JEE Main 2025 28th January Morning Shift Question Paper with Solutions

Given, L = 5 mm, B = 2.5 mm

We know, least count of a screw gauge,

$L.C. = {{Pitch\,length} \over {No.\,of\,division\,on\,circular\,scale}}$

$\Rightarrow L.C. = {{0.75} \over {15}} = 0.05\,mm$

We know, $A = LB$

By taking $ln$ on both sides,

$\Rightarrow \ln A = \ln L + \ln B$

by differentiating both sides,

$\Rightarrow {{dA} \over A} = {{dL} \over L} + {{dB} \over B}$

Here, $dL = dB = 0.05\,mm$ (L.C.)

So fractional error,

${{dA} \over A} = {{0.05} \over 5} + {{0.05} \over {2.5}}$

$= {1 \over {100}} + {2 \over {100}} = {3 \over {100}} = {x \over {100}}$ (given)

Hence, $x = 3$.

When mass is at position A,

Elongation of spring, ${x_1} = 2R - R = R$

(As equilibrium length of spring is R)

${h_1} = 2R$

${v_1} = 0$ (As the bead is released from zero initial speed)

When mass is at position B,

${x_2} = R - R = 0$

${h_2} = R/2,{v_2} = v$

So using energy conservation,

${E_1}^\circ = {E_f}$

$\Rightarrow {1 \over 2}kx_1^2 + {1 \over 2}mv_1^2 + mg{h_1} = {1 \over 2}kx_2^2 + {1 \over 2}mv_2^2 + mg{h_2}$

$\Rightarrow {1 \over 2}k{R^2} + mg(2R) = {1 \over 2}k(o) + {1 \over 2}m{v^2} + mg{R \over 2}$

$\Rightarrow {1 \over 2}k{R^2} + 2mgR - {{mgR} \over 2} = {1 \over 2}m{v^2}$

$\Rightarrow {1 \over 2}m{v^2} = {1 \over 2}k{R^2} + {{3mgR} \over 2}$

$\Rightarrow {v^2} = {{k{R^2}} \over m} + 3gR$

$\Rightarrow v = \sqrt {{{k{R^2}} \over m} + 3gR}$

We know, terminal velocity of a sphere of radius R in a liquid of viscosity $\eta$,

$v = {2 \over 9}{{{R^2}} \over \eta }(\sigma - \rho )$ .... (1)

where, $\sigma$ = mass of density of sphere

$\rho$ = density of liquid

we can see, $v \propto {R^2}$ (for constant $\eta,\sigma$ & $\rho$)

Hence, graph between v and R is parabola.

As v depends on R so the terminal velocities of different diameter balls will be different.

We know, the viscosity of a liquid usually decreases as the temperature increases and $v \propto {1 \over \eta }$

So terminal velocity depends on the temperature. $T \uparrow \Rightarrow \eta \downarrow \Rightarrow v \uparrow$

As the equation $v = {2 \over 9}{{{R^2}} \over v}(\sigma - \rho )$ involves density of liquid $\rho$. So the experiment can be utilized to asses it.

From (1), $\eta = {2 \over 9}{{{R^2}} \over v}(\sigma - \rho )$

Here, $\eta$ does not depend on initial speed of the sphere. Hence, option 3 is correct.

Given, ${I_1} = 2.5{I_2}$ and ${I_3} = n{I_2}$

we know, ${I_1} = {{MR_1^2} \over 4},{I_2} = {{MR_2^2} \over 2},{I_3} = {2 \over 5}MR_3^2$

$\Rightarrow {{MR_1^2} \over 4} = 2.5{{MR_2^2} \over 2} \Rightarrow R_1^2 = 5R_2^2$

Now, ${I_3} = n{I_2}$

$\Rightarrow {2 \over 5}MR_1^2 = n{{MR_2^2} \over 2}$ (As ${R_3} = {R_1}$)

$\Rightarrow {2 \over 5} \times 5R_2^2 = {{nR_2^2} \over 2} \Rightarrow n = 4$

We know, speed of sound in a medium $v = \sqrt {{B \over \rho }}$ where B = bulk modulus, $\rho$ = density of the medium

Since, solids and liquids are much more difficult to compress than gases so they have much higher values of bulk modulus.

i.e., ${B_{solid}} > {B_{liquid}} > {B_{gas}}$

Generally solids and liquids have higher mass densities ($\rho$) than gases. But corresponding increase in bulk modulus is much higher.

So, ${v_{solid}} > {v_{liquid}} > {v_{gas}}$

Hence, option 3 is correct.

