JEE Main 2026 8th April Evening Shift Question Paper with Solutions

Given:

new unit of length α = speed of light c

so,

1 α = c (m)

distance between Earth and Venus:

distance = speed × time = c × t

time:

6 min 40 s = 6×60 + 40 = 400 s

distance:

= c × 400

but since 1 α = c,

distance = 400 α

Given:

$H=(x^pε^qE^r)/t^s$

step 1: dimensions

H(magnetic\ field)=A/m=AL⁻^1

E(electric\ field)=V/m=MLT⁻^3A⁻^1

ε(permittivity)=M⁻^1L⁻^3T⁴A^2

x = L
t = T

step 2: write RHS dimensions

$x^p→L^p$

$ε^q→M^{-q}L^{-3q}T^{4q}A^{2q}$

$E^r→M^rL^rT^{-3r}A^{-r}$

$t^s→T^s(in\ deno\min ator→T⁻s)$

combine:

M: −q + r
L: p − 3q + r
T: 4q − 3r − s
A: 2q − r

step 3: equate with H=AL⁻^1

M⁰L⁻^1T⁰A^1

equations:

−q + r = 0 ⇒ r = q

p − 3q + r = −1 ⇒ p − 2q = −1

4q − 3r − s = 0 ⇒ q − s = 0 ⇒ s = q

2q − r = 1 ⇒ q = 1

step 4: find others

q = 1 ⇒ r = 1, s = 1

p − 2 = −1 ⇒ p = 1

Solution :

Speed of car :  $54\text{ km h}^{-1}$  $= \frac{54 \times 1000}{3600}$   $= 15\text{ m s}^{-1}$

Radius of turn : $r = 20\text{ m}$

Centripetal acceleration : $a_c = \frac{v^2}{r}$ $= \frac{15^2}{20}$  $= \frac{225}{20}$  $= 11.25\text{ m s}^{-2}$

For pendulum in turning car : $\tan\theta = \frac{a_c}{g}$  $= \frac{11.25}{10}$  $= 1.125$

Solution :

Given :

Initial height of stone,

$u_y = 75\text{ m}$

Initial velocity of stone upward,

$u = 10\text{ m s}^{-1}$

Acceleration due to gravity,

$a = -10\text{ m s}^{-2}$

When stone is released, it has same upward velocity as balloon.

Taking upward direction positive,

Equation of motion :

$y = ut + \frac{1}{2}at^2$

When stone hits ground :

$y = -75\text{ m}$

Therefore,

$-75 = 10t - 5t^2$

$5t^2 - 10t - 75 = 0$

$t^2 - 2t - 15 = 0$

$(t-5)(t+3)=0$

$t = 5\text{ s}$

During this time, balloon continues moving upward with constant speed :

$v = 10\text{ m s}^{-1}$

Distance moved upward by balloon :

$s = vt$

$= 10 \times 5$

$= 50\text{ m}$

Therefore, height of balloon when stone hits ground :

$= 75 + 50$

$= 125\text{ m}$

Final Answer :

$125\text{ m}$

First find focal length of the lens:

$\frac{1}{f}=(μ-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=0.5\left(\frac{1}{20}+\frac{1}{20}\right)=\frac{1}{20}$

so f=20 cm

Now system: lens → reflection (silvered surface) → lens again

This behaves like a combination equivalent to a mirror.

For image to coincide with object, the object must be at the centre of curvature of equivalent mirror.

For this system, effective focal length becomes:

$f_{eq}=\frac{f}{2}=10\text{ cm}$

So centre of curvature:

$R=2f_{eq}=20\text{ cm}$

Solution :

For projectile motion,

$T = \frac{2u\sin\theta}{g}$

Given time of flights :

$T_1 = 5\text{ s}$

$T_2 = 10\text{ s}$

For same horizontal range with same speed, projection angles are complementary.

Hence,

$\sin\theta_2 = \cos\theta_1$

Now,

$T_1T_2=\frac{2u\sin\theta_1}{g}\times\frac{2u\cos\theta_1}{g}$

$=\frac{4u^2\sin\theta_1\cos\theta_1}{g^2}$

Using,

$2\sin\theta\cos\theta = \sin2\theta$

$T_1T_2=\frac{2u^2\sin2\theta_1}{g^2}$

Range of projectile :

$R = \frac{u^2\sin2\theta}{g}$

Therefore,

$T_1T_2 = \frac{2R}{g}$

Substituting values :

$5 \times 10 = \frac{2R}{10}$

$50 = \frac{R}{5}$

$R = 250\text{ m}$

Final Answer :

$250\text{ m}$

Take rightward velocity as positive.

