JEE Main 2021 27th July Evening Shift Question Paper with Solutions

${{\Delta y} \over y} = {{2\Delta m} \over m} + {{4\Delta r} \over r} + {{x\Delta g} \over g} + {3 \over 2}{{\Delta l} \over l}$

$18 = 2(1) + 4(0.5) + xp + {3 \over 2}(4)$

$\Rightarrow$ 8 = xp

By checking from options.

$x = {{16} \over 3},p = \pm {3 \over 2}$

At t = 0, u = 0

$a = {{{F_0}} \over M} - {{{F_0}} \over {M{T^2}}}{(t - T)^2} = {{dv} \over {dt}}$

$\int\limits_0^v {dv = \int\limits_{t = 0}^{2T} {\left( {{{{F_0}} \over M} - {{{F_0}} \over {M{T^2}}}{{(t - T)}^2}} \right)dt} }$

$V = \left[ {{{{F_0}} \over M}t} \right]_0^{2T} - {{{F_0}} \over {M{T^2}}}\left[ {{{{t^3}} \over 3} - {t^2}T + {T^2}t} \right]_0^{2T}$

$\Rightarrow$ $V = {{4{F_0}T} \over {3M}}$

Graph for wire of radius R :


As b > a

B_a > B_b

${B_a} = {{{\mu _0}i} \over {2\pi a}}$

${B_b} = {{{\mu _0}i} \over {2\pi b}}$

${W_A} = 1 - {{{Q_2}} \over {{Q_1}}} = 1 - {T \over {{T_1}}} \Rightarrow {{{Q_2}} \over {{Q_1}}} = {T \over {{T_1}}}$

${W_B} = 1 - {{{Q_3}} \over {({Q_2}/2)}} = 1 - {{{T_3}} \over T} \Rightarrow {{2{Q_3}} \over {{Q_2}}} = {{{T_3}} \over T}$

Now, W_A = W_B

${Q_1} - {Q_2} = {{{Q_2}} \over 2} - {Q_3}$

$\Rightarrow {{2{Q_1}} \over {{Q_2}}} + {{2{Q_3}} \over {{Q_2}}} = 3$

$\Rightarrow {{2{T_1}} \over T} + {{{T_3}} \over T} = 3$

$\Rightarrow$ ${{2{T_1}} \over 3} + {{{T_3}} \over 3} = T$

Since, each vibrational mode, corresponds to two degrees of freedom, hence, f = 3 (trans.) + 3 (rot.) + 4 $\times$ 2 (vib.) = 14

& $\gamma = 1 + {2 \over f}$

$\gamma = 1 + {2 \over {14}} = {8 \over 7}$

$W = {{nR\Delta T} \over {\gamma - 1}} = - 582$

As W < 0. work is done on the gas.

$\left| \in \right| = {{d\phi } \over {dt}}$ = 20t + 20 mV

$\left| i \right| = {{\left| \in \right|} \over R}$ = 10t + 10 mA

at t = 5

$\left| i \right|$ = 60 mA

Both velocity vectors are of same magnitude therefore resultant would pass exactly midway through them

$\theta$ = 30$^\circ$

${E_{{k_\alpha }}} = {E_k} - {E_L}$

${{hc} \over {{\lambda _{{k_\alpha }}}}} = {E_k} - {E_L}$

${E_L} = {E_k} - {{hc} \over {{\lambda _{{k_\alpha }}}}}$

= 27.5 KeV $- {{12.42 \times {{10}^{ - 7}}eVm} \over {0.071 \times {{10}^{ - 9}}m}}$

E_L = (27.5 $-$ 17.5) keV

= 10 keV

E_mech. = 8J

(A) at x > x₄

U = constant = 6J

K = E_mech. $-$ U = 2J = constant

(B) at x < x₁

U = constant = 8J

K = E_mech $-$ U = 8 $-$ 8 - 0 J

Particle is at rest.

(C) At x = x₂,

U = 0 $\Rightarrow$ E_mech. = K = 8 J

KE is greatest, and particle is moving at fastest speed.

(D) At x = x₃

U = 4J

U + K = 8J

K = 4J

C = 0.1 $\mu$F = 10^$-$7 F

Resonant frequency = 60 Hz.

${\omega _0} = {1 \over {\sqrt {LC} }}$

$2\pi {f_0} = {1 \over {\sqrt {LC} }} \Rightarrow L = {1 \over {4{\pi ^2}f_0^2C}}$

by putting values $L \simeq 70.3$ Hz.