We know, for thin prism,

$\mu = {{A + \delta } \over A}$

where, $\mu$ = refractive index

A = Angle of prism

$\delta$ = Angle of deviation

We can write, $\mu A = A + \delta$

$\Rightarrow \delta = (\mu - 1)A$

given, ${\delta _{net}} = 0$

$\Rightarrow ({\mu _1} - 1){A_1} - ({\mu _2} - 1){A_2} = 0$

$\Rightarrow (1.54 - 1)4 - (1.72 - 1){A_2} = 0$

$\Rightarrow {A_2} = {{54 \times 4} \over {72}} \Rightarrow {A_2} = 3^\circ$

Given, measured quantity $= {{Modulus\,of\,elasticity} \over {torque}}$

$= {\sigma \over {\varepsilon \tau }} = {F \over {A\varepsilon \tau }}$ where, $\sigma$ = stress, $\varepsilon$ = strain, $\tau$ = torque

$= {F \over {A\varepsilon FL}}$

So, dimension $= {1 \over {[{L^2}][L]}}$ ($\varepsilon$ is dimensionless)

$= [{L^{ - 3}}] = [{M^0}{L^{ - 3}}{T^0}] = [{M^a}{L^b}{T^c}]$

$\Rightarrow c = 0$

Let surface mass density = $\sigma$

So, ${M_1} = \sigma \times \pi R_1^2$

${M_2} = \sigma \times \pi R_2^2 = \sigma \pi {(2{R_1})^2}$ (As ${R_2} = 2{R_1}$

$= 4\sigma \pi R_1^2$

$\Rightarrow {M_2} = 4{M_1}$

${{{I_1}} \over {{I_2}}} = {{{{{M_1}R_1^2} \over 2}} \over {{{{M_2}R_2^2} \over 2}}} = \left( {{{{M_1}} \over {{M_2}}}} \right){\left( {{{{R_1}} \over {{R_2}}}} \right)^2}$

$\Rightarrow {{{I_1}} \over {{I_2}}} = \left( {{{{M_1}} \over {4{M_1}}}} \right){\left( {{{{R_1}} \over {2{R_1}}}} \right)^2} = {1 \over 4} \times {1 \over 4} = {1 \over {16}} = {1 \over x}$

Hence, $x = 16$

We know, theoretical equation for $\beta^-$ decay,

Hence, option 2 is correct.

Electric field due to infinitely long wire

$\overrightarrow E = - {{2k\lambda } \over r}\widehat r$

We know, $V = - \int {\overrightarrow E \,.\,\overrightarrow {dr} }$

$= - \int {{{ - 2k\lambda } \over r}\widehat r\,.\,\overrightarrow {dr} }$

$v = \int {{{2k\lambda } \over r}dr = 2k\lambda \,\ln r + c}$

Net potential due to all wires,

$V = {v_1} + {v_2} + {v_3}$

$= 2k\lambda \ln \left( {\sqrt {{x^2} + {y^2}} } \right) + 2k\lambda \ln \left( {\sqrt {{y^2} + {z^2}} } \right) + 2k\lambda \ln \left( {\sqrt {{z^2} + {x^2}} } \right) + c$

for equipotential surface, $v = c$

$\Rightarrow 2k\lambda \left[ {\ln \left( {\sqrt {{x^2} + {y^2}} \,.\,\sqrt {{y^2} + {z^2}} \,.\,\sqrt {{z^2} + {x^2}} } \right)} \right] = c$

$\Rightarrow \sqrt {({x^2} + {y^2})\,.\,({y^2} + {z^2})\,.\,({z^2} + {x^2})} = c$

$\Rightarrow ({x^2} + {y^2})({y^2} + {z^2})({z^2} + {x^2}) = c$

where, c = constant.

We know, efficiency of a carnot engine, $\eta$ or $e = 1 - {{{T_C}} \over {{T_H}}}$

where, T$_C$ = temperature of cold sink

T$_H$ = temperature of hot source

So, ${\eta _{12}} = 1 - {{273} \over {473}} = {{200} \over {473}} = 0.423$

${\eta _1} = 1 - {{373} \over {473}} = {{100} \over {473}} = 0.211$

${\eta _2} = 1 - {{273} \over {373}} = {{100} \over {373}} = 0.268$

Here, ${\eta _1} + {\eta _2} = 0.211 + 0.268 = 0.479 > 0.423$

So, ${\eta _{12}} < {\eta _1} + {\eta _2}$

In Young's double slit experiment, the distance of $n^{th}$ maxima from central maxima is given by,

$Y=\frac{n\lambda D}{d}$

Here, $n,D$ & $d$ are same for both the cases.

