JEE Main 2025 29th January Evening Shift Question Paper with Solutions

We know, for a plane electromagnetic wave,

$\widehat E \times \widehat B = \widehat C$

Given, $\widehat C = \widehat i$ (+ x direction)

So, $\widehat E \times \widehat B = \widehat i$

i.e. $\widehat j \times \widehat k = \widehat i$

So, $\overrightarrow E = {E_y}$

$\overrightarrow B = {B_2}$

As, $\widehat i,\,\widehat j$ and $\widehat k$ are unit vectors in the directions of x, y and z axes respectively.

Hence, option 3 is correct.

Given, ${c_1} = 6\mu F,{c_2} = 12\mu F,v = 5v$

We know, for a capacitor $Q = CV$

Initial charge on ${c_1}\,.\,Q = {c_1}v$

$\Rightarrow Q = 6 \times 5$

$\Rightarrow Q = 30\mu C$ .... (1)

when switch 'S' is closed, let q charge flows from C$_1$ to C$_2$

So final charges are $Q-q$ and q on C$_1$ and C$_2$ respectively.

Now, when equilibrium condition is reached potential on C$_1$ and C$_2$ becomes equal.

Hence, ${{Q - q} \over {{c_1}}} = {q \over {{c_2}}}$ (As $v = {Q \over c}$)

$\Rightarrow {Q \over {{c_1}}} = q\left( {{{{c_1} + {c_2}} \over {{c_1}{c_2}}}} \right)$

$\Rightarrow q = \left( {{{{c_2}} \over {{c_1} + {c_2}}}} \right)Q$

$\Rightarrow q = \left( {{{12} \over {6 + 12}}} \right) \times 30\mu C$ (From (1))

$\Rightarrow q = \left( {{{12} \over {18}}} \right) \times 30 = 20\mu c$

and $Q - q = 30 - 20 = 10\mu c$

Hence, $q = 10\mu c$ and ${q_2} = 20\mu c$

Hence, option 4 is correct.

Using lens maker formula,

${{{\mu _2}} \over v} - {{{\mu _1}} \over 4} = {{{\mu _2} - {\mu _1}} \over R}$

For the image formed by the surface on the right side.

${\mu _1} = 1,\,{\mu _2} = 1.5,\,u = {{ - R} \over 2},{R_1} = - R$

so, ${{1.5} \over v} - {1 \over { - R/2}} = {{1.5 - 1} \over { - R}}$

$\Rightarrow {{1.5} \over v} + {2 \over R} = {{ - 1} \over {2R}}$

$\Rightarrow {{1.5} \over v} = {{ - 5} \over {2R}} \Rightarrow v = {{ - 3R} \over 5}$

Here '$-$' sign tells that image is formed left to B.

So, the distance of image from $P, = {{R - 3R} \over \Delta } = {{2R} \over 5}$

Now, for the surface on the left side,

${\mu _1} = 1,{\mu _2} = 1.5,\mu = {{3R} \over 2},{R_2} = R$

So, ${{1.5} \over v} - {1 \over {{{3R} \over 2}}} = {{1.5 - 1} \over R}$

$\Rightarrow {{1.5} \over v} - {2 \over {3R}} = {1 \over {2R}}$

$\Rightarrow {{1.5} \over v} = {1 \over R}\left( {{1 \over 2} + {2 \over 3}} \right)$

$\Rightarrow {{1.5} \over v} = {7 \over {6R}}$

$\Rightarrow v = {{9R} \over 7}$

So distance from $P = {{9R} \over 7} - R = {{2R} \over 7}$

So the distance between two images,

$= {{2R} \over 5} - {{2R} \over 7} = 0.4R - 0.2857R$

$\approx 0.4R - 0.286R = 0.114R$.

