JEE Main 2022 30th June Morning Shift Question Paper with Solutions

$I=\left(m_{A}+m_{B}+m_{C}+m_{D}\right) r^{2}$

$$\begin{aligned} &\mathrm{I}_{\text {net }}=(1+2+3+4) \cdot\left(\frac{a}{\sqrt{2}}\right)^{2} \text {, where a = side of square } \\ &=10 \times \frac{1^{2}}{2}=5 \mathrm{~kg} \mathrm{~m}^{2} \end{aligned}$$

Here volume is constant.

$\therefore$ ${{{P_1}} \over {{T_1}}} = {{{P_2}} \over {{T_2}}}$

$\Rightarrow {{100} \over {273}} = {{180} \over {{T_2}}}$

$\Rightarrow {T_2} = {{180} \over {100}} \times 273$

$= 1.8 \times 273$

$= 491\,K$

Change in energy

$$\begin{aligned} &\Delta \mathrm{E}=\mathrm{E}_{\mathrm{f}}-\mathrm{E}_{2} \Rightarrow 12.09=\mathrm{E}_{\mathrm{f}}-(-13.6) \Rightarrow \mathrm{E}_{\mathrm{f}}=-1.51 \mathrm{eV} \\ &\Rightarrow \frac{-13.6}{n^{2}}=-1.51 \Rightarrow \mathrm{n}^{2}=9 \Rightarrow \mathrm{n}=3 \\ &\text { So, } \Delta \mathrm{L}=\mathrm{L}_{\mathrm{f}}-\mathrm{L}_{\mathrm{i}} \\ &=\frac{h}{2 \pi}(3-1)=\frac{2 h}{2 \pi}=\frac{h}{\pi}=\frac{6.63 \times 10^{-34}}{3.14}=2.11 \times 10^{-34} \mathrm{Js} \end{aligned}$$

Given the electric field expression:

$E_z = 300 \sin(5\pi \times 10^3 x - 3\pi \times 10^{11} t) \, \text{Vm}^{-1}$

The amplitude of the electric field ($E_0$) is $300 \, \text{V/m}$.

The velocity ($v$) of the wave in the medium is given by the ratio of the coefficients of time and displacement in the wave equation, which can be calculated as:

$v = \frac{\text{Coefficient of } t}{\text{Coefficient of } x} = \frac{3\pi \times 10^{11}}{5\pi \times 10^3} = \frac{3}{5} \times 10^8 \, \text{m/s}$

The relationship between the electric field amplitude and the magnetic field amplitude in an electromagnetic wave is given by:

$B_0 = \frac{E_0}{v}$

Substituting the values for $E_0$ and $v$ into this formula:

$B_0 = \frac{300}{\frac{3}{5} \times 10^8}$

$B_0 = \frac{300 \times 5}{3 \times 10^8}$

$B_0 = \frac{1500}{3 \times 10^8}$

$B_0 = 5 \times 10^{-6} \, \text{T}$

Therefore, the amplitude of the magnetic field ($B_0$) is $5 \times 10^{-6}$ T.

We know,

Maximum height of a projectile $(H) = {{{u^2}{{\sin }^2}\theta } \over {2g}}$

Given, Ratio of initial velocity of two projectile

${{{u_1}} \over {{u_2}}} = {{\sqrt 3 } \over {\sqrt 2 }}$

Both projectile reach the same maximum height.

$\therefore$ ${H_1} = {H_2}$

$\Rightarrow {{u_1^2{{\sin }^2}{\theta _1}} \over {2g}} = {{u_2^2{{\sin }^2}{\theta _2}} \over {2g}}$

$\Rightarrow u_1^2{\sin ^2}{\theta _1} = u_2^2{\sin ^2}{\theta _2}$

$\Rightarrow {\left( {{{{u_1}} \over {{u_2}}}} \right)^2} = {\left( {{{\sin {\theta _2}} \over {\sin {\theta _1}}}} \right)^2}$

$\Rightarrow {\left( {{{\sqrt 3 } \over {\sqrt 2 }}} \right)^2} = {\left( {{{\sin 60^\circ } \over {\sin {\theta _1}}}} \right)^2}$

