JEE Main 2023 29th January Evening Shift Question Paper with Solutions

$m = 20$ kg

$t = 20$ sec.

Acceleration $= {F \over {20}}$ m/s$^2$

$\therefore$ $v = u + at$

$v = 0 + \left( {{F \over {20}}} \right)(20)$

$= F$ ms$^{-1}$

Now for next 10 sec.

$S=ut$

$50=F(10)$

$F=5$

$w = \int {F\,.\,ds}$

$\overrightarrow F = q\overrightarrow E = 2 \times {10^{ - 2}} \times 30\widehat i = 0.6N\widehat i$

$w = \overrightarrow F .\overrightarrow d = (0.6\widehat i)\,.\,( - \widehat i, - 2\widehat j)$

$= - 0.6$ J

$= - 600$ mJ

${n_A} = 20$ moles

${n_B} = 10$ moles

$N = {N_0}{e^{ - \lambda t}}$

${N_A} = (20\,N){e^{ - \left( {{{\ln 2} \over 1} \times 2} \right)}}$

$= {{20\,N} \over 4} = 5\,N$ (N = Avogadro's Number)

${N_B} = 10\,N\,{e^{ - 4\ln 2}}$

$= \left( {{{10\,N} \over {16}}} \right)$

${N_A} + {N_B} = 5\,N + {{10\,N} \over {16}} = \left( {{{90\,N} \over {16}}} \right) = 3.38 \times {10^{24}}$

As per given information

${{{R_1} + {R_2}} \over {10}} = {{0.6} \over {0.4}}$ ...... (1)

& ${{{{{R_1}{R_2}} \over {{R_1} + {R_2}}}} \over 3} = {{0.4} \over {0.6}}$ ..... (2)

$\Rightarrow \left. \matrix{ {R_1} + {R_2} = 15 \hfill \cr \& \,{R_1}{R_2} = 30 \hfill \cr} \right] \Rightarrow {R_1}{R_2} = 30\,{\Omega ^2}$

By given information

$32 = 4 \times {{{\mu _0}i} \over {2r}}$ ..... (i)

Also, $r' = 4r$ ...... (ii)

and $B' = 1 \times {{{\mu _0}i} \over {2r'}}$ .... (iii)

$\Rightarrow B' = {{{\mu _0}i} \over {2(4r)}} = {{{\mu _0}i} \over {8r}} = {1 \over 8} \times 16 = 2\,T$

Here, At is distance, so dimensions of

$[At] = [x] = [L]$

Given. The dimensions of t is $[\mathrm{T^{-1}]}$

${\left[A \times \mathrm{T}^{-1}\right]=[\mathrm{L}] \Rightarrow[A]=[\mathrm{LT}]}$

$\left[ {{t \over B}} \right] = [y] = [L]$

$\Rightarrow \frac{\mathrm{T}^{-1}}{[B]}=[L] \Rightarrow B=\left[\mathrm{L}^{-1} \mathrm{~T}^{-1}\right]$

We know that

$\tan {\theta _0} = {{{\mu _2}} \over {{\mu _1}}}$

$\tan {\theta _0} = \sqrt {{{6.8} \over {2.8}}} = \sqrt {{{17} \over 7}}$

${\theta _0} = {\tan ^{ - 1}}\sqrt {1 + {{10} \over 7}} \Rightarrow \theta = 7$

Statement I is correct as photon do not carry any charge, hence cannot feel force from either fields.

Statement II is wrong as $E_0=cB_0$

${E_0} = {{{B_0}} \over {\sqrt {{\mu _0}{\varepsilon _0}} }}$

$\mu=\frac{\mathrm{real\,depth}\,(l_1)}{\mathrm{apparent\,depth}\,(l_2)}$

$=\frac{5.25}{5}=1.05$

$\frac{d\mu}{\mu}=\frac{dl_1}{l_1}+\frac{dl_2}{l_2}$

$d\mu = \left( {{{d{l_1}} \over {{l_1}}} + {{d{l_2}} \over {{l_2}}}} \right)\mu$

$= \left( {{{0.01} \over {5.25}} + {{0.01} \over {5.00}}} \right) \times 1.05$

$= {{41} \over {10}} \times {10^{ - 3}}$

so $x = 41$

As the block moves with constant speed, the horizontal force is balanced by viscous force thus

$F = \eta A{{\Delta v} \over {\Delta z}}$

$0.1 = 5 \times {10^{ - 3}} \times 0.2 \times {v \over {.25 \times {{10}^{ - 3}}}}$