Given:
v₀ = 49 m/s
ω₀ = v₀ / (4R)

For pure rolling:
v = ωR

step 1: friction effects

kinetic friction:

f = μmg = 0.25 × mg

translational deceleration:

a = −f/m = −μg = −0.25 × 9.8 = −2.45 m/s²

step 2: angular acceleration

torque = fR

I (solid cylinder) = (1/2)mR²

$\alpha=\frac{fR}{I}=\frac{μmgR}{(1/2)mR^2}=\frac{2μg}{R}$

$\alpha=\frac{2\times0.25\times9.8}{R}=\frac{4.9}{R}$

step 3: write v and ω

$v=49−2.45t$

$\omega=\frac{49}{4R}+\frac{4.9}{R}t$

step 4: rolling condition

$v=\omega R$

$49−2.45t=\frac{49}{4}+4.9t$

step 5: solve

$49−\frac{49}{4}=4.9t+2.45t$

$\frac{147}{4}=7.35t$

$t=\frac{147}{4\times7.35}=\frac{147}{29.4}=5$

Solution :

Given :

Density of liquid :

$\rho = 600\text{ kg m}^{-3}$

Area at point $A$ :

$A_A = 1.0\text{ cm}^2$

$= 10^{-4}\text{ m}^2$

Area at point $B$ :

$A_B = 20\text{ mm}^2$

$= 20 \times 10^{-6}\text{ m}^2$

Speed at point $A$ :

$v_A = 10\text{ cm s}^{-1}$

$= 0.1\text{ m s}^{-1}$

Using equation of continuity :

$A_Av_A = A_Bv_B$

$10^{-4}\times 0.1 = 20\times10^{-6}\times v_B$

$v_B = \frac{10^{-5}}{20\times10^{-6}}$

$= 0.5\text{ m s}^{-1}$

Using Bernoulli’s equation :

$P_A + \frac{1}{2}\rho v_A^2=P_B + \frac{1}{2}\rho v_B^2$

Therefore,

$P_A - P_B=\frac{1}{2}\rho(v_B^2-v_A^2)$

$=\frac{1}{2}(600)(0.5^2-0.1^2)$

$=300(0.25-0.01)$

$=300 \times 0.24$

$= 72\text{ Pa}$

Final Answer :

$72$

Solution :

Terminal velocity of a spherical drop is given by :

$v \propto r^2$

Initial drop radius :

$R$

Initial terminal velocity :

$v_1 \propto R^2$

When the drop breaks into 64 identical droplets :

Volume is conserved.

Therefore,

$\frac{4}{3}\pi R^3 = 64 \times \frac{4}{3}\pi r^3$

$R^3 = 64r^3$

$R = 4r$

$r = \frac{R}{4}$

Terminal velocity of each small droplet :

$v_2 \propto r^2$

$\propto \left(\frac{R}{4}\right)^2$

$= \frac{R^2}{16}$

Therefore,

$\frac{v_1}{v_2}=\frac{R^2}{R^2/16}$

$= 16$

Final Answer :

$16$

Solution :

For one mole of diatomic gas with rotational modes only :

$C_v = \frac{5R}{2}$

Therefore,

$C_p = C_v + R$

$= \frac{5R}{2} + R$

$= \frac{7R}{2}$

Given :

$n = 1$

Temperature rise :

$\Delta T = 1.2^\circ\text{C} = 1.2\text{ K}$

Cross-sectional area :

$A = 4\text{ cm}^2$

$= 4 \times 10^{-4}\text{ m}^2$

Displacement of piston :

$x = 25\text{ mm}$

$= 25 \times 10^{-3}\text{ m}$

Change in volume :

$\Delta V = Ax$

$= (4 \times 10^{-4})(25 \times 10^{-3})$

$= 10^{-5}\text{ m}^3$

Atmospheric pressure :

$P = 100\text{ kPa}$

$= 10^5\text{ Pa}$

Work done :

$W = P\Delta V$

$= 10^5 \times 10^{-5}$

$= 1\text{ J}$

Heat supplied :

$Q = nC_v\Delta T + W$

$= 1 \times \frac{5 \times 8.3}{2} \times 1.2 + 1$

$= 24.9 + 1$

$= 25.9\text{ J}$

Final Answer :

$25.9\text{ J}$

Solution :

