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

When the Vernier zero is to the right of the main scale zero → positive zero error.

Given:
1 MSD = 1 mm
10 VSD

Least count:

LC = 1 MSD − 1 VSD
Since 10 VSD = 9 MSD ⇒ 1 VSD = 0.9 mm

So:

LC = 1 − 0.9 = 0.1 mm

Now, 7th Vernier division coincides:

Zero error = 7 × LC = 7 × 0.1 = 0.7 mm

Since shift is to the right → positive error

final answer:
+0.7 mm

Dimension given

L → M L² T⁻² A⁻²
C → M⁻¹ L⁻² T⁴ A²
R → M L² T⁻³ A⁻²

now check option A:

$R/\sqrt{\ LC}$

first find $\sqrt{\ LC}$:

LC = (M L² T⁻² A⁻²)(M⁻¹ L⁻² T⁴ A²)
= T²

so:

$\sqrt{\ LC}=T$

now:

R/\sqrt{\ LC}=(ML^2T⁻^3A⁻^2)/T
= M L² T⁻⁴ A⁻²

this matches exactly

Use variation of g:

• above surface:
$gₕ\approx g(1−2h/R)$

• below surface:
$g_d\approx g(1−d/R)$

given:

h = d = 16 km
R = 6400 km

step 1: values

$g_{below}=g(1−16/6400)=g(1−1/400)=0.9975g$

$g_{above}=g(1−2\times16/6400)=g(1−32/6400)=g(1−1/200)=0.995g$

step 2: change in g

$Δg=0.9975g−0.995g=0.0025g$

step 3: percentage change

$0.25\%\frac{0.0025g}{g}\times100=0.25\%$

Solution :

Let acceleration of masses $m_2$ and $m_3$ downward be $a$.

Then mass $m_1$ moves upward with same acceleration $a$.

Given :

$m_1 = 4\text{ kg}$

$m_2 = 4\text{ kg}$

$m_3 = 6\text{ kg}$

For mass $m_1$ :

$T_1 - m_1g = m_1a$

$T_1 - 40 = 4a$

$T_1 = 40 + 4a \quad ...(1)$

For mass $m_2$ :

$m_2g + T_2 - T_1 = m_2a$

$40 + T_2 - T_1 = 4a \quad ...(2)$

For mass $m_3$ :

$m_3g - T_2 = m_3a$

$60 - T_2 = 6a$

$T_2 = 60 - 6a \quad ...(3)$

Substituting (1) and (3) into (2) :

$40 + (60-6a) - (40+4a) = 4a$

$60 - 10a = 4a$

$14a = 60$

$a = \frac{30}{7}\text{ m s}^{-2}$

Now,

$T_1 = 40 + 4\left(\frac{30}{7}\right)$

$= \frac{400}{7}\text{ N}$

and,

$T_2 = 60 - 6\left(\frac{30}{7}\right)$

$= \frac{240}{7}\text{ N}$

Therefore,

$\frac{T_1}{T_2} = \frac{400/7}{240/7}$

$= \frac{5}{3}$

Final Answer :

$\frac{5}{3}$

Let’s solve step by step.

step 1: acceleration of wedge

total mass = 10 + 2 = 12 kg

external force = 24 N

$a_{wedge}=\frac{24}{12}=2m/s^2$

step 2: go to wedge frame

pseudo force on block (mass 2 kg):

F_p=ma=2×2=4 N(towards left)

step 3: resolve along incline

incline angle = 37°

components:

gravity down the plane:

$mg\sin⁡37^{\circ\ }=2\times10\times\frac{3}{5}=12N$

pseudo force component up the plane:

$4\cos⁡37^{\circ\ }=4\times\frac{4}{5}=3.2N$

net force down the plane:

12−3.2=8.8 N

step 4: acceleration of block along plane

$a=\frac{8.8}{2}=4.4m/s^2$

step 5: motion along incline

distance = 8.8 m
initial velocity = 0

$s=\frac{1}{2}at^2$

$8.8=\frac{1}{2}\times4.4\times t^2=2.2t^2$

$2t=4⇒t=2s$

Young’s modulus:

Y = (stress) / (strain)

$Y=\frac{F/A}{\Delta L/L}=\frac{FL}{A\Delta L}$

Now if:

radius → 2r ⇒ area $A\propto r^2$ becomes 4A
length → 2L

Substitute in formula:

$Y'=\frac{F(2L)}{(4A)(\Delta L')}$

But extension $ΔL′$ also changes such that:

$\Delta L'=\frac{F(2L)}{Y(4A)}=\frac{1}{2}\Delta L$

Putting back:

Y′=Y

Let’s check both statements clearly.

