JEE Main 2021 20th July Morning Shift Question Paper with Solutions

Consider the given diagram,


Here, r₁ = 10 $\times$ 10³ m and r₂ = 10 $\times$ 10^-3 m

We know that for a complete cyclic process, change in internal energy, ($\Delta$U) = 0 .... (i)

According to 1st law of thermodynamics,

$\Delta$Q = $\Delta$U + W ..... (ii)

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

$\Rightarrow$ $\Delta$Q = 0 + W

$\Rightarrow$ $\Delta$Q = W .... (iii)

$\because$ W = Area = $\pi$r₁r₂

= $\pi$ $\times$ (10 $\times$ 10³) $\times$ (10 $\times$ 10^$-$3)

W = 100 $\pi$ J .... (iv)

From Eqs. (iii) and (iv), we get

$\Delta$Q = 100 $\pi$ J

The given situation can be shown as


Here, radioactive material X is decayed into two particles Y and Z with their respective decay constant, $\lambda$_a and $\lambda$_b. It means that

$\because$ $\lambda = {{\ln 2} \over {{t_{1/2}}}}$

where, t_1/2(a) = 700 yr

and t_1/2(b) = 1400 yr

${\lambda _a} = {{\ln 2} \over {700}}y{r^{ - 1}}$ and ${\lambda _b} = {{\ln 2} \over {1400}}y{r^{ - 1}}$

$\because$ ${\lambda _{total}} = {\lambda _a} + {\lambda _b}$

$= \left( {{{\ln 2} \over {700}} + {{\ln 2} \over {1400}}} \right)y{r^{ - 1}} = \ln 2\left( {{1 \over {700}} + {1 \over {1400}}} \right)y{r^{ - 1}}$

$= \left( {{{3\ln 2} \over {1400}}} \right)y{r^{ - 1}}$

Suppose the initial number of radioactive nuclei was N₀.

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

where, N = number of nuclei present at time = t and N₀ = number of nuclei present at time = 0

$\Rightarrow {{{N_0}} \over 3} = {N_0}{e^{ - \lambda t}}$ or ${{{N_0}} \over 3} = {N_0}{N_0}{e^{ - {\lambda _{total}}t}} \Rightarrow {1 \over 3} = {e^{ - {\lambda _{total}}t}}$

Taking log on both the sides of above equation, we get

$\ln \left( {{1 \over 3}} \right) = \ln ({e^{ - {\lambda _{total}}t}})$

$\Rightarrow \ln \left( {{1 \over 3}} \right) = - {\lambda _{total}}t$

$\Rightarrow 1.1 = {{3 \times 0.693} \over {1400}} \times t$

$\Rightarrow$ t $\approx$ 740 yr

The given situation can be shown as


Before collision,


As the collision is perfectly elastic, therefore momentum is conserved, i.e.

p_initially = p_finally

$\Rightarrow$ mu = Mv ..... (i)

Angular momentum will also be conserved about point O.

$\Rightarrow mv\,.\,{L \over 2} = {{M{L^2}} \over {12}}\omega$

$\Rightarrow \omega = {{6mv} \over {ML}}$ .... (ii)

$\because$ Coefficient of restitution,

$e = {{{\mathop{\rm Relative}\nolimits} \,velocity\,after\,collision} \over {{\mathop{\rm Relative}\nolimits} \,velocity\,before\,collision}}$

$\Rightarrow 1 = {{v + {{\omega L} \over 2}} \over u}$

$\Rightarrow v + {{\omega L} \over 2} = u$ .... (iii)

From Eqs. (ii) and (iii), we get

$v + {{3mu} \over M} = u$

$\Rightarrow {{mu} \over M} + {{3mu} \over M} = u$ [using Eq. (i)]

$\Rightarrow {{4mu} \over M} = u \Rightarrow {m \over M} = {1 \over 4}$ .... (iv)

According to question,

ratio of masses $\left( {{m \over M}} \right) = {1 \over x}$.

Comparing it with Eq. (iv), we get x = 4.