For a monoatomic ideal gas,

$\gamma = \frac{5}{3}$

In adiabatic process :

$PV^\gamma = \text{constant}$

Given :

Initial pressure and volume :

$P_i = P,\quad V_i = V$

Final volume :

$V_f = 27V$

Therefore,

$P_iV_i^\gamma = P_fV_f^\gamma$

$PV^{5/3} = P_f(27V)^{5/3}$

Since,

$27^{5/3} = (3^3)^{5/3} = 3^5 = 243$

$PV^{5/3} = 243P_fV^{5/3}$

$P_f = \frac{P}{243}$

Internal energy of monoatomic gas :

$U = \frac{3}{2}PV$

Initial internal energy :

$U_i = \frac{3}{2}PV$

Final internal energy :

$U_f = \frac{3}{2}P_fV_f$

$= \frac{3}{2}\times \frac{P}{243}\times 27V$

$= \frac{3}{2}\times \frac{PV}{9}$

$= \frac{PV}{6}$

Change in internal energy :

$\Delta U = U_f-U_i$

$= \frac{PV}{6}-\frac{3PV}{2}$

$= \frac{PV-9PV}{6}$

$= -\frac{8PV}{6}$

$= -\frac{4PV}{3}$

Final Answer :

$-\frac{4PV}{3}\text{ J}$

For a spring-mass system:

$\nu=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$

For a spring:

$k\propto\frac{1}{L}$

So if length is halved:

$k_2=2k_1$

Now frequency:

$nu_2=\frac{1}{2\pi}\sqrt{\frac{2k_1}{m}}=\sqrt{2}$

final answer:

$\frac{\nu_2}{\nu_1}=\sqrt{2}$

Solution :

Power emitted by source :

$P = 15\text{ kW}$

$= 1.5 \times 10^4\text{ W}$

Number of photons emitted per second :

$n = 2.5 \times 10^{22}\text{ photons/s}$

Energy of one photon :

$E = \frac{P}{n}$

$=\frac{1.5 \times 10^4}{2.5 \times 10^{22}}$

$= 0.6 \times 10^{-18}$

$= 6 \times 10^{-19}\text{ J}$

Using photon energy relation :

$E = \frac{hc}{\lambda}$

Therefore,

$\lambda = \frac{hc}{E}$

$=\frac{(6.6 \times 10^{-34})(3 \times 10^8)}{6 \times 10^{-19}}$

$=\frac{19.8 \times 10^{-26}}{6 \times 10^{-19}}$

$= 3.3 \times 10^{-7}\text{ m}$

$= 330\text{ nm}$

Wavelength 330nm lies in ultraviolet region.

Final Answer :

Ultraviolet region.

Solution :

Radius of loop :

$r = 2\text{ cm}$

$= 2 \times 10^{-2}\text{ m}$

Area of loop :

$A = \pi r^2$

$= 3.14 \times (2 \times 10^{-2})^2$

$= 12.56 \times 10^{-4}\text{ m}^2$

Current in loop :

$I = 100\sqrt2\text{ A}$

Unit normal vector :

$\hat n = \frac{\hat i + \hat k}{\sqrt2}$

Magnetic field :

$\vec B = B_0(3\hat i + 2\hat k)$

Given,

$B_0 = 4 \times 10^{-3}\text{ T}$

Magnetic moment :

$\vec m = IA\hat n$

Torque on loop :

$\tau = |\vec m \times \vec B|$

$= IA|\hat n \times \vec B|$

Now,

$\hat n \times \vec B=\frac{\hat i + \hat k}{\sqrt2}\times4 \times 10^{-3}(3\hat i + 2\hat k)$

$=\frac{4 \times 10^{-3}}{\sqrt2}[(\hat i + \hat k)\times(3\hat i + 2\hat k)]$

Using,

$\hat i \times \hat i = 0$

$\hat k \times \hat k = 0$

$\hat i \times \hat k = -\hat j$

$\hat k \times \hat i = \hat j$

Therefore,

$\hat n \times \vec B=\frac{4 \times 10^{-3}}{\sqrt2}(-2\hat j + 3\hat j)$

$=\frac{4 \times 10^{-3}}{\sqrt2}\hat j$

Magnitude :

$|\hat n \times \vec B|=\frac{4 \times 10^{-3}}{\sqrt2}$

Hence,

$\tau=(100\sqrt2)(12.56 \times 10^{-4})\left(\frac{4 \times 10^{-3}}{\sqrt2}\right)$

