JEE Main 2021 25th July Evening Shift Question Paper with Solutions

no. of moles is conserved

n₁ + n₂ = n₃

P₁V₁ + P₂V₂ = P₃V

${{4S} \over {{r_1}}}\left( {{4 \over 3}\pi r_1^3} \right) + {{4S} \over {{r_2}}}\left( {{4 \over 3}\pi r_2^3} \right) = {{4S} \over {{r_3}}}\left( {{4 \over 3}\pi r_3^3} \right)$

$r_1^2 + r_2^2 = r_3^2$

${r_3} = \sqrt {r_1^2 + r_2^2}$

F = (5y + 20)$\widehat j$

$W = \int {Fdy = \int\limits_0^{10} {(5y + 20)dy} }$

$= \left( {{{5{y^2}} \over 2} + 20y} \right)_0^{10}$

$= {5 \over 2} \times 100 + 20 \times 10$

$= 250 + 200 = 450$ J

${V_s} = hv - \phi$

$4.8 = {{hc} \over \lambda } - \phi$ ..... (i)

$1.6 = {{hc} \over {2\lambda }} - \phi$ ..... (ii)

Using above equation (i) - (ii)

$3.2 = {{hc} \over \lambda } - {{hc} \over {2\lambda }}$

$3.2 = {{hc} \over {2\lambda }}$ ..... (iii)

$\left[ {\lambda = {{hc} \over {6.4}}} \right]$

Put in equation (ii)

$\phi$ = 1.6

${{hc} \over {{\lambda _{th}}}} = 1.6$

${\lambda _{th}} = {{hc} \over {1.6}}$

$= \left( {{{hc} \over {6.4}}} \right) \times 4 = 4\lambda$

For figure (a)

${P_{avg}} = {{v_{rms}^2} \over R}$

${{v_{rms}^2} \over {{Z^2}}} \times R = {{v_{rms}^2} \over R} \times 1$

${R^2} = {Z^2}$

$25 = {\left( {\sqrt {{{({x_C} - {x_L})}^2} + {5^2}} } \right)^2}$

$= 25{({x_C} - {x_L})^2} + 25$

${x_C} = {x_L} \Rightarrow {1 \over {\omega C}} = \omega L$

${\omega ^2} = {1 \over {LC}} = {{{{10}^6}} \over {0.1 \times 40}}$

$\omega = 500$

${n_e}{n_h} = {n_i}^2$

${n_e} = {{{n_i}^2} \over {{n_h}}} = {{{{(1.5 \times {{10}^{16}})}^2}} \over {4.5 \times {{10}^{22}}}}$

$= {{1.5 \times 1.5 \times {{10}^{32}}} \over {4.5 \times {{10}^{22}}}}$

$= 5 \times {10^9}$/m³

$V = \alpha t + \beta {t^2}$

${{ds} \over {dt}} = \alpha t + \beta {t^2}$

$\int\limits_{{S_1}}^{{S_2}} {ds = \int\limits_1^2 {(\alpha t + \beta {t^2})dt} }$

${S_2} - {S_1} = \left[ {{{\alpha {t^2}} \over 2} + {{\beta {t^3}} \over 3}} \right]_1^2$

As particle is not changing direction

So distance = displacement

Distance = $\left[ {{{\alpha [4 - 1]} \over 2} + {{\beta [8 - 1]} \over 3}} \right]$

$= {{3\alpha } \over 2} + {{7\beta } \over 3}$

When a dielectric is inserted in a capacitor

Due to free charge $\overrightarrow E = {\overrightarrow E _0}$ only

After dielectric $E' = {{{E_0}} \over k}$

${q_B} = {q_f}\left( {1 - {1 \over k}} \right)$

Density is same

$M = {4 \over 3}\pi {R^3}\rho ,2m = {4 \over 3}\pi R{'^3}\rho$

$R' = {2^{1/3}}R$

$\omega = {{GMm} \over {{R^2}}}$

${\omega _2} = {{G2Mm} \over {R{'^2}}}$

${\omega _2} = {2^{1/3}}\omega$

Given X = Y

$\sqrt {{X^2} + {Y^2} - 2 \times Y\cos \theta }$

$= n\sqrt {{X^2} + {Y^2} + 2 \times Y\cos \theta }$

Square both sides

$2{X^2}(1 - \cos \theta ) = {n^2}.2{X^2}(1 + \cos \theta )$

$1 - \cos \theta = {n^2} + {n^2}\cos \theta$

$\cos \theta = {{1 - {n^2}} \over {1 + {n^2}}}$

$\theta = {\cos ^{ - 1}}\left[ {{{{n^2} - 1} \over { - {n^2} - 1}}} \right]$

By photoelectric equation

${{hc} \over \lambda } - \phi = {k_{\max }}$

${k_{\max }} = {{1240} \over {500}} - 1.25 \approx 1.25$

$r = {{\sqrt {2mk} } \over {eB}}$

$B = {{\sqrt {2mk} } \over {er}}$

$= 125 \times {10^{ - 7}}T$

At maximum angle $\theta$ ray at point B goes in gazing emergence, at all less values of $\theta$, TIR occurs.