A_max = A_c + A_m = 12

A_min = A_c $-$ A_m = 3

$\Rightarrow$ A_c = ${{15} \over 2}$ & A_m = ${9 \over 2}$

modulation index = ${{{A_m}} \over {{A_c}}} = {{9/2} \over {15/2}} = 0.6$

$\Rightarrow$ x = 1

x(t) = A sin($\omega$t + $\phi$)

v(t) = A$\omega$ cos ($\omega$t + $\phi$)

2 = A sin$\phi$ ...... (1)

2$\omega$ = A$\omega$ cos$\phi$ ....... (2)

From (1) and (2)

tan$\phi$ = 1

$\phi$ = 45$^\circ$

Putting value of $\phi$ in equation (1),

$2 = A\left\{ {{1 \over {\sqrt 2 }}} \right\}$

$A = 2\sqrt 2$

$\therefore$ x = 2

Thickness t = n$\lambda$

So, n $\lambda$_vac = (n + 1) $\lambda$_air

n $\lambda$ = (n + 1) ${\lambda \over {{\mu _{air}}}}$

n = ${1 \over {{\mu _{air}} - 1}} = {{{{10}^4}} \over 3}$

t = n$\lambda$

$= {{{{10}^4}} \over 3} \times 6000\mathop A\limits^o$

= 2 mm

(A) True, atom of each element emits characteristic spectrum.

(B) True, according to Bohr's postulates $mvr = {{nh} \over {2\pi }}$ and hence electron resides into orbits of specific radius called stationary orbits.

(C) False, density of nucleus is constant.

(D) False, A free neutron is unstable decays into proton and electron and antineutrino.

(E) True, unstable nucleus show radioactivity.

Standard graph between angle of deviation and incident angle.

q = CV

$[C] = \left[ {{q \over V}} \right] = {{(A \times T)} \over {M{L^2}{T^{ - 2}}}}$

$= {M^{ - 1}}{L^{ - 2}}{T^4}{A^2}$

$[E] = \left[ {{F \over q}} \right] = {{ML{T^{ - 2}}} \over {AT}}$

$= ML{T^{ - 3}}{A^{ - 1}}$

$F = {{{q_1}{q_2}} \over {4\pi { \in _0}{r^2}}}$

$[{ \in _0}] = {M^{ - 1}}{L^{ - 3}}{T^4}{A^2}$

Speed of light $c = {1 \over {\sqrt {{\mu _0}{ \in _0}} }}$

${\mu _0} = {1 \over {{ \in _0}{c^2}}}$

$[{\mu _0}] = {1 \over {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}]{{[L{T^{ - 1}}]}^2}}}$

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

P = const.

$P = Fv = {{m{v^2}dv} \over {dx}}$

$\int\limits_0^x {{P \over m}dx} = \int\limits_0^v {{v^2}dv}$

${{Px} \over m} = {{{v^3}} \over 3}$

${\left( {{{3Px} \over m}} \right)^{1/3}} = v = {{dx} \over {dt}}$

${\left( {{{3P} \over m}} \right)^{1/3}}\int\limits_0^t {dt} = \int\limits_0^x {{x^{ - 1/3}}} dx$

$\Rightarrow x = {\left( {{{8P} \over {9m}}} \right)^{1/2}}{t^{3/2}}$

16 = R₀ [1 + $\alpha$ (15 $-$ T₀)]

20 = R₀ [1 + $\alpha$ (100 $-$ T₀)]

Assuming T₀ = 0$^\circ$C, as a general convention.

$\Rightarrow$ ${{16} \over {20}} = {{1 + \alpha \times 15} \over {1 + \alpha \times 100}}$

$\Rightarrow$ $\alpha$ = 0.003$^\circ$C^$-$1

initially, energy of electron = + 3eV

finally, in 2^nd excited state,

energy of electron = $- {{(13.6eV)} \over {{3^2}}}$

$= - 1.51eV$

Loss in energy is emitted as photon,

So, photon energy ${{hc} \over \lambda } = 4.51eV$

Now, photoelectric effect equation

$K{E_{\max }} = {{hc} \over \lambda } - \phi = 4.51 - \left( {{{hc} \over {{\lambda _{th}}}}} \right)$