$\Rightarrow {3 \over 2} = {3 \over {4{{\sin }^2}{\theta _1}}}$

$\Rightarrow \sin _{{\theta _1}}^2 = {1 \over 2}$

$\Rightarrow \sin {\theta _1} = {1 \over {\sqrt 2 }} = \sin 45^\circ$

$\Rightarrow {\theta _1} = 45^\circ$

For maximum power

$$\begin{aligned} &\text { power factor }=\cos \theta=1\\ & \therefore \frac{R}{Z}=1 \\ &R^{2}=Z^{2} \\ &R^{2}=\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^{2}+\mathrm{R}^{2} \\ &\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}} \\ &\frac{70}{11}=\frac{1}{100 \pi \times C} \\ &\Rightarrow C=\frac{11}{7000 \pi}=500 \times 10^{-6} F=500 \mu F \end{aligned}$$

$\mathrm{A}_{\min }=\mathrm{A}_{\mathrm{C}}-\mathrm{A}_{\mathrm{m}}$

$=(44-11) \text { volt }=33 \text { volt }$

From law of conservation of charge

$$\begin{aligned} &q_{i}=\mathrm{q}_{\mathrm{f}} \Rightarrow 8 \times\left(4 \pi \mathrm{E}_{0} \mathrm{R}\right) \times 12=\left(4 \pi E R^{1}\right) \times \mathrm{V}_{f} \\ &\Rightarrow 96 \mathrm{R}=\mathrm{V}_{\mathrm{f}} \mathrm{R}^{1} \\ &\text { And, } 8 \times \frac{4}{3} \pi R^{3}=\frac{4}{3} \pi R^{1} 3 \\ &8=\left(\frac{R^{1}}{R}\right)^{3} \end{aligned}$$

$$\begin{aligned} & \mathrm{R}^{1}=2 \mathrm{R} \end{aligned}$$

From (i) & (ii), we get

So, $96 \mathrm{R}=\mathrm{V}_{\mathrm{f}} \times 2 \mathrm{R} \Rightarrow \mathrm{V}_{\mathrm{f}}=48$ Volt

$$\begin{aligned} &V_{f}=\frac{1}{2} C_{f} V_{f}^{2}=\frac{1}{2} \times\left(4 \pi \varepsilon_{0} \mathrm{R}^{1}\right) \mathrm{V}_{f}^{2} \\ &=\frac{1}{2} \times\left(4 \pi \varepsilon_{0} \times 2 \mathrm{R}\right) \times 48^{2} \\ &=\left(\frac{1}{2} \times 4 \pi \varepsilon_{0} R \times 12^{2}\right) \times \frac{48^{2} \times 2}{12^{2}}=32 \,E \end{aligned}$$

We know,

Quality factor (Q factor)

${Q_1} = {{{w_1}} \over {\Delta w}}$

$= {1 \over {\sqrt {LC} }} \times {L \over R}$

$= {1 \over R}\sqrt {{L \over C}}$

Now, when $L' = 2L$ and $C' = 2C$ then ${Q_2} = {1 \over R}\sqrt {{{2L} \over {2C}}} = {1 \over R}\sqrt {{L \over C}} = {Q_1}$

$\therefore$ Q₂ remains same as Q₁.

Also, as ${w_1} = {1 \over {\sqrt {LC} }}$

$\Rightarrow 2\pi {f_1} = {1 \over {\sqrt {LC} }}$

$\Rightarrow {f_1} = {1 \over {2\pi \sqrt {LC} }}$

$\therefore$ When $L' = 2L$ and $C' = 2C$ then new resonating frequency

${f_2} = {1 \over {2\pi \sqrt {2L \times 2C} }} = {1 \over {2\pi \times 2\sqrt {LC} }} = {1 \over 2} \times {f_1}$

Capacitance of Fig. 1, ${C_0} = {{{\varepsilon _0}A} \over d}$

In Fig. 2, C₁ and C₂ are parallel with each other,

$\therefore$ Equivalent capacitance,

${C_{eq}} = {C_1} + {C_2}$

$= {{{\varepsilon _0}A/2} \over d} + {{{\varepsilon _0}{\varepsilon _r}A/2} \over d}$

$= {{{\varepsilon _0}A} \over {2d}} + {{{\varepsilon _0}{\varepsilon _r}A} \over {2d}}$

$= {{A{\varepsilon _0}} \over {2d}}\left( {1 + {\varepsilon _r}} \right) = {{{C_0}} \over 2}\left( {1 + {\varepsilon _r}} \right)$

Initially, copper wire radius (r₁) = 9 mm

Resistance (R) = 14 $\Omega$

We know, $R = {{\rho L} \over A} = {{\rho L} \over {\pi r_1^2}} = 14$

Now this copper wire is replaced by 7 parallel copper wire of resistance R₁.