The force equation in vertical direction is

Mg $-$ R = Ma


where,

M = collective mass of block and all three iron cylinders = 10 + 3 $\times$ 20 = 70 kg

a = acceleration of block = 0.2 ms^$-$2

g = 10 ms^$-$2 and $\mu$_s = 0.2

and R = normal reaction

Force along vertical axis Mg $-$ R = Ma

$\therefore$ 70g $-$ R = 70 $\times$ 0.2

$\Rightarrow$ R = 70 $\times$ 10 $-$ 14

= 700 $-$ 14 = 686 N

Given,

q = 8$\mu$C/g = 8 $\times$ 10^$-$6 C/g = 8 $\times$ 10^$-$3 C/kg

s = 10 cm = 0.1 m $\Rightarrow$ E = 100 V/m

We know that, acceleration, a = ${{force(F)} \over {mass(m)}}$

$\Rightarrow$ a = ${{qE} \over m}$ [$\because$ F = qE]

= ${{8 \times {{10}^{ - 6}} \times 100} \over {{{10}^{ - 3}}}}$ = 0.8 ms^$-$2

As per question, when electric field is switched on, the body strikes to the wall and then returns back.

For one oscillation,

s = ut + ${1 \over 2}$ at²

$\Rightarrow$ 0.1 = ${1 \over 2}$ $\times$ 0.8 t² [$\because$ u = 0]

$\Rightarrow$ 0.2 = 0.8 t²

$\Rightarrow$ ${2 \over 8}$ = t² $\Rightarrow$ t² = ${1 \over 4}$ $\Rightarrow$ t = ${1 \over 2}$

$\therefore$ Time period = 2 $\times$ ${1 \over 2}$ = 1 s

Therefore, if the collision of the body is perfectly elastic, the time period of motion will be 1s.

Let's say the charge q and (Q $-$ q) are at r distance from each other. This can be shown as


According to Coulomb's law, force between both the parts can be given as

$F = {{kq(Q - q)} \over {{r^2}}}$

$F = {k \over {{r^2}}}(qQ - {q^2})$

As we know that ${{dF} \over {dq}} = 0$, for maximum force.

$\Rightarrow {{dF} \over {dq}} = {d \over {dq}}\left[ {{k \over {{r^2}}}(qQ - {q^2})} \right] = 0$

$\Rightarrow {k \over {{r^2}}}(Q - 2q) = 0 \Rightarrow Q = 2q$

As we know that, the equation of refraction at spherical surface is :

${{{\mu _{II}}} \over v} - {{{\mu _I}} \over u} = {{{\mu _{II}} - {\mu _I}} \over R}$

where,

$\mu$_II = refractive index of region II = 1.4

$\mu$_I = refractive index of region I = 1.25

R = radius of curvature = $-$25 cm

u = object distance = $-$40 cm

and v = image distance.

$\Rightarrow {{1.4} \over v} - {{1.25} \over { - 40}} = {{1.4 - 1.25} \over { - 25}}$

$\Rightarrow {{1.4} \over v} = {{ - 0.15} \over {25}} - {{1.25} \over {40}}$ $\Rightarrow$ v = $-$ 37.58 cm

where, negative sign indicate real image.

Given, current, I = 5A

Area of cross-section of wire, A = 0.04 m²

We know that, $J = {I \over A}$

$\Rightarrow I = JA$

or $I = J\,.\,A$ or $I = JA\cos \theta$

where, J = current density.

$\Rightarrow 5 = J\left( {{4 \over {100}}} \right) \times \cos (60^\circ )$ [$\because$ Given, $\theta$ = 60$^\circ$]

$J = 500 \times {1 \over 2}$ [$\because$ cos60$^\circ$ = ${1 \over 2}$]

$\Rightarrow$ J = 250 Am^$-$2

The relation between electric field, current density and resistivity can be given as,

E = $\rho$ . J

= 44 $\times$ 10^$-$8 $\times$ 250 [$\because$ Resistivity, $\rho$ = 44 $\times$ 10^$-$8 $\Omega$-m]

= 11 $\times$ 10^$-$5 V/m

Suppose, I₀ be the actual length of metal wire and Y be its Young's modulus.