Statement I

$ΔU=nCᵥ(Tf−Ti)$ is correct.

Also, using relation:

$C_v=\frac{R}{\gamma-1}$

so,

$\Delta U=n\frac{R}{\gamma-1}(T_f-T_i)$

So Statement I is correct.

Statement II

Relation:

$\gamma=1+\frac{2}{f}$

This is correct for an ideal gas (from kinetic theory).

Now relation between them

Statement II helps derive relations between Cₚ, Cᵥ, and γ, which is used in Statement I.

So II correctly explains I.

final answer:
Both statements are correct and II is the correct explanation of I

A. Zeroth law gives concept of temperature
correct. it states that if two systems are each in thermal equilibrium with a third system, they are in equilibrium with each other. this allows us to define a common property called temperature.

B. First law gives concept of internal energy
correct. the first law relates heat, work and internal energy through ΔU = Q − W. this establishes internal energy as a state function of the system.

C. In isothermal expansion of ideal gas, ΔQ ≠ ΔW
wrong. in isothermal process, temperature remains constant so ΔU = 0. hence heat supplied is exactly equal to work done, so ΔQ = ΔW.

D. Product of intensive and extensive variables is extensive
correct. an intensive quantity does not depend on system size, while an extensive one does. multiplying them (like pressure × volume) gives a quantity that scales with size, so it is extensive.

E. Ratio of any extensive variable to mass is extensive
wrong. dividing an extensive quantity by mass gives a “specific” property (like density or specific heat), which does not depend on size, so it is intensive.

Solution :

Take bottom vertex potential as :

$V_B = 0$

Since the right side battery is 5V

Top vertex potential :

$V_A = 5\text{ V}$

Let left node potential be $V_C$ and middle node potential be $V_D$.

Applying KCL at node $C$ :

$\frac{V_C-5}{4} + \frac{V_C}{4} + (V_C-10-V_D) = 0$

$V_C - \frac{5}{4} + V_C + 4V_C - 40 - 4V_D = 0$

$6V_C - 4V_D = 45 \quad ...(1)$

Applying KCL at node $D$ :

$\frac{V_D-5}{2} + \frac{V_D}{2} + (V_D -V_C +10)=0$

$V_D - \frac{5}{2} + 2V_D - 2V_C +20 =0$

$-2V_C +4V_D = -15 \quad ...(2)$

Adding (1) and (2) :

$4V_C = 30$

$V_C = 7.5\text{ V}$

Substituting in (2) :

$-2(7.5)+4V_D = -15$

$V_D = 0\text{ V}$

Current $I_1$ :

$I_1 = \frac{V_D-(V_C-10)}{1}$

$= 0-(7.5-10)$

$= 2.5\text{ A}$

Current $I_3$ :

$I_3 = \frac{V_C-0}{4}$

$= \frac{7.5}{4}$

$= 1.875\text{ A}$

Current $I_2$ :

$I_2 = \frac{5-7.5}{4} + \frac{5-0}{2}$

$= -0.625 + 2.5$

$= 1.875\text{ A}$

Final Answer :

$I_1 = 2.5\text{ A}$

$I_2 = 1.875\text{ A}$

$I_3 = 1.875\text{ A}$

Solution :

Electric field :

$\vec E = -2E_0\hat i$

Force on electron :

$\vec F = q\vec E$

Since charge of electron is :

$q = -e$

Therefore,

$\vec F = (-e)(-2E_0\hat i)$

$= 2eE_0\hat i$

Acceleration of electron :

$a = \frac{F}{m}$

$= \frac{2eE_0}{m}$

Velocity after time $t$ :

$v = v_0 + at$

$= v_0 + \frac{2eE_0t}{m}$

de Broglie wavelength :

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

$= \frac{h}{m\left(v_0+\frac{2eE_0t}{m}\right)}$

$= \frac{h}{mv_0+2eE_0t}$

Given,

$\lambda_0 = \frac{h}{4mv_0}$

Therefore,

$h = 4mv_0\lambda_0$

Substituting :

$\lambda=\frac{4mv_0\lambda_0}{mv_0+2eE_0t}$

$=\frac{4\lambda_0}{1+\frac{2eE_0t}{mv_0}}$

Final Answer :