At point B

${4 \over 3} \times \sin \theta '' = 1 \times \sin 90^\circ$

$\theta '' = {\sin ^{ - 1}}\left( {{3 \over 4}} \right)$

$\theta ' = \left( {{\pi \over 2} - \theta ''} \right)$

At point A

$1 \times \sin \theta = {4 \over 3} \times \sin \theta '$

$\sin \theta = {4 \over 3} \times \sin \left( {{\pi \over 2} - \theta ''} \right)$

$\sin \theta = {4 \over 3}\cos \left[ {{{\cos }^{ - 1}}{{\sqrt 7 } \over 4}} \right]$



$\sin \theta = {4 \over 3} \times {{\sqrt 7 } \over 4}$

$\theta = {\sin ^{ - 1}}\left( {{{\sqrt 7 } \over 3}} \right)$

$\Rightarrow$ i = i₀sin$\omega$t

when i = i₀

i₀ = i₀sin$\omega$t₁ $\Rightarrow$ $\omega$t₁ = ${\pi \over 2}$ ..... (i)

When i = ${{{i_1}} \over {\sqrt 2 }}$

${{{i_1}} \over {\sqrt 2 }}$ = i₀sin$\omega$t₂ $\Rightarrow$ $\omega$t₂ = ${\pi \over 4}$ ...... (ii)

Time taken by current from maximum value to rms value

$\Rightarrow ({t_1} - {t_2}) = {\pi \over {2\omega }} - {\pi \over {4\omega }} = {\pi \over {4\omega }} = {\pi \over {4 \times 2\pi f}}$

$= {1 \over {8 \times 50}}$

${1 \over {400}}$ sec

= 2.5 ms

${{20 - {V_0}} \over 5} + {{0 - {V_0}} \over 5} + {{20 - {V_0}} \over 2} = 0$

$4 + 10 = {{2{V_0}} \over 5} + {{{V_0}} \over 2}$

$14 = {{4{V_0} + 5{V_0}} \over {10}}$

${V_0} = {{140} \over 9}$ Volt

Potential difference across 2$\Omega$ resistor is 20 $-$ V₀

That is $\left( {20 - {{140} \over 9}} \right)$ Volt

Hence answer is $\left( {{{40} \over 9}} \right)$ Volt

${{d\overrightarrow v } \over {dt}} = \overrightarrow a = {{\overrightarrow F } \over m} = (8\widehat i + 2\widehat j)m/{s^2}$

${{d\overrightarrow r } \over {dt}} = \overrightarrow v = (8t\widehat i + 2t\widehat j)m/s$

$\overrightarrow r = (8\widehat i + 2\widehat j){{{t^2}} \over 2}m$

At t = 10 sec

$\overrightarrow r = \left[ {(8\widehat i + 2\widehat j)50} \right]m$

$\Rightarrow \overrightarrow r = (400\widehat i + 100\widehat j)m$

Here assume current as



By KVL in outer loop

9 $-$ 12i $-$ 4i = 0

16i = 9

8i = ${9 \over 2}$ = 4.5

= 45 $\times$ 10^-1

${E_1} = k{l_1}$ .... (i)

${E_1} + {E_2} = k{l_2}$ .... (ii)

${{{E_1}} \over {{E_1} + {E_2}}} = {{{l_1}} \over {{l_2}}} = {{250} \over {400}} = {5 \over 8}$

$8{E_1} = 5{E_1} + 5{E_2}$

$3{E_1} = 5{E_2}$

${{{E_1}} \over {{E_2}}} = {5 \over 3}$

$\tan 37^\circ = {3 \over 4} = {{{{k{P_2}} \over {{r^3}}}} \over {{{2k{P_1}} \over {{r^3}}}}} = {{{P_2}} \over {2{P_1}}} = {3 \over 4}$