$\Rightarrow v = 25 \times {10^{ - 3}}$ m/s

$\overrightarrow v = 4\widehat j$ (m/s)

$a = {{{v^2}} \over R} = {{16} \over 2} = 8$ m/s$^2$

$\overrightarrow a = 8\left( {m/{s^2}} \right)\left( {\widehat i} \right)$

Heat required to raise the temperature of ice to 0$^\circ$C is

$= {{60} \over {1000}}(2222.3)(12)$

$= 16000.5$ J

$\approx 16$ kJ

Heat required to melt ice completely

$= \left( {{{600} \over {1000}}} \right)(336)$ kJ

$= 201.6$ kJ

Energy left $= (184 - 16) = 168$ kJ

$\therefore$ Partial ice will melt

$\therefore$ $168 = ({m_{ice\,melted}})336$

$0.5$ kg $= ({m_{ice\,melted}})$

$\therefore$ ${m_{ice}}:{m_{water}} = 1:5$

Let the emf is E and internal resistance is r of this secondary cell so

${{RE} \over {r + R}} \propto l$

so ${{{R_1}E} \over {r + {R_1}}} \propto {l_1}$

& ${{{R_2}E} \over {r + {R_2}}} \propto {l_2}$

$\Rightarrow {{{R_1}(r + {R_2})} \over {{R_2}(r + {R_1})}} = {{{l_1}} \over {{l_2}}}$

or ${{5(r + 15)} \over {15(r + 5)}} = {{200} \over {300}}$

$\Rightarrow r = 5\,\Omega$

$V = 2a{r^2} + b$

$\Rightarrow E = - {{dV} \over {dr}} = - 4ar$

$\Rightarrow {1 \over {4\pi \varepsilon }}{Q \over {{r^2}}} = - 4ar$

$\Rightarrow {Q \over {{4 \over 3}\pi {r^3}}} = 3 \times \varepsilon \times ( - 4a) = - 12a\varepsilon$

$\Rightarrow \lambda = 12$

When a man lifts a bucket out of a well using a rope, work is done by the man and the gravitational force. The work done by the man is positive as he has to exert an upward force to lift the bucket. The work done by the gravitational force is negative because the direction of the force is opposite to the direction of displacement.

Therefore, the statement (A) "Work done by a man in lifting a bucket out of a well by means of rope tied to the bucket is negative." is incorrect.

Therefore, the statement (B) "Work done by gravitational force in lifting a bucket out of a well by a rope tied to the bucket is negative." is correct.

Work is defined as the product of force and displacement in the direction of the force. When a body slides down an inclined plane, the force of friction acts against the motion of the body, opposing its descent.

The direction of the force of friction is opposite to the direction of the displacement of the body, which is downwards. Hence, the work done by the force of friction is negative.

Therefore, the statement (C) "Work done by friction on a body sliding down an inclined plane is positive" is incorrect.

If the body is moving on a rough horizontal plane, there will be friction present, which will act in the opposite direction to the applied force. The force of friction will oppose the motion of the body, reducing its velocity. As a result, the net work done on the body will not be zero, as the force of friction and the applied force will not cancel each other out completely.

Therefore, the statement (D) "Work done by an applied force on a body moving on a rough horizontal plane with uniform velocity is zero." is incorrect.

Statement E: "Work done by the air resistance on an oscillating pendulum is negative."

This statement refers to the work done by the air resistance on an oscillating pendulum, which is a physical system that swings back and forth under the influence of gravity.

As the pendulum oscillates, it experiences air resistance, which opposes its motion and slows it down. The direction of the air resistance force is opposite to the direction of the displacement of the pendulum, which is back and forth.

Hence, the work done by the air resistance force is negative, as the direction of the force and the displacement are opposite.

Therefore, the statement (E) "Work done by the air resistance on an oscillating pendulum is negative" is correct.

The modulation index (m) of an amplitude-modulated wave is defined as the ratio of the amplitude of the modulating signal to the amplitude of the carrier wave. It is a measure of the degree of modulation of the carrier wave and is given by the following equation:

From energy conservation

Work done to pull the loop out = Energy is lost in the resistance

Emf in the loop $= {{d\phi } \over {dt}} = {{B \times A} \over t} = {{40 \times 25 \times {{10}^{ - 4}}} \over {1s}} = 0.1\,V$

Energy lost $= {{em{f^2}} \over R} = {{{{(0.1)}^2}} \over {10}} = {10^{ - 3}}\,J$

A potentiometer is an electrical instrument used to measure the electromotive force (EMF) of a cell.

The sensitivity of a potentiometer is defined as the change in potential difference per unit length of the wire.