So, $Y \propto \lambda$

$\Rightarrow {{{Y_1}} \over {{Y_2}}} = {{{\lambda _1}} \over {{\lambda _2}}} \Rightarrow {{10} \over {{Y_2}}} = {{600} \over {660}} = {1 \over {11}}$

$\Rightarrow {Y_2} = 11\,mm$.

$\sigma = {{{\sigma _0}x} \over {ab}}$

Let's consider a strip of infinitesimal width dx at x distance away from O.

As $\sigma$ depends only on x i.e. it's constant for a specific value of x so centre of mass of the strip will be $\left( {x,{b \over 2}} \right)$.

Now, we know,

$COM = \left( {{{\int {xdm} } \over {\int {dm} }},{{\int {ydm} } \over {\int {dm} }}} \right)$

$= \left( {{{\int {x\sigma dA} } \over {\int {\sigma dA} }},{{\int {{b \over 2}\sigma dA} } \over {\int {\sigma dA} }}} \right)$

$= \left( {{{\int {xbdx{{{\sigma _0}x} \over {ab}}} } \over {\int {{{{\sigma _0}x} \over {ab}}bdx} }},{{{b \over 2}\int {\sigma dA} } \over {\int {\sigma dA} }}} \right)$

For whole plate, x varies from o to a.

So, $= \overleftarrow C OM = \left( {{{{{{\sigma _0}} \over a}\int_0^a {{x^2}dx} } \over {{{{\sigma _0}} \over a}\int_0^a {xdx} }},{b \over 2}} \right)$

$= \left( {{{{1 \over 3}\left[ {{x^3}} \right]_0^a} \over {{1 \over 2}\left[ {{x^2}} \right]_0^a}},{b \over 2}} \right)$

$= \left( {{2 \over 3}{{{a^3}} \over {{a^2}}},{b \over 2}} \right)$

$\Rightarrow COM = \left( {{2 \over 3}a,{b \over 2}} \right)$

In $\Delta OMA,{{OM} \over {OA}} = \cos 60^\circ = {1 \over 2}$

$\Rightarrow OM = {l \over 2}$ (as $OA = l$)

$MN = ON - OM = l - {l \over 2} = {l \over 2}$

Using work $-$ energy theorem,

${W_{all}} = \Delta k$

$\Rightarrow {W_c} = {k_f} - {k_i}$

$\Rightarrow qE{l \over 2} = {1 \over 2}m{v^2} - o$ (As ${v_i} = o$ and $w = FS$)

$\Rightarrow {v^2} = {{qEl} \over m}$

$\Rightarrow v = \sqrt {{{qEl} \over m}}$

For $A \to B$

$\overrightarrow V = - {v_x}\widehat i + {v_y}\widehat j$

$\overrightarrow B = {{{\mu _0}I} \over {2\pi r}}\left( { - \widehat k} \right)$

$\because$ $\overrightarrow F = q\left( {\overrightarrow v \times \overrightarrow B } \right)$

So, $\overrightarrow F = {{q{\mu _0}I} \over {2\pi r}}\left[ {\left[ {\left( { - {v_x}} \right)\widehat i + {v_y}\widehat j} \right] \times \left( { - \widehat k} \right)} \right]$

$= {{q{\mu _0}I} \over {2\pi r}}\left[ {{v_x}\left( {\widehat i \times \widehat k} \right) - {v_y}\left( {\widehat j \times \widehat k} \right)} \right]$

$\overrightarrow F = {{q{\mu _0}I} \over {2\pi r}}\left[ { - {v_x}\widehat j - {v_y}\widehat i} \right]$

$\Rightarrow {F_x} = {{ - {\mu _0}Iq} \over {2\pi r}}{v_y}$

$\Rightarrow m{a_x} = {{ - {\mu _0}Iq} \over {2\pi r}}{v_y}$

$\Rightarrow {v_x}{{d{v_x}} \over {dr}} = {{ - {\mu _0}Iq} \over {2\pi m}}{{{v_y}} \over r}$

$\Rightarrow {{{v_x}d{v_x}} \over {{v_y}}} = {{ - {\mu _0}Iq} \over {2\pi m}}{{dr} \over r}$