$\lambda = \frac{4\lambda_0}{1+\frac{2eE_0t}{mv_0}}$

For hydrogen atom:

$radius\ r_n\propto n^2$

given:

$r_i:r_f=16:4=4:1$

so:

$n_i^2:n_f^2=4:1\Rightarrow n_i:n_f=2:1$

 choose quantum numbers

Since electron moves from higher to lower orbit:

nᵢ > n_f

Smallest integers satisfying ratio 2:1 are:

$n_i​=4,n_f​=2$

use Rydberg formula:

$\frac{1}{\lambda}=R\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$

$=1.0973\times10^7\left(\frac{1}{4}-\frac{1}{16}\right)$

$=1.0973\times10^7\cdot\frac{3}{16}\approx2.057\times10^6$

$lambda=\frac{1}{2.057\times10^6}\approx4.86\times10^{-7}m$ 

=486 nm

start with Q = C V

differentiating with respect to time:
dQ/dt = C dV/dt

since dQ/dt = I, we get:
I = C dV/dt

given:

I = 4 A
C = 6 μF = 6 × 10⁻⁶ F

$\frac{dV}{dt}=\frac{I}{C}=\frac{4}{6\times10^{-6}}=\frac{2}{3}\times10^6$

$\approx0.67\times10^6\text{ V/s}$

Step -1

upper 3  branches have 3 batteries of emf 5 volt each and 3 resistors each of 3 ohm.

so by parallel combination it will be effectively 15 volts and 3 ohm which is connect to last branch ( 3 resistor of 3 ohm each) which had equivalent resistance 9 ohm.

so current = 15/(3+9) = 1.25 A

fringe width β = λD / d

in a medium of refractive index n, wavelength becomes λ/n

so new fringe width:

$β'=(λ/n)D/d=β/n$

given:

$β=2.4μm$
$n=1.2$

$β′=\frac{2.4}{1.2}=2μm$

1. Given parameters:

Angle of an equilateral prism: $A = 60^\circ$
Speed of light in the prism: $v = 2.12 \times 10^8 \text{ m/s}$
Speed of light in vacuum: $c = 3 \times 10^8 \text{ m/s}$

2. Refractive index calculation:

The refractive index $\mu$ of the prism is given by:

&#x27; in math mode at position 1:̲\mu = \frac{c}{…" style="color:#cc0000">$\mu = \frac{c}{v} = \frac{3 \times 10^8}{2.12 \times 10^8} \approx 1.414 \approx \sqrt{2}</span>$

3. Minimum angle of deviation calculation:

Using the prism formula:

&#x27; in math mode at position 1:̲\mu = \frac{\si…" style="color:#cc0000">$\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}</span>$

Substitute $A = 60^\circ$ and $\mu = \sqrt{2}$ into the formula:

&#x27; in math mode at position 1:̲\sqrt{2} = \fra…" style="color:#cc0000">$\sqrt{2} = \frac{\sin\left(\frac{60^\circ + \delta_m}{2}\right)}{\sin(30^\circ)}</span>$

Since $\sin(30^\circ) = \frac{1}{2}$:

&#x27; in math mode at position 1:̲\sqrt{2} \cdot …" style="color:#cc0000">$\sqrt{2} \cdot \frac{1}{2} = \sin\left(\frac{60^\circ + \delta_m}{2}\right)</span>$

&#x27; in math mode at position 1:̲\frac{1}{\sqrt{…" style="color:#cc0000">$\frac{1}{\sqrt{2}} = \sin\left(\frac{60^\circ + \delta_m}{2}\right)</span>$

Since $\sin(45^\circ) = \frac{1}{\sqrt{2}}$:

&#x27; in math mode at position 1:̲\frac{60^\circ …" style="color:#cc0000">$\frac{60^\circ + \delta_m}{2} = 45^\circ</span>$

&#x27; in math mode at position 1:̲60^\circ + \del…" style="color:#cc0000">$60^\circ + \delta_m = 90^\circ</span>$

&#x27; in math mode at position 1:̲\delta_m = 30^\…" style="color:#cc0000">$\delta_m = 30^\circ</span>$

The minimum angle of deviation is 30 degrees.

Use photoelectric equation:

work function = hν − eV₀

step 1: find photon energy

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

$=\frac{6.62\times10^{-34}\times3\times10^8}{331\times10^{-9}}$

$J=\frac{1.986\times10^{-25}}{3.31\times10^{-7}}\approx6\times10^{-19}$

step 2: subtract stopping energy

$eV_0=1.6\times10^{-19}\times0.2=0.32\times10^{-19}$

step 3: work function

$\phi=6\times10^{-19}-0.32\times10^{-19}=5.68\times10^{-19}$

Key idea: cutting a symmetric biconvex lens into two plano-convex halves doubles its focal length.