$\therefore$ Equivalent resistance of 7 parallel copper wire,

${1 \over {{R_{eq}}}} = {1 \over {{R_1}}} + {1 \over {{R_1}}} + {1 \over {{R_1}}} + {1 \over {{R_1}}} + {1 \over {{R_1}}} + {1 \over {{R_1}}} + {1 \over {{R_1}}}$

$\Rightarrow {1 \over {{R_{eq}}}} = {7 \over {{R_1}}}$

$\Rightarrow {R_{eq}} = {{{R_1}} \over 7}$

$= {1 \over 7} \times {{\rho L} \over {\pi r_2^2}}$

$= {1 \over 7} \times {{\rho L} \over {\pi {{\left( {{{{r_1}} \over 3}} \right)}^2}}}$ [as ${r_2} = {{{r_1}} \over 3}$ ; ${r_1} = 9$ m and ${r_2} = 3$]

$= {1 \over 7} \times {{\rho L} \over {\pi r_1^2}} \times 9$

$= {1 \over 7} \times 14 \times 9$

$= 18\,\Omega$

Here F_max force is trying to move the block upward so friction force will be applied towards downward.

Along horizontal direction,

N = 50 N

Along vertical diretion,

F_max = mg + f_max

= 2 $\times$ 10 + $\mu$N

= 20 + 0.5 $\times$ 50

= 20 + 25

= 45 N

As we know

$$\begin{aligned} &I=\frac{E}{A t}=\frac{n h v}{A t} \Rightarrow \frac{n}{t}=\frac{I A \lambda}{h C} \Rightarrow \frac{n}{t}=\frac{\rho \lambda}{h c} \Rightarrow \frac{n}{t}=\rho \lambda \\ &\Rightarrow\left(\frac{n}{t}\right)_{2}=\left(\frac{n}{t}\right)_{1} \times \frac{p_{2} \lambda_{2}}{p_{1} \lambda 1}=9 \times 10^{20} \times \frac{P}{P} \times \frac{800}{600}=12 \times 10^{20} \end{aligned}$$

5 parts of main scale division = 1 cm

$\therefore$ 1 part of main scale division = ${1 \over 5}$ cm

$\therefore$ 1 M.S.D. = ${1 \over 5}$ cm

(n + 1) vernier scale division = n main scale division.

$\therefore$ 1 V.S.D. = ${n \over n+1}$ M.S.D.

= ${n \over n+1}$ $\times$ 1 M.S.D.

= ${n \over n + 1}$ $\times$ ${1 \over 5}$ cm

We know,

L.C. = 1 M.S.D. $-$ 1 V,S.D.

= ${1 \over 5}$ cm $-$ ${n \over {5(n + 1)}}$ cm

= ${{n + 1 - n} \over {5(n + 1)}}$ cm

= ${1 \over {5(n + 1)}}$ cm

$\mathrm{P}=\mathrm{P}_{0}+\frac{2 T}{R}$

$\Rightarrow P-P_{0}=\frac{2 T}{R}$

$$\begin{aligned} &500=\frac{2 \times T}{2 \times 10^{-3}} \\ &T=500 \times 10^{-3} \\ &\text { So, } x=500 \end{aligned}$$

In the width of ${r_2} - {r_1}$ total n turns presents.

$\therefore$ In 1 unit width ${n \over {{r_2} - {r_1}}}$ turns presents.