It is directly proportional to the length of the potentiometer wire (Option A), because a longer wire has a larger potential difference that can be measured.

It is inversely proportional to the potential gradient of the wire (Option C), because a smaller potential gradient results in a smaller change in potential difference per unit length.

Therefore, the correct answer is (C) A and C only.

Horizontal displacement $x = v\cos \theta t$

$= 10\sqrt 2 t$

So torque of weight about point of projection is

$\tau = mgx\,.\,( - \widehat k)$

${{d\overrightarrow L } \over {dt}} = mgx( - \widehat k)$

$\int_0^{\overrightarrow L } {d\overrightarrow L = 0.1 \times 10 \times 10\sqrt 2 \int_0^2 {t\,dt( - \widehat k)} }$

$\overrightarrow L = - 20\sqrt 2 \widehat k$

$|\overrightarrow L | = 20\sqrt 2 = \sqrt {800}$ kg m$^2$/s

$\because {T^2} \propto {R^3}$

$\therefore$ ${{T_1^2} \over {T_2^2}} = {{R_1^3} \over {R_2^3}}$

${{{{24}^2}} \over {T_2^2}} = {{R_1^3} \over {{{\left( {{{{R_1}} \over 4}} \right)}^3}}}$

${{{{24}^2}} \over {T_2^2}} = {4^3}$

${T_2} = {{24} \over {{2^3}}} = 3$ hours

Resolving power of microscope $= \left( {{{2n\sin \theta } \over \lambda }} \right)$

$n\sin \theta$ = Numerical aperture

$n$ is the refractive index of medium.

Smooth case:

$a = g\sin 45^\circ = {9 \over {\sqrt 2 }}$

${t_1} = \sqrt {{{2L} \over a}} = \sqrt {{{2L} \over {g/\sqrt 2 }}} = \sqrt {{{2\sqrt 2 L} \over g}}$ ..... (1)

Rough case:

$a = g\sin 45^\circ - \mu g\cos 45^\circ$

$= {g \over {\sqrt 2 }}(1 - \mu )$

${t_2} = \sqrt {{{2L} \over a}} = \sqrt {{{2\sqrt 2 L} \over {g(1 - \mu )}}}$ ..... (2)

From (1) to (2) and ${t_1} = {{{t_2}} \over n}$ we have

$\sqrt {{{2\sqrt 2 L} \over g}} = {1 \over n}\sqrt {{{2\sqrt 2 L} \over {g(1 - \mu )}}} \Rightarrow \mu = 1 - {1 \over {{n^2}}}$

Velocity of aircraft = 1050 km/h = 300 m/s

Now, weight of aircraft = $\Delta PA$

$\Delta P = {{5.4 \times {{10}^5} \times g} \over {500}} = 10800$ $\mathrm{Pa}$

From Bernoulli's principle

$\Delta P = {1 \over 2}\rho \left[ {V_{upper}^2 - V_{lower}^2} \right]$

$10800 = {1 \over 2} \times 1.2 \times V_{lower}^2\left[ {{{\left( {{{{V_{upper}}} \over {{V_{lower}}}}} \right)}^2} - 1} \right]$

${\left( {{{{V_{upper}}} \over {{V_{lower}}}}} \right)^2} = 1 + {{10800 \times 2} \over {1.2 \times {{(300)}^2}}} = 1.2$

${{{V_{upper}}} \over {{V_{lower}}}} = 1.096$

$\Rightarrow$ Fractional increases = 9.6%

${{dv} \over {dt}} = {{{v^2}} \over R} \Rightarrow {{{v^2}} \over R} = v{{dv} \over {ds}}$

$\Rightarrow {{dv} \over v} = {{ds} \over R} \Rightarrow \left. {\ln v} \right|_{15}^v = {s \over R}$

$\Rightarrow v = 15{e^{\Delta /R}} = {{ds} \over {dt}} \Rightarrow dt = {1 \over {15}}{e^{ - \Delta /R}}ds$

$\Delta t = {R \over {15}}[1 - {e^{ - \Delta /R}}]$

$= 40[1 - {e^{ - \pi /2}}]$ seconds

$\Rightarrow t = 40$

${v_{rms}} = \sqrt {{{\alpha + 5} \over \alpha }} {v_{avg}}$

$\sqrt {{{3RT} \over m}} = \sqrt {{{\alpha + 5} \over 5}} \sqrt {{{8RT} \over {\pi m}}}$