$=100 \times 12.56 \times 4 \times 10^{-7}$

$=5.024 \times 10^{-4}\text{ N m}$

Final Answer :

$5.024 \times 10^{-4}$

Solution :

Self inductance of solenoid :

$L = \mu_0 n^2Al$

Given :

Length of solenoid :

$l = 30\text{ cm}$

$= 0.3\text{ m}$

Turns per cm :

$10\text{ turns/cm}$

Therefore,

$n = 1000\text{ turns/m}$

Area :

$A = 5\text{ cm}^2$

$= 5 \times 10^{-4}\text{ m}^2$

Using,

$\mu_0 = 4\pi \times 10^{-7}$

Therefore,

$L=(4\pi \times 10^{-7})(1000)^2(5 \times 10^{-4})(0.3)$

$=4\pi \times 10^{-7}\times 10^6 \times 1.5 \times 10^{-4}$

$=6\pi \times 10^{-5}\text{ H}$

Induced emf :

$e = L\frac{dI}{dt}$

Given,

$\frac{dI}{dt} = \frac{4-2}{3.14}$

$= \frac{2}{3.14}$

Therefore,

$e=6\pi \times 10^{-5}\times \frac{2}{3.14}$

Using,

$\pi \approx 3.14$

$e = 12 \times 10^{-5}\text{ V}$

Hence,

$\alpha = 12$

Final Answer :

$12$

Solution :

Position vectors of charges :

$\vec r_1 = 2\hat i + 3\hat j + 3\hat k$

$\vec r_2 = \hat i + \hat j + \hat k$

Vector from $q_1$ to $q_2$ :

$\vec r = \vec r_2 - \vec r_1$

$= (\hat i + \hat j + \hat k) - (2\hat i + 3\hat j + 3\hat k)$

$= -\hat i -2\hat j -2\hat k$

Magnitude of separation :

$r = \sqrt{(-1)^2+(-2)^2+(-2)^2}$

$= \sqrt{1+4+4}$

$= 3$

Therefore,

$r^2 = 9$

Magnitude of electrostatic force :

$F = \frac{1}{4\pi\epsilon_0}\frac{|q_1q_2|}{r^2}$

Given :

$q_1 = 3 \times 10^{-6}\text{ C}$

$q_2 = -4 \times 10^{-6}\text{ C}$

Substituting values :

$F = 9 \times 10^9 \times \frac{(3 \times 10^{-6})(4 \times 10^{-6})}{9}$

$= 9 \times 10^9 \times \frac{12 \times 10^{-12}}{9}$

$= 12 \times 10^{-3}$

$= 1.2 \times 10^{-2}\text{ N}$

Since charges are unlike, force on $q_2$ is towards $q_1$.

Unit vector along direction :

$\hat r = \frac{\hat i +2\hat j +2\hat k}{3}$

Therefore,

$\vec F = 1.2 \times 10^{-2}\left(\frac{\hat i +2\hat j +2\hat k}{3}\right)$

$= 4 \times 10^{-3}(\hat i +2\hat j +2\hat k)\text{ N}$

Final Answer :

$\vec F = 4 \times 10^{-3}(\hat i +2\hat j +2\hat k)\text{ N}$

Use Snell’s law:

n₁ sinθ₁ = n₂ sinθ₂
here n₁ = 1, n₂ = 1.5

Angles are measured w.r.t. the normal (vertical axis).

So use horizontal components of direction vectors.

Incident ray AO = (2, −3)

$\sin\theta_1=\frac{|x|}{\sqrt{x^2+y^2}}=\frac{2}{\sqrt{13}}$

Refracted ray OB = (C, −4)

$\sin\theta_2=\frac{|C|}{\sqrt{C^2+16}}$

Apply Snell’s law:

$\frac{2}{\sqrt{13}}=1.5\cdot\frac{C}{\sqrt{C^2+16}}$

Solve:

$\frac{2}{\sqrt{13}}=\frac{3}{2}\cdot\frac{C}{\sqrt{C^2+16}}$

$\frac{4}{\sqrt{13}}=\frac{3C}{\sqrt{C^2+16}}$

Square:

$\frac{16}{13}=\frac{9C^2}{C^2+16}$

$16(C^2+16)=117C^2$

$16C^2+256=117C^2$

$101C^2=256$

$C^2=\frac{256}{101}$

$C=\frac{16}{\sqrt{101}}\approx1.6$

Use Einstein’s photoelectric equation:

K_{\max}=\frac{hc}{\lambda}$</p> <p>For the two դեպ:</p> <p>$K_1=\frac{hc}{\lambda_1}-\phi,\quad $</p> <p>$K_2=\frac{hc}{\lambda_2}$</p> <p>Given:</p> <p>$\lambda_1=2\lambda_2$</p> <p>So,</p> <p>$\frac{hc}{\lambda_1}=\frac{hc}{2\lambda_2}=\frac{1}{2}\frac{hc}{\lambda_2}$</p> <p>Now rewrite:</p> <p>$K_1=\frac{1}{2}\frac{hc}{\lambda_2}$</p> <p>$K_2=\frac{hc}{\lambda_2}$</p> <p>Let:</p> <p>$X=\frac{hc}{\lambda_2}$</p> <p>Then:</p> <p>$K_1=\frac{X}{2}-\phi,\quad K_2=X-\phi$</p> <h3><br/></h3> <p>$K_2-K_1=(X-\phi)-\left(\frac{X}{2}-\phi\right)=\frac{X}{2}$</p> <p>$\Rightarrow X=2(K_2-K_1)$</p> <h3><br/></h3> <p>$\phi=X-K_2$</p> <p>Substitute:</p> <p>$\phi=2(K_2-K_1)-K_2$$

Solution :

In $\alpha$-decay :

$Q = K_\alpha + K_D$

where :

$K_\alpha$ = kinetic energy of $\alpha$-particle

$K_D$ = kinetic energy of daughter nucleus

Using conservation of momentum :

$p_\alpha = p_D$

Therefore,

$\frac{K_\alpha}{K_D} = \frac{M_D}{M_\alpha}$

Hence,

$K_\alpha = Q\left(\frac{M_D}{M_D+M_\alpha}\right)$

For substance $A$ :

Mass number of parent nucleus :

$200$

Daughter nucleus mass number :

$196$

Therefore,

$K_A = Q\left(\frac{196}{196+4}\right)$

$= Q\left(\frac{196}{200}\right)$

$= \frac{49Q}{50}$

For substance $B$ :

Mass number of parent nucleus :

$212$

Daughter nucleus mass number :

$208$

Therefore,

$K_B = Q\left(\frac{208}{208+4}\right)$

$= Q\left(\frac{208}{212}\right)$

$= \frac{52Q}{53}$

Required ratio :

$\frac{K_A}{K_B} = \frac{49/50}{52/53}$

$= \frac{49 \times 53}{50 \times 52}$

$= \frac{2597}{2600}$

Final Answer :

$\frac{2597}{2600}$

Solution :

From the logic circuit :

Top AND gate receives inputs :

$A \text{ and } B$

Hence output of top gate :

$AB$

Bottom AND gate receives inputs :

$\bar A \text{ and } B$

Hence output of bottom gate :

$\bar AB$

These two outputs are connected to OR gate.

Therefore,

$Y = AB + \bar AB$

Taking $B$ common :

$Y = B(A+\bar A)$

Using Boolean identity :

$A+\bar A = 1$

Therefore,

$Y = B$

Hence output waveform follows input $B$ exactly.

From given waveform of $B$ :

$B = 1 \quad \text{for } 0<t<2$

$B = 0 \quad \text{for } 2<t<3$

Therefore,

$Y = 1 \quad \text{for } 0<t<2$

$Y = 0 \quad \text{for } 2<t<3$

Final Answer :

Output waveform is same as waveform of $B$.

Solution :

For a capacitor filled with dielectric :

$C = \frac{\epsilon_0K A}{d}$

Resistance of dielectric slab :

$R = \frac{\rho d}{A}$

Eliminating $A$ :

$A = \frac{\rho d}{R}$

Substituting in capacitance formula :

$C=\frac{\epsilon_0K}{d}\times\frac{\rho d}{R}$

$C = \frac{\epsilon_0K\rho}{R}$

Therefore,

$K = \frac{CR}{\epsilon_0\rho}$

Given :

$C = 1\mu F = 10^{-6}\text{ F}$

$R = 17.7 \times 10^{14}\Omega$

$\rho = 10^{13}\Omega\text{m}$

$\epsilon_0 = 8.85 \times 10^{-12}\text{ F/m}$

Substituting :

$K=\frac{(10^{-6})(17.7 \times 10^{14})}{(8.85 \times 10^{-12})(10^{13})}$

$=\frac{17.7 \times 10^8}{8.85 \times 10^1}$

$=2 \times 10^7$

Given,

$K = a \times 10^7$

Therefore,

$a = 2$

Final Answer :

$2$