So,
new objective focal length $f_o'=2f_o$

For a compound microscope:

magnification $M\propto\frac{L}{f_o}$

To keep magnification same:

$\frac{L'}{f_o'}=\frac{L}{f_o}$

$\frac{L'}{2f_o}=\frac{L}{f_o}\Rightarrow L'=2L$

final answer:
tube length must be doubled

Breaking stress = $12\times10^8N/m^2$

Areas:
$upper=0.008cm^2=8\times10^{-7}m^2$
$lower=0.004cm^2=4\times10^{-7}m^2$

step 1: max tension each wire can bear

upper wire:

$T_u=σA=12\times10^8\times8\times10^{-7}=960N$

lower wire:

$T_l=12\times10^8\times4\times10^{-7}=480N$

step 2: tensions in wires

let mass added = mmm

lower wire carries:

$T_l=(10+m)g=(10+m)\times10$

=100+10m

limit:

$100+10m\le480\Rightarrow m\le38$

upper wire carries total mass:

$30+10+m=40+m$

$T_u=(40+m)\times10=400+10m$

limit:

$400+10m\le960\Rightarrow m\le56$

step 3: final

smaller limit governs → m=38

For series RC circuit:

current $I=\frac{V}{Z}$, where

$Z=\sqrt{R^2+X_C^2},\quad X_C=\frac{1}{\omega C}$

At frequency ω\omegaω:

$Z_1=\sqrt{R^2+X^2}$

$(where\ X=\frac{1}{\omega C}​)$

At frequency $\omega/4$:

$X'=\frac{1}{(\omega/4)C}=4X$

$Z_2=\sqrt{R^2+(4X)^2}=\sqrt{R^2+16X^2}$

Given:

$\frac{I_2}{I_1}=\frac{1/3}{1}=\frac{1}{3}\Rightarrow\frac{Z_1}{Z_2}=\frac{1}{3}$

$\frac{\sqrt{R^2+X^2}}{\sqrt{R^2+16X^2}}=\frac{1}{3}$

Square:

$\frac{R^2+X^2}{R^2+16X^2}=\frac{1}{9}$

$9(R^2+X^2)=R^2+16X^2$

$9R^2+9X^2=R^2+16X^2$

$8R^2=7X^2$

$\frac{R}{X}=\sqrt{\frac{7}{8}}$

Let’s decode the circuit step-by-step.

step 1: identify gates

• first gate (A, B) → OR gate
→ output = A + B

This output goes to:
• LED-1
• one input of next gate

step 2: second gate

inputs:
• (A + B)
• C

this is an AND gate

so output = (A + B) · C

step 3: final gate

inputs:
• A (direct wire from top)
• (A + B)C

this is AND

so final output:

= A · (A + B) · C

but:

A · (A + B) = A

so:

final output = A · C

step 4: LED conditions

LED-1 glows when:
A + B = 1

LED-2 glows when:
A · C = 1

step 5: combine conditions

for both LEDs ON:

A · C = 1 ⇒ A = 1, C = 1

then A + B = 1 is already satisfied (since A = 1)

so B can be 0 or 1

final answer:
(A, B, C) = (1, 0, 1) or (1, 1, 1)

Use shear relation:

shear modulus $G=\frac{\text{stress}}{\text{strain}}$

$G=\frac{F/A}{x/L}\Rightarrow x=\frac{FL}{AG}$

given:

side = 5 cm = 0.05 m
area $A=(0.05)2=2.5\times10^{-3}m^2$
L=0.05m
F=10
$G=10^5$

$x=\frac{10\times0.05}{2.5\times10^{-3}\times10^5}$

$=\frac{0.5}{250}=0.002m$

=2 mm

For a freely falling body:

v² = 2g(h₀ − h)

Here initial height h₀ = 18 m

So:

v² = 2g(18 − h)

Given condition:

|v| = g

So:

v² = g²

equate:

g² = 2g(18 − h)

divide by g:

g = 2(18 − h)

10 = 36 − 2h

2h = 26

h = 13 m

Given wave:

$y=3\sin(36t+0.018x+\pi/4)$

Compare with standard form:

$y=A\sin(ωt\pm kx+φ)$

So,

k = 0.018 rad/cm

Distance between two successive crests = wavelength:

$\lambda=\frac{2\pi}{k}$

$\lambda=\frac{2\times3.14}{0.018}=\frac{6.28}{0.018}$

$\lambda\approx348.9\text{ cm}$

$\lambda\approx349$

In a magnetic field, acceleration is perpendicular to the magnetic field:

a ⟂ B ⇒ a · B = 0

Given:

$B=(3i+2j)$
$a=(4i−\frac{x}{2}j)$

dot product:

$(4)(3)+\left(\frac{−x}{2}\right)(2)=0$

12 − x = 0

x = 12

Let the inductance of each inductor be $L$ and the total current entering the circuit be $I$.

Since $L_2$ and $L_3$ are identical and connected in parallel, the current divides equally between them.

$I_2=I_3=\frac{I}{2}$

The magnetic energy stored in an inductor carrying current $I$ is  $U=\frac{1}{2}LI^2$

For $L_1$,  $U_1=\frac{1}{2}LI^2$

For $L_2$,  $U_{L_2}=\frac{1}{2}L\left(\frac{I}{2}\right)^2$

$U_{L_2}=\frac{1}{8}LI^2$

Similarly, for $L_3$,  $U_{L_3}=\frac{1}{8}LI^2$

Therefore, the total magnetic energy stored in the circuit is  $U_t=U_1+U_{L_2}+U_{L_3}$

$U_t=\frac{1}{2}LI^2+\frac{1}{8}LI^2+\frac{1}{8}LI^2$

$U_t=\frac{4+1+1}{8}LI^2$

$U_t=\frac{3}{4}LI^2$

Hence,  $\frac{U_t}{U_{L_2}}=\frac{\frac{3}{4}LI^2}{\frac{1}{8}LI^2}$

$\frac{U_t}{U_{L_2}}=6$

Hence, the correct answer is 6.

Silver Carbonate decomposes into Silver $(Ag)$ on heating:

$Ag_2CO_3\longrightarrow\ 2Ag+CO_2+\frac{1}{2}O_2$

The obtained residue is Silver ($Ag$).

Molar Mass of $Ag_{2}CO_3$:

$2\left(108\right)+12+3\left(16\right)=276gmol^{-1}$

From Reaction,

276g\ \ Ag_2CO_3\longrightarrow\ 216g\ Ag\

therefore,

$2.76g\ Ag_2CO_3\longrightarrow\ \frac{216}{276}\times\ 2.76=2.16g$

Therefore, Option B is correct $i.e$ $2.16g$

For multi-electron atoms, the energy of an atomic orbital is determined using Bohr-Bury's $n+l$ rule:

Orbitals with a lower $n+l$ value have lower energy.

If two orbitals have the same $n+l$ value, the one with the lower principal quantum number $n$ has lower energy.

Let's Calculate each separately:

A. $n=3,\ l=2\ \left(3d\ orbital\right)\longrightarrow\ n+l=3+2=5$

B. $n=4,\ l=0\ \left(4s\ orbital\right)\longrightarrow\ n+l=4+0=4$

C. $n=6,\ l=1\ \left(6p\ orbital\right)\longrightarrow\ n+l=6+1=7$

D. $n=5,\ l=1\ \left(5p\ orbital\right)\longrightarrow\ n+l=5+1=6$

E. $n=2,\ l=1\ \left(2p\ orbital\right)\longrightarrow\ n+l=2+1=3$

Therefore, by rules:

$E<B<A<D<C$

Therefore, Option D is correct.

Statement A: Heisenberg uncertainty principle is applicable to electrons.

Correct,

The Heisenberg uncertainty principle applies to microscopic particles like electrons.

Statement B: The size of $2p_x$ orbital is less than the size of $3p_x$ orbital.

Correct,

Sizes of Orbitals increases with increase in Principal Quantum Number $(n)$

so, 

$3p_x>2p_x$

Statement C: The energy of $2s$ orbital of $H$ atom is equal to the energy of $2s$ orbital of $Li$.

Incorrect,

In hydrogen-like atoms, energy depends mainly on $n$, but in multi-electron atoms like lithium, electron-electron interactions change orbital energies.

So, $2s$ of $H$ and $2s$ of $Li$ don't have same energy.

Statement D: The electronic configuration of Cr is [Ar] $3d^5 4s^1$.

Correct,

Chromium shows higher stability due to half-filled $3d^5$ configuration.

Therefore, Option B is correct.