We know, Young's modulus, $E = {\sigma \over \varepsilon }$

$\Rightarrow E = {{{F \over A}} \over {{{\Delta L} \over L}}} = {F \over A}\left( {{L \over {\Delta L}}} \right)$

$\Rightarrow [E] = {{[F]} \over {[A]}} = {{\left[ {{M^1}{L^1}{T^{ - 2}}} \right]} \over {\left[ {{L^2}} \right]}}$

(A) $\Rightarrow [E] = [{M^1}{L^{ - 1}}{T^{ - 2}}]$

(B) we know, Torque = $r\times F$

$[\tau ] = [r][F]$

$= [L][{M^1}{L^1}{T^{ - 2}}]$

$[\tau ] = [{M^1}{L^2}{T^{ - 2}}]$

(C) We know, $F = \mu A{u \over y}$

where, F = force, $\mu$ = coefficient of viscosity

A = area $\frac{u}{y}$ = rate of shear deformation

$\Rightarrow \mu = {{Fy} \over {Au}}$

$\Rightarrow [\mu ] = {{[{M^1}{L^1}{T^{ - 2}}][L]} \over {[{L^2}][L{T^{ - 1}}]}}$

$\Rightarrow [\mu ] = [{M^1}{L^{ - 1}}{T^{ - 1}}]$

(D) We know, $F = {{G{m_1}{m_2}} \over {{r^2}}}$ (Newton's law of gravitation)

$\Rightarrow G = {{F{r^2}} \over {{m_1}{m_2}}}$<$\Rightarrow G = {{F{r^2}} \over {{m_1}{m_2}}}$

$\Rightarrow [G] = {{[{M^1}{L^1}{T^{ - 2}}][{L^2}]} \over {[{M^2}]}}$

$\Rightarrow [G] = [{M^{ - 1}}{L^3}{T^{ - 2}}]$

Hence, option D is correct.

Here, we will use the concept of relative motion.

Let initial (at $t = 0$) position of both cars is x = 0 (As they cross each other at t = 0 for Ist time)

given : For P

${a_P} \propto t$

For Q,

${a_Q} = a'$ (Let) (constant)

$\Rightarrow {a_{{p^{(t)}}}} = kt$ where, k = constant

$\Rightarrow {{d{v_p}(t)} \over {dt}} = kt$ (as $a = {{dv} \over {dt}}$)

$\Rightarrow \int_0^{{v_p}(t)} {d{v_p}(t) = \int_0^t {ktdt} }$ [let p and Q both starts from rest]

$\Rightarrow {v_p}(t) = {{k{t^2}} \over 2}$

$\Rightarrow {{d{x_p}(t)} \over {dt}} = {{k{t^2}} \over 2}$ [As $v = {{dx} \over {dt}}$]

$\Rightarrow \int_0^{{x_p}(t)} {d{x_{p(t)}} = \int_0^t {{{k{t^2}} \over 2}dt} }$

$\Rightarrow {x_p}(t) = {{k{t^3}} \over 6}$ .... (1)

Now, for Q,

${x_Q}(t) = {1 \over 2}{a^1}{t^2}$ .... (2) [using Newton's 2nd equation of motion]

So, relative position of P w.r.t. Q,

${x_{P{Q^{(t)}}}} = {x_p}(t) - {x_Q}(t)$

$\Rightarrow {x_{P{Q^{(t)}}}} = {{k{t^3}} \over 6} - {1 \over 2}{a^1}{t^2}$ (From (1) and (2))

When both cars cross each other, $X_{PQ}=O$

As $X_{PQ}(t)$ is a cubic polynomial, so it has maximum 3 roots.

Hence, the maximum number of crossing = 3

We know, ideal gas equation,

$\mathrm{Pv = nRT}$

For isothermal process, T = constant

$\mathrm{\Rightarrow PV = const.}$

$\mathrm{\Rightarrow P \propto \frac{1}{V}}$

For an adiabatic process,

$\mathrm{PV^r=constant}$

We can see, for the same pressure incresement, ${P_2} - {P_1}$, $\left( {{v_4} - {v_3}} \right) > \left( {{v_2} - {v_1}} \right)$

i.e. the volume falls off more rapidly in an isothermal process in comparison to the adiabatic process. Hence, option 4 is corret.

Given an electric flux $\phi = -2 \times 10^4 \, \text{Nm}^2\text{C}^{-1}$ through a spherical Gaussian surface with a radius of 8.0 cm (r = 0.08 m), we are to find the value of the point charge $Q$. According to Gauss's Law:

$\phi = \frac{Q}{\varepsilon_0}$

Where $\varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2\text{N}^{-1}\text{m}^{-2}$.