From Hooke's law,

$Y = {{T{I_0}} \over {A\Delta I}}$

where, $\Delta I = I - {I_0}$

$\Rightarrow Y = {{T{I_0}} \over {A(I - {I_0})}}$ or $I - I = {{T{I_0}} \over {AY}}$

$\therefore$ ${{{I_1} - {I_0}} \over {{I_2} - {I_0}}} = {{{T_1}{I_0}} \over {AY}} \times {{AY} \over {{T_2}{I_0}}} = {{{T_1}} \over {{T_2}}}$

$\Rightarrow {I_1}{T_2} - {I_0}{T_2} = {I_2}{T_1} - {I_0}{T_1}$

$\Rightarrow {I_0} = {{{I_1}{T_2} - {I_2}{T_1}} \over {{T_2} - {T_1}}} = {{{T_1}{I_2} - {T_2}{I_1}} \over {({T_1} - {T_2})}}$

Since, entropy of the system is given by

$S = {\alpha ^2}\beta \ln \left[ {{{\mu kR} \over {J{\beta ^2}}} + 3} \right]$ .... (i)

As, $S = {Q \over {\Delta T}}$ [given]

$\Rightarrow [S] = {{[M{L^2}{T^{ - 2}}]} \over {[K]}}$ .... (ii)

$\because$ Dimensions of Q = [ML²T^$-$2]

Dimension of T = [K]

Boltzmann constant, $k = {{energy} \over T}$ [$\because$ Dimensions of energy = [ML²T^$-$2]]

$\Rightarrow [k] = {{[M{L^2}{T^{ - 2}}]} \over {[K]}}$ ..... (iii)

From Eqs. (ii) and (iii), we can write,

$[S] = [k] = {{[M{L^2}{T^{ - 2}}]} \over {[K]}}$ ..... (iv)

$\because$ Gas constant, $[R] = {{[Energy]} \over {[nT]}} = {{[M{L^2}{T^{ - 2}}]} \over {[mol\,K]}}$ .... (v)

and mechanical equivalent of heat

[J] = [M⁰L⁰T⁰] .... (vi)

As, [$\mu$kR] = [J$\beta$]²

Using Eqs. (iii), (v) and (vi), we get

$\Rightarrow [mol] \times {{[M{L^2}{T^{ - 2}}]} \over {[K]}} \times {{[M{L^2}{T^{ - 2}}]} \over {[mol\,K]}} = [{\beta ^2}]$

$\Rightarrow [\beta ] = [M{L^2}{T^{ - 2}}{K^{ - 1}}]$ ..... (vii)

Using Eq. (i), we can write,

$[{\alpha ^2}] = {{[S]} \over {[\beta ]}} = {{[M{L^2}{T^{ - 2}}{K^{ - 1}}]} \over {[M{L^2}{T^{ - 2}}{K^{ - 1}}]}} \Rightarrow \alpha = [{M^0}{L^0}{T^0}]$ .... (viii)

So, from Eqs. (iii) and (viii), we can say that $\alpha$ and k have different dimensions.

Given, magnification, M = 6

Since we know that magnifying power of a simple microscope is given by

$M = 1 + {D \over {{f_0}}}$

where, D = least distance of distinct vision = 25 cm

and f₀ = focal length of objective lens.

$\Rightarrow 6 = 1 + {D \over {{f_0}}} \Rightarrow 6 = 1 + {{25} \over {{f_0}}} \Rightarrow 5 = {{25} \over {{f_0}}} \Rightarrow {f_0} = 5$ cm

For compound microscope, magnifying power is given by

$M = {{I\,.\,D} \over {{f_0}{f_e}}} = 2$M_{simple microscope}

where, f₀ = f_e are the focal lengths of the objective lens and eye piece respectively

and l = length of the given tube = 0.6 m

$\Rightarrow 12 = {{60 \times 25} \over {5\,.\,{f_e}}}$ [$\because$ magnification is doubled]

$\Rightarrow$ f_e = 25 cm

This is the required focal length of eyepiece.

Given,

AC voltage, V(t) = 20 sin $\omega$t volt.

