = \frac{v}{\frac{4}{5}v} = \frac{5}{4} = 1.25$

$\lambda_m = 1.25\lambda_0$

$\alpha = 1.25$

$\Delta m = [Z \cdot m_p + (A - Z) \cdot m_n] - M_{\text{actual}}$

$m_{\text{nucleons}} = (6 \times 1.00727 \text{ u}) + (6 \times 1.00866 \text{ u})$

$m_{\text{nucleons}} = 6.04362 \text{ u} + 6.05196 \text{ u} = 12.09558 \text{ u}$

$\Delta m = 12.09558 \text{ u} - 12 \text{ u} = 0.09558 \text{ u}$

$\Delta m = 0.09558 \times 931.5 \text{ MeV}/c^2$

$\Delta m \approx 89.03277 \text{ MeV}/c^2 \approx 89.03 \text{ MeV}/c^2$

$\text{Width } (x) \propto \frac{1}{\text{Doping concentration}}$

$N_D \gg N_A \quad (10^{18} > 10^{15})$

Hence, the depletion region must extend much deeper into the lightly doped p-side than into the heavily doped n-side.

$U = \frac{1}{2} \times \text{Stress} \times \text{Strain} \times \text{Volume}$

$U = \frac{1}{2} \times Y \times (\text{Strain})^2 \times \text{Volume}$

$\text{Strain} = \frac{\Delta L}{L} = \frac{3 \times 10^{-3}\text{ m}}{3\text{ m}} = 10^{-3}$

$U = \frac{1}{2} \times \left(1.1 \times 10^{11}\right) \times \left(10^{-3}\right)^2 \times \left(600 \times 10^{-6}\right)$

$U = \frac{1}{2} \times 1.1 \times 10^{11} \times 10^{-6} \times 600 \times 10^{-6}$

$U = 330 \times 10^{-1} = 33\text{ J}$

$\text{Heat Lost} = \text{Heat Gained}$

$Q = mc\Delta T$ (temperature change without phase change)

$Q = mL$ (Phase change)

The first part of the water ($x$ grams) is cooled from $50^\circ\text{C}$ to $0^\circ\text{C}$.

$Q_{\text{extracted}} = m_1 \cdot c \cdot \Delta T$

$Q_{\text{extracted}} = \left(x \times 10^{-3}\right) \times 4200 \times 50$

$Q_{\text{extracted}} = x \times 10^{-3} \times 210,000 = 210x\text{ J}$

The remaining water, $(1000-x)$ grams, is initially at $50^\circ\text{C}$. To evaporate it, it must first be heated to its boiling point ($100^\circ\text{C}$) and then converted into steam at $100^\circ\text{C}$.

The total heat required ($Q_{\text{required}}$) consists of two parts:

Heating from $50^\circ\text{C}$ to $100^\circ\text{C}$ ($\Delta T_2 = 100 - 50 = 50\text{ K}$): $Q_{\text{heating}} = m_2 \cdot c \cdot \Delta T_2$

Evaporating at $100^\circ\text{C}$: $Q_{\text{phase\_change}} = m_2 \cdot L$

$Q_{\text{required}} = m_2 \left(c\Delta T_2 + L\right)$

$Q_{\text{required}} = (1000 - x) \times 10^{-3} \times \left[(4200 \times 50) + \left(2256 \times 10^3\right)\right]$

$Q_{\text{required}} = 2466(1000 - x)\text{ J}$

$Q_{\text{extracted}} = Q_{\text{required}}$

$210x = 2466(1000 - x)$

$x = \frac{2,466,000}{2676}$

$x \approx 921.52 \approx 922$ 

$\omega_r = \frac{1}{\sqrt{LC}}$

$\omega_r = \frac{1}{\sqrt{1.6 \times 40 \times 10^{-6}}}$

$\omega_r = \frac{1}{\sqrt{64 \times 10^{-6}}} = \frac{1}{8 \times 10^{-3}} = 125 \text{ rad/s}$

$X_L = \omega_r L$

$X_L = 125 \times 1.6 = 200 \ \Omega$

$I = \frac{E}{R + r}$

$E = 0.25(5 + r) \quad \text{--- (Equation 1)}$

$E = 0.5(2 + r) \quad \text{--- (Equation 2)}$

Since the cell remains the same, its $\text{emf}$ ($E$) is constant: $0.25(5 + r) = 0.5(2 + r)$

$5 + r = 4 + 2r$

$r = 1\ \Omega$

$\Phi_B = B \cdot A = (2t^2 + 2t + 3) \cdot (0.04\pi)$

$\varepsilon = \frac{d\Phi_B}{dt} = 0.04\pi \cdot \frac{d}{dt}(2t^2 + 2t + 3)$

$\varepsilon = 0.04\pi \cdot (4t + 2)$

$\varepsilon_{t=3} = 0.04\pi \cdot (4(3) + 2) = 0.04\pi \cdot (12 + 2) = 0.04\pi \cdot 14$

$\varepsilon_{t=3} = 0.56\pi\text{ V}$

$I = \frac{\varepsilon}{R} = \frac{0.56\pi}{2} = 0.28\pi\text{ A}$

$I = 0.28 \times \frac{22}{7} = 0.04 \times 22 = 0.88\text{ A}$

$\frac{\alpha}{50} = 0.88$

$\alpha = 0.88 \times 50 = 44$

Step 1:

Chemical Equation:

$Zn+H_2SO_4\longrightarrow\ ZnSO_4+H_2$