$= 4.51eV - {{12400eV\mathop A\limits^o } \over {4000\mathop A\limits^o }}$

$= 1.41eV$

$R = \sqrt {2gh} \times \sqrt {{{(12 - h) \times 2} \over g}}$

$\sqrt {4h(12 - h)} = R$

For maximum R

${{dR} \over {dh}} = 0$

$\Rightarrow$ h = 6 m

At terminal speed

a = 0

F_net = 0

mg = F_v = 6$\pi$ $\eta$Rv

$v = {{mg} \over {6\pi \eta Rv}}$

$v = {{{\rho _w}{{4\pi } \over 3}{R^3}g} \over {6\pi \eta R}}$

$= {{2{\rho _w}{R^2}g} \over {9\eta }}$

$= {{400} \over {81}}$ m/s

= 4.94 m/s

The problem describes two identical particles of mass m=1kg each moving in a circle of radius R under the action of their mutual gravitational attraction. This means that the particles are moving around a common center, and the distance between the particles is 2R (as the diameter of the circle).

In this case, the force providing the centripetal force for each particle to move in a circular path is the gravitational force between the particles.

The gravitational force between two masses m1 and m2 separated by a distance r is given by Newton's law of universal gravitation:

$F = G \frac{{m_1 m_2}}{{r^2}}$

Since the two particles are identical, m1=m2=m=1kg. And the distance between them r is 2R. Substituting these into the above equation gives the gravitational force between the two particles:

$F = G \frac{{m^2}}{{(2R)^2}} = G \frac{{1}}{{4R^2}}$

This gravitational force is also equal to the centripetal force needed for each particle to move in a circular path of radius R. The centripetal force is given by:

$F = m R ω^2$

Setting these two equations equal to each other gives:

$G \frac{{1}}{{4R^2}} = 1 \cdot R \cdot ω^2$

Rearranging this to solve for ω (the angular speed) gives:

$ω = \frac{1}{2} \sqrt{\frac{G}{R^3}}$

So, the angular speed of each particle is: $\frac{1}{2} \sqrt{\frac{G}{R^3}}$.

Y = A . B + $\overline B$

From graph voltage at t = 3.2 sec is 6 volt.


i = ${{6 - 5} \over 1}$

$\Rightarrow$ i = 1 A

$mg\cos \theta = {{m{v^2}} \over R}$ .... (1)

$\cos \theta = {h \over R}$

Energy conservation

$mg\{ R - h\} = {1 \over 2}m{v^2}$ ..... (2)

from (1) & (2)

$\Rightarrow$ $mg\left\{ {{h \over R}} \right\} = {{2mg\{ R - h\} } \over R}$

$h = {{2R} \over 3}$ = 2m

Option D is correct.

${T^2} = {{4{\pi ^2}} \over {GM}}.{r^3}$

$M = {{4{\pi ^2}} \over G}.{{{r^3}} \over {{T^2}}}$

by putting values

$M = 6 \times {10^{23}}$

$T = 2\pi \sqrt {{m \over k}}$

$0.2 = 2\pi \sqrt {{{0.5} \over k}}$

k = 50$\pi$²

$\approx$ 500

x = A sin ($\omega$t + $\phi$)

= 5 cm sin $\left( {{{\omega T} \over 4} + 0} \right)$

= 5 cm sin $\left( {{\pi \over 2}} \right)$

= 5 cm

$PE = {1 \over 2}k{x^2}$

$= {1 \over 2}(500){\left( {{5 \over {100}}} \right)^2}$

= 0.6255

Let E be electric field in air

T sin$\theta$ = qE

T cos$\theta$ = mg

tan$\theta$ = ${{qE} \over {mg}}$


$Q = \left[ {{{{C_1}{C_2}} \over {{C_1} + {C_2}}}} \right][{V_1} + {V_2}]$

$E = {Q \over {A{ \in _0}}} = \left[ {{{{C_1}{C_2}} \over {{C_1} + {C_2}}}} \right]{{[{V_1} + {V_2}]} \over {A{ \in _0}}}$

${C_1} = {{{ \in _0}A} \over {d - t}} \Rightarrow E = {{{C_2}[{V_1} + {V_2}]} \over {({C_1} + {C_2})(d - t)}}$

Now, $\theta = {\tan ^{ - 1}}\left[ {{{q.E} \over {mg}}} \right]$

$\theta = {\tan ^{ - 1}}\left[ {{q \over {mg}} \times {{{C_2}({V_1} + {V_2})} \over {({C_1} + {C_2})(d - t)}}} \right]$

n_eq KMnO₄ = n_eq H₂C₂O₄ . 2H₂O

or, ${{10 \times 0.05} \over {1000}} \times 5 = {{10 \times M} \over {1000}} \times 2$

$\therefore$ Conc. of oxalic acid solution = 0.125 M

= 0.125 $\times$ 125 g/L = 15.75 g/L

= 1575 $\times$ 10^$-$2 g/L

${n_{{H^ + }}} = {{400 \times 0.2} \over {1000}} \times 2 = 0.16$

${n_{O{H^ - }}} = {{600 \times 0.1} \over {1000}} = 0.06$ (L.R.)