Rearranging the equation to solve for $Q$, we have:

$Q = \phi \varepsilon_0$

Substituting the given values:

$Q = -2 \times 10^4 \times 8.85 \times 10^{-12}$

Calculating this yields:

$Q = -17.7 \times 10^{-8} \, \text{C}$

Thus, the value of the point charge is $Q = -17.7 \times 10^{-8} \, \text{C}$.

Given, $Q = {{a{b^4}} \over {cd}}$

$a = (60 \pm 3)\,Pa \Rightarrow a = 60\,Pa,\,\Delta a = 3Pa$

$b = (20 \pm 0.1)\,m \Rightarrow b = 20m,\,\Delta b = 0.1\,m$

$c = (40 \pm 0.2)\,Ns{m^{ - 2}} \Rightarrow c = 40\,Ns{m^{ - 2}},\,\Delta c = 0.2\,Ns{m^{ - 2}}$

$d = (50 \pm 0.1)\,m \Rightarrow \,d = 50\,m,\,\Delta d = 0.1m$

As, $Q = {{a{b^4}} \over {cd}}$

by taking ln on both sides,

$\ln Q = \ln a + u\ln b - \ln c - \ln d$

Now, by differentiating,

${{dQ} \over Q} = {{da} \over a} + 4{{db} \over b} - {{dc} \over c} - {{dd} \over d}$

So, maximum fractional error in Q is given by,

${{\Delta Q} \over Q} = {{\Delta a} \over a} + 4{{\Delta b} \over b} + {{\Delta c} \over c} + {{\Delta d} \over d}$

$\Rightarrow {{\Delta Q} \over Q} = {3 \over {60}} + 4\left( {{{0.1} \over {20}}} \right) + {{0.2} \over {40}} + {{0.1} \over {40}}$

$= {1 \over {20}} + {1 \over {50}} + {1 \over {200}} + {1 \over {500}}$

$\Rightarrow {{\Delta Q} \over Q} = {{50 + 20 + 5 + 2} \over {1000}} = {{77} \over {1000}}$

Hence the % error in $Q = {{\Delta Q} \over Q} \times 100\%$

$= {{7700} \over {1000}}\% = {x \over {1000}}$ (given)

So, $x = 7700$

As A to B is can adiabatic path,

So, $\Delta {Q_{A \to B}} = o$ .... (1)

So, now we need to find the heat absorbed for the isothermal path (B to C)

We know, change in internal energy

$\Delta u = m{c_v}\Delta T$

$\Rightarrow \Delta u = o$ (for isothermal as $\Delta T = o$)

Now, using Ist law of thermodynamics

$\Delta Q = \Delta u + w$

$\Rightarrow \Delta Q = o + w \Rightarrow \Delta Q = w$

So, $\Delta {Q_{B \to C}} = {W_{B \to C}}$

$= nRT\ln \left( {{{{v_C}} \over {{v_B}}}} \right)$

$\Rightarrow \Delta {Q_{B \to C}} = nR(450)\ln \left( {{4 \over 3}} \right)$

$\Rightarrow \Delta {Q_{B \to C}} = 450nR\ln \left( {{4 \over 3}} \right)$ .... (2)

So net heat absorbed,

$Q = \Delta {Q_{A \to B}} + \Delta {Q_{B \to C}}$

$\Rightarrow Q = 450nR\ln \left( {{4 \over 3}} \right)$ ... (from 1 and 2)

Net heat absorbed per mole,

$\Rightarrow {Q \over n} = 450R(\ln 4 - \ln 3)$.

Option 2 is correct.

When an e transition from n$^{th}$ to ground state then $\sum {(n - 1)}$ spectral lines are produced.

