JEE Main 2021 16th March Evening Shift Question Paper with Solutions

We know that, on an inclined plane

Acceleration, $a = {{g\sin \theta } \over {1 + {I \over {m{R^2}}}}}$

$\Rightarrow a = {{g\sin \theta } \over {1 + {1 \over 2}}}$ [$\because$ For disc, $I = {{m{R^2}} \over 2}$]

$\Rightarrow a = {2 \over 3}g\sin \theta$ ....... (i)

As per question, acceleration of the disc will be ${2 \over b}g\sin \theta$.

Comparing it with Eq. (i), we get

$b = 3$

First overtone of open pipe $= {{{v_2}} \over {{L_2}}}$

First overtone of closed pipe at one end $= {{3v} \over {4L}}$

As per question,

${{3V} \over {4L}} = {{{V_2}} \over L}$

$\Rightarrow \sqrt {{B \over {{\rho _1}}}} \,.\,{3 \over {4L}} = \sqrt {{B \over {{\rho _2}}}} \,.\,{1 \over {{L_2}}}$ ($\because$ $V = \sqrt {{B \over \rho }}$)

$\Rightarrow {L_2} = {{4L} \over 3}\sqrt {{{{\rho _1}} \over {{\rho _2}}}}$ ..... (i)

According to question, the length of the open pipe is

${x \over 3}L\sqrt {{{{\rho _1}} \over {{\rho _2}}}}$ ..... (ii)

Comparing Eqs. (i) and (ii), we get

$x = 4$

Voltage across secondary coil,

= ${{Power\,across\,load} \over {Current\,pas\sin g\,through\,load}}$

$\Rightarrow {V_2} = {P \over {{I_L}}} = {{60} \over {0.11}} \Rightarrow {V_2} = 545.45$ V

Voltage across primary coil, V₁ = 220V

$\Rightarrow$ V₂ > V₁

It means that step-up transformer is used.

The situation is depicted in the following figure.

where, V_MR = velocity of man = 12 km/h

and v_R = velocity of water flow in river = 6 km/h

As, v_MR should be along CD.

$\Rightarrow {v_R} - {v_{MR}}\sin \theta = 0$

$\Rightarrow 6 - 12\sin \theta = 0 \Rightarrow \sin \theta = {6 \over {12}}$

$\Rightarrow \sin \theta = {1 \over 2}$

$\Rightarrow \theta = {\sin ^{ - 1}}\left( {{1 \over 2}} \right) = {\sin ^{ - 1}}(\sin 30^\circ )$ [$\because$ $\sin 30^\circ = {1 \over 2}$]

$\Rightarrow \theta = 30^\circ$

$\therefore$ $\alpha = 90^\circ + \theta = 90^\circ + 30^\circ = 120^\circ$

$\Rightarrow \alpha = 120^\circ$

Given, energy dissipated by a resistor, H = 10 mJ = 10 $\times$ 10^$-$3 J

Time, t = 1 s

Electric current, I = 2 mA = 2 $\times$ 10^$-$3 A

Resistance, R = ?

According to Joule's law of heating,

H = I²Rt

$\Rightarrow R = {H \over {{I^2}T}}$ ....... (i)

Substituting the given values in Eq. (i), we get

$R = {{10 \times {{10}^{ - 3}}} \over {{{(2 \times {{10}^{ - 3}})}^2} \times 1}}$

$\Rightarrow R = {{{{10}^{ - 2}}} \over {4 \times {{10}^{ - 6}}}} \Rightarrow R = 0.25 \times {10^4}$

$\Rightarrow R = 2500\,\Omega$

Given, temperature of source, T_H = 127$^\circ$C

= 273 + 127 = 400 K

Efficiencies, $\eta$ = 60% = 0.6

The efficiency of Carnot (ideal) heat engine is given by

$\eta = \left( {1 - {{{T_L}} \over {{T_H}}}} \right)$

where, T_L = temperature of sink,

$0.6 = \left( {1 - {{{T_L}} \over {{T_H}}}} \right) \Rightarrow {{{T_L}} \over {{T_H}}} = 1 - 0.6$

${{{T_L}} \over {{T_H}}} = 0.4 \Rightarrow {T_L} = 0.4 \times {T_H}$

$= 0.4 \times 400 = 160K$

$= 160 - 273 = - 113^\circ C$

We know that binding energy of earth,

$BE = - {3 \over 5}{{G{M^2}} \over R}$

$\therefore$ Energy required to break the earth into pieces

$= - BE = {3 \over 5}{{G{M^2}} \over R}$ ...... (i)

According to question, the amount of energy that needs to be supplied is ${x \over 5}{{G{M^2}} \over R}$.

Comparing it with value in Eq. (i), we get,

$x = 3$

(A) ${}_{58}Ce \to [Xe]4{f^2}5{d^0}6{s^2}$

In complex $C{e^{ + 4}} \to [Xe]4{f^0}5{d^0}6{s^0}$

There is no unpaired electron, so, $\mu$_m = 0 BM

(B) ${}_{64}Gd \to [Xe]4{f^7}5{d^1}6{s^2}$

In complex, ${}_{64}G{d^{2 + }} \to [Xe]4{f^7}5{d^1}$

There are 8 unpaired electrons.

$\therefore$ ${\mu _m} = \sqrt {8(8 + 2)} = \sqrt {80} = 8.94BM$

(C) ${}_{63}Eu \to [Xe]4{f^9}5{d^0}6{s^0}$

In complex ${}_{63}E{u^{ + 3}} \to [Xe]4{f^6}5{d^0}6{s^0}$

contain six unpaired electrons.

So, ${\mu _m} = \sqrt {6(6 + 2)} = \sqrt {48} BM = 6.93BM$

Hence, order of spin only magnetic moment

$B > C > A$.