Frequency, f = 50Hz

Separation between the plates, d = 2 mm = 2 $\times$ 10^$-$3 m

Area, A = 1 m²

As, $C = {{{\varepsilon _0}A} \over d}$

where, ${{\varepsilon _0}}$ = absolute electrical permittivity of free space = 8.854 $\times$ 10^$-$12 N^$-$1 kg²m^$-$2

$C = {{{\varepsilon _0} \times 1} \over {2 \times {{10}^{ - 3}}}}$ .... (i)

Capacitive reactance $({X_C}) = {1 \over {\omega C}}$ .... (ii)

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

${X_C} = {{2 \times {{10}^{ - 3}}} \over {2 \times 50\pi \times {\varepsilon _0}}}$ ($\because$ $\omega$ = 2$\pi$f)

$= {{2 \times {{10}^{ - 3}}} \over {25 \times 4\pi {\varepsilon _0}}}$

$\Rightarrow {X_C} = {{2 \times {{10}^{ - 3}}} \over {25}} \times 9 \times {10^9}$

$\Rightarrow {X_C} = {{18} \over {25}} \times {10^6}\,\Omega$

By using Ohm's law,

As, ${I_0} = {{{V_0}} \over {{X_C}}} = {{20 \times 25} \over {18}} \times {10^{ - 6}} = 27.78 \times {10^{ - 6}}$

$\Rightarrow$ I₀ = 27.78$\mu$A

$\therefore$ The amplitude of the oscillating displacement current for applied AC voltage will be approximately 27.79 $\mu$A.

Consider the given figure and draw the direction of current as follows


Here, current I is flowing through resistance R_s and I₁ is the current flowing through R = 2000 $\Omega$

$\therefore$ Current, $I = {{{V_i} - {V_z}} \over {{R_s}}} \Rightarrow I = {{100 - 50} \over {1000}}$

= 50 mA .... (i)

[$\because$ Resistance and Zener diode are in parallel, therefore voltages across them are same]

and current, ${I_1} = {{50} \over {2000}} = 25$ mA .... (ii)

By using Kirchhoff's current law,

$\because$ I = I₁ + I_z

$\Rightarrow$ I_z = I $-$ I₁ ..... (iii)

From Eq. (i) and (ii) and, we get

$\Rightarrow$ I_z = (50 $-$ 25) mA

I_z = 25 mA

According to first law of thermodynamics,

$\Delta$Q = $\Delta$U + $\Delta$W ..... (i)

where, $\Delta$Q = quantity of heat energy supplied to the system, $\Delta$U = change in the internal energy of a closed system and $\Delta$W = work done by the system on its surroundings.

As per question, no work is done

$\therefore$ $\Delta$W = 0 ..... (iii)

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

$\Delta$Q = 0 + $\Delta$U $\Rightarrow$ $\Delta$Q = $\Delta$U

or $\Delta$Q = $\Delta$U = nC_V$\Delta$T

where,

C_V = specific heat capacity at constant volume for diatomic gas = ${{5R} \over 2}$

$\Delta$T = change in temperature = (50 $-$ 0) = 50$^\circ$C

n = number of moles = 4

$\Rightarrow$ $\Delta$Q = nC_V$\Delta$T

= $4 \times {{5R} \over 2} \times (50)$ = 500 R = 500 R

The given situation can be represented as


Equating forces perpendicular to the inclined plane,

$N = mg\cos 30^\circ + {{m{v^2}} \over R}\sin 30^\circ$

$\Rightarrow N - mg\cos 30^\circ = {{m{v^2}} \over R}\sin 30^\circ$ .... (i)

Equating forces along the inclined plane,

$mg\sin 30^\circ + {\mu _s}N = {{m{v^2}} \over R}\cos 30^\circ$ .... (ii)

On dividing Eq. (i) by Eq. (ii), we get

${{N - mg\cos 30^\circ } \over {mg\sin 30^\circ + {\mu _s}N}} = \tan 30^\circ$ [$\because$ $\cos 30^\circ = {{\sqrt 3 } \over 2}$ and $\sin 30^\circ = {1 \over 2}$]

$\Rightarrow$ ${{N - mg(\sqrt3/2) } \over {mg(1/2) + (0.2)N}} = {1 \over {\sqrt 3 }}$

$N\sqrt 3 - mg{{\sqrt 3 } \over 2}.\sqrt 3 = {{mg} \over 2} + 0.2N$

$\Rightarrow (\sqrt 3 - 0.2)N = mg{{(1 + 3)} \over 2} = 2mg$

$\Rightarrow N = {{2mg} \over {\sqrt 3 - 0.2}} = {{2 \times 800 \times 10} \over {1.532}}$

$= 10.44 \times {10^3}V$

Therefore, $N = 10.2 \times {10^3}$ kg-m/s²

As the rectangular conductor moves in field area, so flux is increasing up to x = b, then flux is generated on return journey from x = b to x = 0. The flux is shown by plot A of the graph.