Given, $n=4$

So, $\sum {(4 - 1) = \sum {3 = 3 + 2 + 1 = 6} }$

Hence, 6 spectral lines are emitted by atomic Hydrogen that is in the $4^{th}$ energy level

Using lens formula,

For Ist case : ${1 \over f} = \left( {{a^{{\mu _g}}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$

where, ${R_1},{R_2}$ = radii of refracting surfaces

$f$ = focal length

${a^{{\mu _g}}} = {{^{{\mu _g}}} \over {^{{\mu _a}}}} = {{1.5} \over 1} = 1.5$

So, ${1 \over {24}} = (1.5 - 1)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$ (As f = 24 cm given)

$\Rightarrow {1 \over {24}} = 0.5\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$

$\Rightarrow {1 \over {12}} = \left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$

$\Rightarrow {1 \over {{R_1}}} - {1 \over {{R_2}}} = {1 \over {12}}$ .... (i)

For IInd case:

${1 \over f} = \left( {w{\mu _g} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$

$\Rightarrow {1 \over f} = \left( {{{1.5} \over {1.33}} - 1} \right)\left( {{1 \over {12}}} \right)$ ..... from (i)

$\Rightarrow {1 \over f} = {{17} \over {133}} \times {1 \over {12}} \Rightarrow f = {{133 \times 12} \over {17}}$

$\Rightarrow f = 93.88$ cm

$\approx$ 94 cm

Hence, option 1 is correct.

Here, we will use the Newton's law of cooling.

We know, ${{{T_1} - {T_2}} \over {\Delta t}} = k\left( {{{{T_1} + {T_2}} \over 2} - {T_s}} \right)$ .... (1)

where, T$_1$ and T$_2$ are initial and final temperatures respectively.

$\mathrm{T_s}$ = surrounding temperature

$\Delta t$ = time taken

$k$ = positive constant

Given,

Ist case :

${T_1} = 90^\circ C$

${T_2} = 80^\circ C$

$\Delta t = t$, ${T_s} = 20^\circ C$

IInd case :

${T_1} = 80^\circ C$

${T_2} = 60^\circ C$

${T_s} = 20^\circ C$, $\Delta t = ?$

Now, for Ist case,

${{90 - 80} \over t} = k\left( {{{90 + 80} \over 2} - 20} \right)$ ...... (from (1))

$\Rightarrow {{10} \over t} = k\left( {85 - 20} \right)$

$\Rightarrow {{10} \over t} = 65k$ .... (2)

For IInd case,

${{80 - 60} \over {\Delta t}} = k\left( {{{80 + 60} \over 2} - 20} \right)$ .... (From (1))

$\Rightarrow {{20} \over {\Delta t}} = k\left( {70 - 20} \right)$

$\Rightarrow {{20} \over {\Delta t}} = 50k$ .... (3)

Now, by (2) $\div$ (3),

${{{{10} \over t}} \over {{{20} \over {\Delta t}}}} = {{13} \over {10}}$

$\Rightarrow {{\Delta t} \over t}\left( {{1 \over 2}} \right) = {{13} \over 5}$

$\Rightarrow \Delta t = {{13} \over 5}t$

Here,

we will use lens maker formula,

${1 \over f} = \left( {\mu - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$

So, the focal length of L$_1$ remains the same as the original lens because the curvature of the surfaces does not change

So, ${\left( f \right)_{{L_1}}} = f$ .... (1)

Now, for L$_3$, L$_3$ is plano-convex lens

${R_1} = R,\,{R_2} = \infty$

So, ${1 \over {{{\left( f \right)}_{{L_3}}}}} = \left( {\mu - 1} \right)\left( {{1 \over R} - {1 \over \infty }} \right)$

$\Rightarrow {1 \over {{{\left( f \right)}_{{L_3}}}}} = \left( {\mu - 1} \right){1 \over R}$ ..... (2)

For whole double convex lens,

${1 \over f} = \left( {\mu - 1} \right)\left( {{1 \over R} - {1 \over { - R}}} \right)$ (As ${R_1} = R$, ${R_2} = - R$)

$\Rightarrow {1 \over f} = \left( {\mu - 1} \right)\left( {{1 \over R} + {1 \over R}} \right)$

$\Rightarrow {1 \over f} = \left( {\mu - 1} \right)\left( {{2 \over R}} \right)$ .... (3)

From (2) and (3),

${1 \over {{{\left( f \right)}_{{L_3}}}}} = {1 \over {2f}} \Rightarrow {\left( f \right)_{{L_3}}} = 2f$

Hence, ${{{{\left( f \right)}_{{L_1}}}} \over {{{\left( f \right)}_{{L_3}}}}} = {f \over {2f}} \quad\text{....(from (1)})\quad= 1:2$

(A) The unit "gauss" is used to measure magnetic induction, also known as magnetic flux density, within the CGS system of units.