Now, heat liberated from reaction = heat gained by solutions

or, 0.06 $\times$ 57.1 $\times$ 10³

= (1000 $\times$ 1.0) $\times$ 4.18 $\times$ $\Delta$T

$\therefore$ $\Delta$T = 0.8196K

= 81.96 $\times$ 10^$-$2 K $\approx$ 82 $\times$ 10^$-$2 K

N_b = No of electrons in bonding molecular orbital

N_a $=$ No of electrons in anti bonding molecular orbital

(1) $\,\,\,\,$ upto 14 electrons, molecular orbital configuration is



Here N_a = Anti bonding electron $=$ 4 and N_b = 10

(2) $\,\,\,\,$ After 14 electrons to 20 electrons molecular orbital configuration is - - -



Here N_a = 10

and N_b = 10

In O atom 8 electrons present, so in O₂, 8 $\times$ 2 = 16 electrons present.

Then in $O_2^ +$ no of electrons = 15

in $O_2^ -$ no of electrons = 17

in $O_2^{2 - }$ no of electrons = 18

Molecular orbital configuration of O $_2^{2 - }$ (18 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^2}^ *$

$\therefore\,\,\,\,$ N_b = 10

and N_a = 8

1, 1, 1-Trichloro ethane [CCl₃-CH₃]

Hybridization of all the given complexes are given in following table.

${[Mn{(CN)_6}]^{3 - }} \Rightarrow M{n^{3 + }}$

CN^$-$ is a strong field ligand.

$\therefore$ Pairing of electron takes place.

${[Fe{(CN)_6}]^{3 - }} \Rightarrow F{e^{3 + }}$

CN^$-$ is a strong ligand.

$\therefore$ Pairing of electron takes place.

${[Co{({C_2}{O_4})_3}]^{3 - }} \Rightarrow C{o^{3 + }}$

${C_2}O_4^{2 - }$ acts as strong field ligand with Co³⁺.

$\therefore$ Pairing of electrons takes place.

${[Fe{F_6}]^{3 - }} \Rightarrow F{e^{3 + }}$

F is a weak field ligand.

$\therefore$ Pairing of electron does not occurs.

${[MnC{l_6}]^{3 - }} \Rightarrow M{n^{3 + }}$

Cl^$-$ is a weak field ligand.

$\therefore$ Pairing of electron does not occurs.

${[Fe{F_6}]^{3 - }}$ have 5 unpaired electrons and ${[MnC{l_6}]^{3 - }}$ have 4 unpaired electrons.

So, both statement I and statement II are true.

(A) Crystalline solids have definite arrangement of constituent particles and have long range order.

(C), (D) Different constituent particles of an amorphous solid have different bond strengths and soften over a range of temperatures.

\matrix{ {Mg} & {Al} & P & {S \to IE} \cr } order $\Rightarrow$ Al < Mg < S < P

Valence [N_e] : \matrix{ {\mathop {3{s^2}}\limits^{Mg} } & {\mathop {3{s^2}3{P^1}}\limits^{Al} } & {\mathop {3{s^2}3{P^3}}\limits^P } & {\mathop {3{s^2}3{P^4}}\limits^S } \cr }

Statement I : Penicillin is bactericidal not bacteriostatic hence given statement is false.

Statement II : Structure of penicillin given is correct.

Silica is used to remove FeO impurity from the ore of copper

$FeO + Si{O_2} \to \mathop {FeSi{O_3}}\limits_{iron\,silicate\,(Slag)}$

2SO₂(g) + O₂(g) $\to$ 2SO₃(g)


$\therefore$ Final total pressure = 625 + 250 = 875 m bar

Li makes alloy with Lead to make white metal bearings for motor engines

Liquid Na metal is used s coolant in fast breeder nuclear reactor

K is a very absorbent of CO₂

Cs is used in making photoelectric cell

By sodium fusion extract we can detect Sulphur, Nitrogen, Phosphorous and Halogens, because they are converted in to their ionic form with sodium metal.

k_p = Po$_2^{1/2}$ = 4

$\therefore$ Po₂ = 16 bar = 16 atm