${{{P_2}} \over {{P_1}}} = {3 \over 2}$

${{{P_1}} \over {{P_2}}} = {2 \over 3}$

Object is projected as shown so as per motion under gravity

$S = ut + {1 \over 2}a{t^2}$

$- 75 = + 10t + {1 \over 2}( - 10){t^2} \Rightarrow t = 5$ sec

Object takes t = 5 s to fall on ground

Height of balloon from ground

H = 75 + ut

= 75 + 10 $\times$ 5 = 125 m

$\because$ mean free path

$\lambda = {1 \over {\sqrt 2 \pi {d^2}n}}$

${{{\lambda _1}} \over {{\lambda _2}}} = {{d_2^2{n_2}} \over {d_1^2{n_1}}}$

$= {\left( {{5 \over {10}}} \right)^2} = 0.25 = 25 \times {10^{ - 2}}$

Modulation index

$\mu = {{{A_m}} \over {{A_c}}} = {{20} \over {20}} = 1$

We know, $A = {A_0}{e^{ - \lambda t}}$

For half life

${{{A_0}} \over 2} = {e^{ - \lambda {t_{1/2}}}}$

$\Rightarrow$ ${\lambda {t_{1/2}}}$ = ln 2 .....(1)

And when radioactive element becomes ${\left( {{1 \over 8}} \right)^{th}}$ of its initial value in 30 years

${{{A_0}} \over 8} = {A_0}{e^{ - \lambda \times 30}} \Rightarrow \lambda \times 30 = \ln 8$

$\Rightarrow$ 30$\lambda = 3\ln 2$

$\Rightarrow$ $\lambda = {{3\ln 2} \over {30}}$ .....(2)

Putting value of $\lambda$ in (1), we get

${{3\ln 2} \over {30}} \times {t_{1/2}}$ = ln 2

$\Rightarrow$ ${t_{1/2}}$ = 10 years

$\eta = 1 - {{{T_L}} \over {{T_H}}}$ ..... (i)

$2\eta = 1 - {{({T_L} - 62)} \over {{T_H}}} = 1 - {{{T_L}} \over {{T_H}}} + {{62} \over {{T_H}}}$

$\Rightarrow \eta = {{62} \over {{T_H}}} \Rightarrow {1 \over 6} = {{62} \over {{T_H}}} \Rightarrow {T_H} = 6 \times 62 = 372$ K

In $^\circ$C $\Rightarrow$ 372 $-$ 273 = 99$^\circ$ C

$\Delta$m = (Zm_p + (A $-$ Z)m_n) $-$ M_Al

= (13 $\times$ 1.00726 + 14 $\times$ 1.00866) $-$ 27.18846

= 27.21562 $-$ 27.18846

= 0.02716 u

E = 27.16 x $\times$ 10^$-$3 J

Number of significant figures = 3

Cheese $\to$ liquid in solid

Pumice stone $\to$ gas in solid

Hair cream $\to$ liquid in liquid

Cloud $\to$ liquid in gas

Ho = [Xe]4f¹¹6s²

Ho³⁺ = [Xe] 4f¹⁰

So number of e^$-$ present in 4f is 10.

The total number of products possible = 2

Complex [Co(en)₃]Cl₂ is most stable complex among the given complex compounds because more number of chelate rings are present in this complex as compare to others.

(1) [Co(en)(NH₃)₄]Cl₂ 1 chelate ring

(2) [Co(en)₃]Cl₂ 3 chelate ring

(3) [Co(en)₂(NH₃)₂]Cl₂ 2 chelate ring

(4) [Co(NH₃)₆]Cl₂ 0 chelate ring

Among SO₃, O₂, SO₂ and CO$_3^{2 - }$, only O₂ and CO$_3^{2 - }$ has only one $\pi$-bond.

Note :

(1) $\,\,\,\,$ Bond strength $\propto$ Bond order

(2) $\,\,\,\,$ Bond length $\propto$ ${1 \over {Bond\,\,order}}$

(3) $\,$ Bond order $= {1 \over 2}$ [N_b $-$ N_a]

N_b = No of electrons in bonding molecular orbital

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

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



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

(5) $\,\,\,\,$ 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

$\therefore\,\,\,\,$ Molecular orbital configuration of O₂ (16 electrons) is

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

$\therefore\,\,\,\,$N_a = 6

N_b = 10

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 6} \right] = 2$

Molecular orbital configuration of O$_2^ +$ (15 electrons) is

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

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

N_a = 5

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 5} \right]$ = 2.5

Molecular orbital configuration of $O_2^ -$ (17 electrons) is

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

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

N_a = 7

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 7} \right]$ = 1.5

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

N_a = 8

$\therefore\,\,\,\,$ BO = ${1 \over 2}$ [ 10 $-$ 8] = 1

As Bond strength $\propto$ Bond order so, correct order is

$O_2^{2 - } < O_2^ - < {O_2} < O_2^ +$