As, emf, $e = - {{d\phi } \over {dt}}$, which is shown by curve B and power dissipated, P = VI which is shown by curve C.

Given, V_C(t) = 160 sin(2$\pi$ $\times$ 10⁶t) V

V_max = 200 V, V_min = 120 V and V_m(t) = A_m sin(2$\pi$ $\times$ 10³t) V.

Since, maximum amplitude,

A_max = A_m + A_c

$\Rightarrow$ V_max = V_m + V_c $\Rightarrow$ 200 = V_m + 160

$\Rightarrow$ V_m = 200 $-$ 160 $\Rightarrow$ V_m = 40

$\therefore$ Peak voltage, A_m = 40

Given,

Acceleration due to gravity on Earth, g_E = 10 m/s²

and acceleration due to gravity on Mars, g_M = 4 m/s²

We know that, g_E (at height h) = ${g_{Earth}}\left( {1 - {{2h} \over R}} \right)$

$\therefore$ Weight at Earth, mg_E = 100 $\times$ 10 = 1000 N

As the spaceship moves far away from Earth, the value of g_E decreases to zero at a point where g_E + g_M = 0 and hence weight will also be zero.

This point is called neutral point and is shown by graph (c) in the given figure.

Then, g_E increases 4ms^$-$2 at Mars surface and weight becomes 400N which is also exhibited by graph (c).

Given, k = 100 N/m

m = 100 g = 0.1 kg

x = 0.05 m and H = 2 m

By energy conservation,

${1 \over 2}k{x^2} = {1 \over 2}m{v^2} \Rightarrow v = x\sqrt {{k \over m}}$

$= 0.05 \times \sqrt {{{100} \over {0.1}}} = 0.5\sqrt {10}$ ms^$-$1 .... (i)

Time of flight of ball, $t = \sqrt {{{2H} \over g}}$

$\Rightarrow t = \sqrt {{{2 \times 2} \over {10}}} = {2 \over {\sqrt {10} }}$s .... (ii)

$\therefore$ Range of ball, d = vt

$= 0.5\sqrt {10} \times \left( {{2 \over {\sqrt {10} }}} \right)$ [From Eqs. (i) and (ii)]

$= 0.5 \times 2 = 1 m$

Given, kinetic energy of $\alpha$-particle (K_$\alpha$) = kinetic energy of deuteron (K_d)

Since, kinetic energy, K = ${1 \over 2}$ mv²

$\Rightarrow$ mv² = 2K $\Rightarrow$ v² = ${{2K} \over m}$

$\Rightarrow$ v = $\sqrt {{{2K} \over m}}$ .... (i)

We know that,

r = ${{mv} \over {Bq}}$ .... (ii)

where, r = radius of curvature of path of a charged particle, m = mass of the charged particle, q = charge of the particle, v = velocity of charged particle and B = magnetic field.

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

$r = {{m\sqrt {{{2K} \over m}} } \over {Bq}}$

$\Rightarrow r = {{\sqrt {2Km} } \over {Bq}}$ .... (iii)

Since, m, K and B are same for both deuteron and $\alpha$-particle.

From Eq. (iii), we get

$\gamma \propto {{\sqrt m } \over q}$

$\therefore$ ${{{r_d}} \over {{r_\alpha }}} = \sqrt {{{{m_d}} \over {{m_\alpha }}}} .{{{q_\alpha }} \over {{q_d}}} = \sqrt {{2 \over 4}} \left( {{2 \over 1}} \right)$

[$\because$ m_d = 2m_p and m_$\alpha$ = 4m_p

q_$\alpha$ = 2e and q_d = e.]