$B = {\phi \over A} = {{\max well} \over {c{m^2}}} = gauss\,(CGS)$

(B) Magnetic intensity (H) is defined as the product of the number of turns per unit length in a coil (n) and the current that it carries (I).

i.e. $H = nI = {N \over L}I$

Unit : A/m or Am$^{-1}$

(C) Magnetic flux, $\phi = BA$

unit : weber

(D) Magnetic moment (M) : magnetic moment of a current loop, $M = NIA$

where, N = number of turns

I = current

A = Area

So, unit = Am$^2$

Hence, option A is correct.

We know, electric field due to uniformly charged infinite plane sheet,

$E = {{{\sigma _0}} \over {2{\varepsilon _0}}}$

perpendicularly directed away from the sheet for positive charge

So, $\overrightarrow E = {{{\sigma _0}} \over {2{\varepsilon _0}}} \uparrow \theta$

note : E is constant

Let length of dipole = d

so dipole moment, $\overrightarrow p = qd \uparrow$

We know,

Torque, $\overrightarrow \tau = \overrightarrow p \times \overrightarrow E$

$\overrightarrow \tau = qd\widehat i \times {{{\sigma _0}} \over {2{\varepsilon _0}}}\widehat i$

$\widehat \tau = O$ (as $\widehat i \times \widehat i = 0$)

force, on charge $- q,\overrightarrow {{F_1}} = - qE\widehat i$

on charge $+ q,\overrightarrow {{F_2}} = qE\widehat i$

So net force on dipole, $\overrightarrow F = \overrightarrow {{F_1}} + \overrightarrow {{F_2}}$

$= - qE\widehat i + qE\widehat i$

$\overrightarrow F = O$

We know, potential energy of dipole

$u(O) = - pE\cos \theta$

here, $O = 0^\circ$ (given)

So, ${u_{\min }} = - pE$

Hence, option B is correct.

Given, $r = 10\,cm = {1 \over {10}}\,m$

$I = 0.15\,A$, and ${{dv} \over {dt}} = 7 \times {10^8}v/s$

We know, for a parallel plate capacitor,

$c = {{{\varepsilon _0}A} \over d} \Rightarrow c = {{{\varepsilon _0}\pi {r^2}} \over d}$ .... (1)

In a capacitor,

$Q = CV$

$\Rightarrow V = {Q \over C}$

by differentiating w.r.t. t

$\Rightarrow {{dv} \over {dt}} = {1 \over C}{{dQ} \over {dt}}$

$\Rightarrow {{dv} \over {dt}} = {d \over {{\varepsilon _0}\pi {r^2}}}I$ (As $I = {{dQ} \over {dt}}$)

$\Rightarrow d = {{{\varepsilon _0}\pi {r^2}} \over I}{{dv} \over {dt}}$ .... (From (1))

$\Rightarrow d = {{9 \times {{10}^{ - 12}}} \over {0.15}} \times {{22} \over 7} \times {1 \over {100}} \times 7 \times 10$

$= {{198} \over {0.15}} \times {10^{ - 6}}$

$= 1320 \times {10^{ - 6}}$ m

$\Rightarrow d = 1320\,\mu m$

We know, for an oscillation,

${V_{\max }} = wA$ .... (1)

where, A = amplitude

w = frequency of oscillation

and $w = \sqrt {{k \over m}}$ .... (2)

from (1) and (2), ${V_{\max }} = \sqrt {{k \over m}} A$

$\Rightarrow {V_{\max }} \propto \sqrt k$ (As m and A are same (or constant))

$\Rightarrow {{{{\left( {{V_{\max }}} \right)}_A}} \over {{{\left( {{V_{\max }}} \right)}_B}}} = \sqrt {{{{k_1}} \over {{k_2}}}}$

Hence, option 2 is correct.

We know that two bodies of the same mass exchange their velocities after an elastic head-on collision.