${{3 \times \pi } \over 8} = {{\alpha + 5} \over \alpha }$

${{33} \over {28}} = {{\alpha + 5} \over \alpha }$

$\alpha = 28$

As per the circuit,

$X=A'B+AB'$

If : $A=0,B=0\Rightarrow X=0$

$A=0,B=1\Rightarrow X=1$

$A=1,B=0\Rightarrow X=1$

$A=1,B=1\Rightarrow X=0$

For (a), $i = {V \over R} = {{220} \over {40}} = 5.5\,A$

for (b), ${X_C} = {1 \over {2\pi fC}} = {1 \over {2\pi \times 50 \times 0.5 \times {{10}^{ - 6}}}} = {{{{10}^6}} \over {50\pi }}\Omega$

${X_L} = 2\pi fL = 2\pi \times 50 \times 50 \times {10^{ - 3}} = 50\pi \,\Omega$

${X_C} > {X_L}$, hence impedance is greater than 40 $\Omega$.

${i_{rms}} = {{220} \over Z}$

$\therefore$ ${\left. {{i_{rms}}} \right|_b} < {\left. {{i_{rms}}} \right|_a}$

Current drawn is maximum when circuit is in resonance.

$\omega = {1 \over {\sqrt {LC} }}$

$2\pi (7000) = {1 \over {\sqrt {2 \times {{10}^{ - 6}}C} }}$

$\Rightarrow C = {1 \over {3872}}F$

$F = - 25x$

$.250{{{d^2}x} \over {d{t^2}}} = - 25x$

${{{d^2}x} \over {d{t^2}}} = - 100x$

$\Rightarrow \omega = 10$ rad/sec

& $\omega A = {v_{\max }}$

$10\,A = 4$

$\Rightarrow A = 0.4$ m

$= 40$ cm

The tetrapeptide code is

FLDY

F $\to$ Phenyl alanine

L $\to$ Leucine

D $\to$ Aspartic acid

Y $\to$ Tyrosine

(A) $\mathrm{\Delta U=q-p\Delta V}$

(B) $\mathrm{\Delta G=\Delta H-T\Delta S}$

(C) $\mathrm{\Delta S=\frac{q_{rev}}{T}}$

(D) $\mathrm{\Delta H=\Delta U+(\Delta nRT)}$

Hence, (B) and (C) relations are correct.

Statement I is correct, Statement II is incorrect as Si forms electron precise hydride.

Statement I is correct as the decrease in first ionization enthalpy from B to Al is indeed much larger than from Al to Ga. This is due to the shielding effect of the electron in the outermost shell and the increased effective nuclear charge as one moves from B to Al and from Al to Ga.

Statement II is also correct, as the d orbitals in Ga are indeed completely filled. This results in the shielding effect being maximum for Ga and thus, the decrease in first ionization enthalpy from Al to Ga is smaller compared to that from B to Al.

$\mathrm{2NH_3(g)\frac{5}{2}O_2(g)\rightleftharpoons 2NO(g)+3H_2O(g)}$

Clearly, $\mathrm{{K_{eq}} = {1 \over {{k_1}}} \times {k_2} \times k_3^3}$

$= {{1.6 \times {{10}^{12}} \times {{10}^{ - 39}}} \over {4 \times {{10}^5}}}$

$= 0.4 \times {10^{ - 32}}$

$= 4 \times {10^{ - 33}}$

(i) $\mathrm{Mn_2O_7\to Mn}$ in (+7) oxidation state

(ii) It is also correct

(iii) Sc only shows +3 oxidation state

(iv) $\mathrm{Cr^{+6}}$ is oxidising in nature

Number of voids

$=0.02\times3\times6.02\times10^{23}$

$\simeq36\times10^{21}$

${C_x}{H_y} + \left( {x + {y \over 4}} \right){O_2} \to xC{O_2} + {y \over 2}{H_2}O$

${y \over 2} = 3$

$y = 6$

$x + {y \over 4} = {{19} \over 2}$

$x = {{19} \over 2} - {3 \over 2} = 8$

So, formula is ${C_8}{H_6}$.

Propanamide $\mathrm{\buildrel {B{r_2}/KOH} \over \longrightarrow}$ Ethylamine

Starch forms blue coloured complex with iodine.

CrO$_5$ is blue in colour in amyl alcohol.

$E_{cell}^o = {{2.303\,RT} \over {2F}}\log k$

$E_{cell}^o = {{0.059} \over 2}\log ({10^{20}})$

$E_{Z{n^{2 + }}/Zn}^o + 0.76 = 0.59$

$E_{Z{n^{2 + }}/Zn}^o = 0.59 - 0.76$

$E_{Zn/Z{n^{2 + }}}^o = 0.17\,V$