${{{r_d}} \over {{r_\alpha }}} = \sqrt 2$

According to question, a circular disc reaches from top to bottom of an inclined plane of length L. This can be shown as


When the disc slips down the inclined plane, it takes time t₁. Therefore, in this case its acceleration, a₁ = g sin$\theta$

$\because$ s = ut₁ + ${1 \over 2}$a₁t$_1^2$

$\Rightarrow$ s = ${1 \over 2}$ g sin$\theta$t$_1^2$ .... (i)

And when the disc rolls down the inclined plane, it takes time t₂. Therefore in this case, its acceleration,

${a_2} = {{g\sin \theta } \over {1 + {{{K^2}} \over {{R^2}}}}} = {{g\sin \theta } \over {1 + {1 \over 2}}} = {2 \over 3}g\sin \theta$ [$\because$ for disc, ${{{K^2}} \over {{R^2}}} = {1 \over 2}$]

$\because$ s = ut₂ + ${1 \over 2}$ a₂t$_2^2$ = ${1 \over 2}$ . ${2 \over 3}$ g sin$\theta$ t$_2^2$ .... (ii)

On dividing Eq. (i) by Eq. (ii), we get

${{{t_2}} \over {{t_1}}} = \sqrt {{3 \over 2}}$ .... (iii)

According to question, value of ${{{t_2}} \over {{t_1}}}$ is $\sqrt {{3 \over x}}$.

Comparing it with Eq. (iii), we get x = 2

The given situation can be represented as


In the above figure, v₁ is the speed of wind and v₂₁ is the speed of butterfly with respect to wind.

So, v₂₁ can be given as

${v_{21}} = 4\sqrt 2 \cos 45^\circ \widehat i + 4\sqrt 2 \sin 45^\circ \widehat j$

$= 4\sqrt 2 \times {1 \over {\sqrt 2 }}\widehat i + 4\sqrt 2 \times {1 \over {\sqrt 2 }}\widehat j = 4\widehat i + 4\widehat j$

and v₁ can be given as

${v_1} = - \widehat j$

$\therefore$ Velocity of butterfly can be given as

${v_2} = {v_1} + {v_{21}} = 4\widehat i + 4\widehat j - \widehat j = 4\widehat i + 3\widehat j$

$\therefore$ Displacement of butterfly, $D = {v_2} \times t$

$= (4\widehat i + 3\widehat j) \times 3 = 12\widehat i + 9\widehat j$

$\therefore$ Magnitude of displacement, $\left| D \right| = \sqrt {{{12}^2} + {9^2}} = 15$ m

$\because$ Since, frequency received by the wall,

$f' = \left( {{{{v_s}} \over {{v_s} - v}}} \right){f_o}$ .... (i)

where, v_s = velocity of sound in air, v = velocity of car and f_o = observed frequency of sound.

Reflected frequency received by man is

$f'' = \left( {{{{v_s} + v} \over {{v_s}}}} \right)f'$ ..... (ii)

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

$f'' = \left( {{{{v_s} + v} \over {{v_s}}}} \right)\left( {{{{v_s}} \over {{v_s} - v}}} \right){f_o} \Rightarrow f'' = \left( {{{{v_s} + v} \over {{v_s} - v}}} \right){f_o}$

$\Rightarrow 500 = \left( {{{330 + v} \over {330 - v}}} \right) \times 400 \Rightarrow {{500} \over {400}} = {{(330 + v)} \over {(330 - v)}}$

$\Rightarrow 5(330 - v) = 4(330 + v) \Rightarrow 1650 - 5v = 1320 + 4v$

$\Rightarrow 9v = 330 \Rightarrow v = {{330} \over 9}$ m/s

or $v = {{330} \over 9} \times {{18} \over 5} = 132$ km/h

As per question,

at t = 0, y = ${1 \over {{{(1 + x)}^2}}}$

and at t = 1 s, y = ${1 \over {1 + {{(x - 2)}^2}}}$ .... (i)

As we know,

At t = t s, y = ${1 \over {1 + {{(x - vt)}^2}}}$

So, at t = 1 s, y = ${1 \over {1 + {{(x - v)}^2}}}$ .... (ii)

On comparing Eqs. (i) and (ii), we get

v = 2 ms^$-$1

Hence, the velocity of the wave will be 2 m/s.