Since, $v_x^2 + v_y^2 = v_0^2$

$\Rightarrow {v_y} = \sqrt {v_0^2 - v_x^2}$

$\Rightarrow \int_0^{{v_0}} {{{{v_x}d{v_x}} \over {\sqrt {v_0^2 - v_x^2} }} = {{ - {\mu _0}Iq} \over {2\pi m}}\int_a^{x'} {{{dr} \over r}} }$

Let $v_0^2 - v_x^2 = {z^2}$

$\Rightarrow - 2{v_x}d{v_x} = 2zdz$

when ${v_x} = 0,\,z = {v_0}$

${v_x} = {v_0},\,z = 0$

$\Rightarrow \int_{{v_0}}^0 {{{zdz} \over z} = - {{{\mu _0}Iq} \over {2\pi m}}\int_a^{x'} {{{dr} \over r}} }$

$\Rightarrow - {v_0} = {{{\mu _0}Iq} \over {2\pi m}}\ln \left( {{{x'} \over a}} \right)$

$\Rightarrow \ln \left( {{{x'} \over a}} \right) = - {{2\pi m{v_0}} \over {{\mu _0}Iq}}$

$\Rightarrow x' = ac{{ - 2\pi m{v_0}} \over {{\mu _0}Iq}}$ .... (1)

Now, for $B \to C$

$\overrightarrow v = - {v_x}\widehat i - {v_y}\widehat j$

$\overrightarrow B = {{{\mu _0}I} \over {2\pi r}}\left( { - \widehat k} \right)$

$\overrightarrow F = q\left( {\overrightarrow v \times \overrightarrow B } \right) = {{q{\mu _0}I} \over {2\pi r}}\left[ {{v_x}\left( {\widehat i \times \widehat k} \right) + {v_y}\left( {\widehat j \times \widehat k} \right)} \right]$

$\Rightarrow \overrightarrow F = {{{\mu _0}Iq} \over {2\pi r}}\left[ { - {v_x}\widehat j + {v_y}\widehat i} \right]$

$\Rightarrow {F_x} = {{{\mu _0}Iq} \over {2\pi r}}{v_y}$

$\Rightarrow {v_x}{{d{v_x}} \over {dr}} = {{{\mu _0}Iq} \over {2\pi m}}{{{v_y}} \over r}$

$\int_{{v_0}}^0 {{{{v_x}d{v_x}} \over {\sqrt {v_0^2 - v_x^2} }} = {{{\mu _0}Iq} \over {2\pi m}}\int_{x'}^x {{{dr} \over r}} }$

Let $v_0^2 - v_x^2 = {z^2}$

$- 2{v_x}d{v_x} = 2zdz$

${v_x} = {v_0} \Rightarrow z = 0$

${v_x} = 0 \Rightarrow z = {v_0}$

$\Rightarrow - \int_0^{{v_0}} {{{zdz} \over z} = {{{\mu _0}Iq} \over {2\pi m}}\left[ {\ln r} \right]_{x'}^x}$

$\Rightarrow - {v_0} = {{{\mu _0}Iq} \over {2\pi m}}\ln \left( {{x \over {x'}}} \right)$

$\Rightarrow x = x'{e^{ - {{2\pi m{v_0}} \over {{\mu _0}Iq}}}}$ .... (2)

From (1) & (2), $x = a{e^{ - {{4\pi m{v_0}} \over {{\mu _0}Iq}}}}$

(A) Surface tension arises due to the extra potential energy of the molecules at the surface of a liquid compared to the molecules in the interior.

(B) When a liquid is heated, the kinetic energy of it molecules increases, causing them to move more freely and overcome the intermolecular forces that contribute to viscosity, thus lowering it.

(C) Higher temperature in gas means faster moving molecules, resulting in more collisions and therefore an increase in viscosity.

(D) Reynold's number $= {{Inertial\,force} \over {Visous\,force}}$

${R_e} = {{\rho VD} \over \mu }$,

v = velocity of flow

D = Diameter of pipe

If, $R_e < 2000$, flow is laminar flow

$2000 < R_e < 3000 \Rightarrow$ transition region

$R_e > 3000 \Rightarrow$ turbulant flow.

(E) Tangent at a point to the streamline gives the direction of net velocity of the flow. If the two streamlines intersect each other. It signifies two directions of velocity which cannot be possible. Thus, two streamlines cannot intersect each other. Hence, option 2 is correct.

from $\Delta OCB,\,\tan \theta = {{CB} \over {OC}} = {R \over R} = 1$

$\Rightarrow \theta = 45^\circ$

We know, $\sin c = {1 \over \mu }$

for minimum $\mu$, sinc maximum $\Rightarrow$ C maximum

Here, ${c_{\max }} = \theta = 45^\circ$

So, $\mu = {1 \over {\sin 45^\circ }} = {1 \over {{1 \over {\sqrt 2 }}}}$

$\Rightarrow \mu = \sqrt 2$

$Y = \overline {A.B + A.\overline B } = \overline {A(B + \overline B )} = \overline {AI}$

$\Rightarrow Y = \overline A$

1.