Given, initially,

Here, ${V_{AB}} = {v_A} - {v_B} = 5 - 2 = 3\,m/s$

${V_{CB}} = {v_C} - {v_B} = 4 - 2 = 2\,m/s$

first A will collide with B and the velocities will be exchanged.

i.e. now, V$_A$ = 2 m/s and V$_B$ = 5 m/s and V$_C$ = 4 m/s

As $V_B > V_C$

So, B will collide and exchange the velocities with C.

So, $V_B=4$ m/s. and $V_C=5$ m/s.

Hence, finally,

\eqalign{ & {V_A} = 2\,m/s \cr & {V_B} = 4\,m/s \cr & {V_C} = 5\,m/s \cr}

So, A is false but R is true.

Let a planet of mass m orbits a star of mass M in circular orbit of radius r with speed v.

For this circular motion, centripetal force is provided by the gravitational force between both masses.

Hence, ${F_C} = {F_4}$

$\Rightarrow {{m{v^2}} \over r} = {{GMm} \over {{r^2}}}$

$\Rightarrow v = \sqrt {{{Gm} \over r}}$

Angular momentum of the planet, $L = mvr$

$\Rightarrow L = m\sqrt {{{Gm} \over r}} r$

$\Rightarrow L = \sqrt {GM} m\sqrt r$

Given, ${m_B} = 4\sqrt 2 {m_A}$ and ${R_B} = 2{R_A}$

M is same for both.

So, ${{{L_B}} \over {{L_A}}} = \left( {{{{m_B}} \over {{m_A}}}} \right)\sqrt {{{{R_B}} \over {{R_A}}}}$

$= \left( {4\sqrt 2 } \right)\sqrt 2 = 4 \times 2$

$\Rightarrow {{{L_B}} \over {{L_A}}} = 8$.

Given, ${{dm} \over {dt}} \propto \sqrt v$

So, ${{dm} \over {dt}} = k\sqrt v$ .... (i)

where, k = constant

We know, Power = $P = FV$

$\Rightarrow P = {{dp} \over {dt}}v$ (As $F = {{dp} \over {dt}}$ where, p = linear momentum)

$\Rightarrow P = {d \over {dt}}(mv)v$

$\Rightarrow P = v{{dm} \over {dt}}\,.\,v$ (As speed is constant i.e. v = constant)

$\Rightarrow P = {v^2}(k\sqrt v )$ [From (i)]

$\Rightarrow P = k{v^{5/2}}$

by squaring both sides,

$\Rightarrow {P^2} = {k^2}{v^5}$

$\Rightarrow {P^2} \propto {v^5}$ (As $\mathrm{k^2}$ = constant)

Hence, option 4 is correct.

We know, magnetic field due to solenoid, $B = {\mu _0}nI$

where, I = current, $n = {N \over L}$ = no. of turns per unit length

$\Rightarrow B = {{{\mu _0}NI} \over L}$

$\Rightarrow L = {{{\mu _0}NI} \over B} = {{4\pi \times {{10}^{ - 7}} \times 200 \times 0.29} \over {2.9 \times {{10}^{ - 4}}}}$

$\Rightarrow L = 8\pi \times {10^{ - 2}}\,m$

$\Rightarrow L = 8\pi \,cm$

Hence, answer = 8.

We know, $e{V_0} = hv - \phi$ .... (i)

$\Rightarrow e{V_0} = K{E_{\max }}$ (where $\phi$ = work function of the metal and $v_0$ = stopping potential)

$\Rightarrow {V_0} = \left( {{1 \over e}} \right)K{E_{\max }}$

Stopping potential depends solely on the frequency of the incident light $+(v)$ & the work function of the metal, not the intensity.

From eq. (i), we can see,

$v \uparrow \Rightarrow {v_0} \uparrow$

$\Rightarrow {e \over {\lambda \downarrow }} \Rightarrow {v_0} \uparrow$ (as $v = {c \over \lambda }$)

So, stopping potential increases with decrease in the wavelength of the incident light.

Hence, option 1 is correct.

To harvest heat energy by converting a temperature difference into electrical energy, a material should have low thermal conductivity and high electrical conductivity. These materials are called thermoelectric materials.

The material should not readily transfer heat, allowing a significant temperature difference to be maintained between the hot and cold sides. So low thermal conductivity is necessary. High electrical conductivity is needed to allow the efficient movement of electrons when a temperature difference is present.

Hence, option 1 is correct.