Given,

Inductance, L = 30 mH

Resistance, R = 1 $\Omega$

Angular frequency, $\omega$ = 300 rad/s

We know that in L-C-R circuit, $\tan \phi = {{{X_C} - {X_L}} \over R}$

where, $\phi$ = phase angle = 45$^\circ$

X_C = capacitive reactance = ${1 \over {\omega C}}$

X_L = inductive reactance = $\omega$L

$\Rightarrow$ $\tan 45^\circ = {{{X_C} - {X_L}} \over R}$

$\Rightarrow {X_C} - {X_L} = R$ [$\because$ tan 45$^\circ$ = 1]

$\Rightarrow {1 \over {\omega C}} - \omega L = R \Rightarrow {1 \over {\omega C}} - 300 \times 30 \times {10^{ - 3}} = 1$

$\Rightarrow {1 \over {\omega C}} = 10 \Rightarrow \omega C = {1 \over {10}}$

$\Rightarrow C = {1 \over {10\omega }} \Rightarrow C = {1 \over {10 \times 300}}$

$\Rightarrow C = {1 \over 3} \times {10^{ - 3}}F$ .... (i)

According to question, the value of capacitance is ${1 \over x} \times {10^{ - 3}}F$. So, on comparing it with Eq. (i), we can say x = 3.

Energy of $\gamma$-ray, E_$\gamma$ = hv

and momentum of $\gamma$-ray, ${p_\gamma } = {h \over \lambda }$ .... (i)

As, ${p_\gamma } = {{{E_\gamma }} \over c} = {{hv} \over c}$ ..... (ii)

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

${p_\gamma } = {{hv} \over c} = {h \over \lambda }$ [$\because$ $\lambda = {c \over v}$]

Since, during the emission of $\gamma$-ray photon, momentum is conserved.

$\therefore$ p_$\gamma$ + p_{decayed nuclei} = 0

$\Rightarrow$ p_$\gamma$ = p_{decayed nuclei}

$\Rightarrow {{hv} \over c}$ = p_{decayed nuclei} ..... (iii)

Kinetic energy of decayed nuclei,

$KE = {1 \over 2}M{v^2} = {{(p_{decayed\,nuclei}^2)} \over {2M}}$ ..... (iv)

From Eqs. (iii) and (iv), we get

$\Rightarrow KE = {1 \over {2M}}{\left[ {{{hv} \over c}} \right]^2}$

$\therefore$ Loss in internal energy = E_$\gamma$ + KE_{decayed nuclei}

$= hv + {1 \over {2M}}{\left[ {{{hv} \over c}} \right]^2} = hv\left[ {1 + {{hv} \over {2M{c^2}}}} \right]$

Energy of photon can be given as

${E_p} = 13.6\left[ {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right]$ eV

where, n₁ = lower energy level and n₂ = higher energy level.

As per question, n₁ = 2, n₂ = 3

$\therefore$ ${E_p} = 13.6\left[ {{1 \over {{{(2)}^2}}} - {1 \over {{{(3)}^2}}}} \right]$

$= 13.6\left[ {{1 \over 4} - {1 \over 9}} \right] = 13.6\left[ {{{9 - 4} \over {36}}} \right] = 1.89$ eV

We know that work-function is the minimum energy required to eject photoelectrons from metal surface.

For gold plate, it will be

$\phi$ = E_P $-$ KE_max .... (i)

[Given, B = 5 $\times$ 10^$-$4 T, r = 7 mm = 7 $\times$ 10^$-$3 m, q = 1.6 $\times$ 10^$-$19 C and m = 9.1 $\times$ 10^$-$31 kg]

Therefore, velocity of photoelectrons will be

$v = {{Bqr} \over m} = {{5 \times {{10}^{ - 4}} \times 1.6 \times {{10}^{ - 19}} \times 7 \times {{10}^{ - 3}}} \over {9.1 \times {{10}^{ - 31}}}} = 6.15 \times {10^5}$ ms^$-$1

Kinetic energy will be

$\therefore$ $KE = {1 \over 2}m{v^2} = {{1 \times 9.1 \times {{10}^{ - 31}} \times {{(6.15 \times {{10}^5})}^2}} \over {2 \times 1.6 \times {{10}^{ - 19}}}}$ eV = 1.075 eV

Now, putting the values, in Eq. (i), we get

$\phi$ = (1.89 $-$ 1.075) eV = 0.82 eV