(As $A\,.\,A = A$

Hence, option 1 is correct.

Here, ${V_A} = {V_C}$ & ${V_B} = {V_D}$

So,

So resistances are in parallel combination.

We know, ${1 \over {{R_p}}} = {1 \over {{R_1}}} + {1 \over {{R_2}}} + {1 \over {{R_3}}}$

$\Rightarrow {1 \over {{R_p}}} = {3 \over r} + {3 \over r} + {3 \over r} = {9 \over r}$

$\Rightarrow {R_p} = {r \over 9}$

Since parallel combination, so v is same for both

we know, $q=cv$

$\Rightarrow q \propto C$ (for constant v)

we can see in the graph,

at final state, $q_2 > q_1$,

so $c_2 > c_1$

Now, as $u = {1 \over 2}c{v^2} \Rightarrow u \propto c$ (for constant v)

$\Rightarrow {u_2} > {u_1}$

Hence, option 1 is correct.

We know, ${E_{proton}} = {{hc} \over \lambda } = E$

${E_{proton}} = E$ (given)

$\Rightarrow K{E_{proton}} = {1 \over 2}{m_P}{v^2} = E$

So, ${{{\lambda _{proton}}} \over {{\lambda _{photon}}}} = {{h/p} \over {hc/E}} = {h \over p} \times {E \over {hc}}$

$= {E \over {\sqrt {2{m_P}E} c}}$ (As $P = \sqrt {2m\,K.E.}$)

${{{\lambda _{proton}}} \over {{\lambda _{photon}}}} = {1 \over c}\sqrt {{E \over {2{m_P}}}}$

In a central force field, the work done is independent of the path chosen, this is true because central forces are conservative forces. Every force encountered in mechanics does not have an associated potential energy this is also true because not all forces are conservative. Hence, option 2 is correct.

We know,

Energy density $= {1 \over 2}{\varepsilon _0}{E^2}C$

$\Rightarrow {{Energy} \over {Volume}} = {1 \over 2}{\varepsilon _0}{E^2}C$

$\Rightarrow Energy = {1 \over 2}{\varepsilon _0}E_c^2x\,vol$

Given that, ${\left( {Energy} \right)_1} = {\left( {Energy} \right)_2}$

$\Rightarrow {1 \over 2}{\varepsilon _0}E_1^2cx\pi R_1^2{L_1} = {1 \over 2}{\varepsilon _0}E_2^2cx\pi R_2^2{L_2}$

$\Rightarrow E_1^2R_1^2{L_1} = E_2^2R_2^2{L_1}\,(as\,{L_1} = {L_2})$

$\Rightarrow E_1^2R_1^2 = E_2^2R_2^2$

$\Rightarrow {E_1}{R_1} = {E_2}{R_2}$

$\Rightarrow {E_1}{R_1} = {E_2}\left( {{{{R_1}} \over 2}} \right)\,\left( {As\,{R_2} = {{{R_1}} \over 2}} \right)$

$\Rightarrow {E_2} = 2{E_1} = 2 \times 100$

$\Rightarrow {E_2} = 200\,N/C$

Hence, option (3) is correct.

We know, $R = \rho {l \over A} \Rightarrow R \propto l$

Side length of triangle $= {l \over 3}$

Side length of square $= {l \over 4}$

So,

$\Rightarrow {\left( {{R_{eq}}} \right)_1} = {{{{2R} \over 3} \times {R \over 3}} \over {{{2R} \over 3} + {R \over 3}}}$ and ${\left( {{R_{eq}}} \right)_2} = {{{{3R} \over 4} \times {R \over 4}} \over {{{3R} \over 4} + {R \over 4}}}$

$= {{2{R^2}} \over 9} \times {3 \over {3R}} = {{2R} \over 9}$ and ${\left( {{R_{eq}}} \right)_2} = {{3{R^2}} \over {16}} - R = {{3R} \over {16}}$

Hence, ${{{{\left( {{R_{eq}}} \right)}_1}} \over {{{\left( {{R_{eq}}} \right)}_2}}} = {{{{2R} \over 9}} \over {{{3R} \over {16}}}} = {{32} \over {27}}$