$\therefore$ In the width of dr number of turns,

$n' = {n \over {{r_2} - {r_1}}} \times dr$

Magnetic field (dB) due to element of dr length is

$dB = {{{\mu _0} \times I \times n'} \over {2r}}$

$= {{{\mu _0}I} \over {2r}} \times {n \over {({r_2} - {r_1})}}$

$\therefore$ Total magnetic field due to entire coil is,

$\int {dB = \int_{{r_1}}^{{r_2}} {{{{\mu _0}I} \over {2r}} \times {n \over {({r_2} - {r_1})}}dr} }$

$\Rightarrow B = {{{\mu _0}In} \over {2({r_2} - {r_1})}}\int_{{r_1}}^{{r_2}} {{{dr} \over r}}$

$= {{{\mu _0}In} \over {2({r_2} - {r_1})}} \times \left[ {\log _e^r} \right]_{{r_2}}^{{r_1}}$

$= {{{\mu _0}In} \over {2({r_2} - {r_1})}} \times \left( {\log _e^{{r_2}} - \log _e^{{r_1}}} \right)$

$= {{{\mu _0}In} \over {2({r_2} - {r_1})}} \times {\log _e}{{{r_2}} \over {{r_1}}}$

Power $\mathrm{P}_{\mathrm{zm}}=\mathrm{I}_{\mathrm{zm}} \mathrm{V}_{z}$

$$\begin{aligned} & \mathrm{I}_{\mathrm{ZM}}=\frac{P_{z M}}{V_{z}}=\frac{0.4}{10}=0.04 \mathrm{~A} \\ & \text { So, } R_{P}=\frac{E_{\max }-V_{z}}{I_{Z M}}=\frac{15-10}{0.04}=\frac{5}{0.04}=125 \Omega \end{aligned}$$

Shortcut Method :

Internal Resistance of Unknown Battery

$r=\left(\frac{\ell_1-\ell_2}{\ell_2}\right) \mathrm{R}$

Where l₁ means balanced length when key K₁ is open

Where l₂ means balanced length when key K₁ is closed

Here l₁ = 120 cm and l₂ = 80 cm and R = 4 $\Omega$

$\therefore$ $r = \left( {{{120 - 80} \over {80}}} \right) \times 4$ = 2 $\Omega$

Normal Method :

Original Potentiometer is $\to$

Here, AB is called potentiometer wire. Let length of AB = l and resistance = R

$\therefore$ (1) Current in the circuit (i) = ${{4} \over {R + {R_h}}}$

(2) Potential difference between A and B is

= V_A $-$ V_B = $\Delta$V_AB = i $\times$ R

(3) ${{\Delta {V_{AB}}} \over {AB}} = {{Potential\,difference} \over {Length}}$ = Potential Gradient

$\Rightarrow {{\Delta {V_{AB}}} \over l} = {{iR} \over l} = {4 \over {R + {R_h}}} \times {R \over l}$ Volt/m

Now, a circuit added to potentiometer wire.

When K₁ is open then circuit look like the following.

When null point is found at C then no current will flow through the circuit added to the potentiometer wire.

Potential difference between A and C is

(1) From potentiometer wire :

${V_A} - {V_C} = \Delta {V_{AC}} = {{{V_{AB}}} \over l} \times 120 = {{4R} \over {(R + {R_h})l}} \times 120$ ...... (1)

(2) From external circuit :

V_A $-$ 1.5 $-$ 0 $\times$ r = V_C

$\Rightarrow$ V_A $-$ V_C = 1.5V = $\Delta$V_AC ...... (2)

Comparing equation (1) and (2), we get

${{4R} \over {(R + {R_h})l}} \times 120 = 1.5$ ..... (3)

When the key K₁ is closed then circuit look like this $\to$

When null point is found at C then no current will flow through the circuit added to the potentiometer wire but i₁ current will flow through circuit created using 1.5V with internal resistance r and 4$\Omega$ resistance.

Potential difference between A and C is

(1) From potentiometer wire :

${V_A} - {V_C} = \Delta {V_{AC}} = {{{V_{AB}}} \over l} \times 80 = {{4R} \over {(R + {R_h})l}} \times 80$ ....... (4)

(2) From external circuit :

${i_1} = {{1.5} \over {4 + r}}$

${V_A} - 1.5 + {i_1}r = {V_C}$

$\Rightarrow {V_A} - {V_C} = 1.5 - {i_1}r = 1.5 - {{1.5r} \over {4 + r}}$ ..... (5)

Or we can use this also,

${V_A} - {i_1} \times 4 = {V_C}$

$\Rightarrow {V_A} - {V_C} = {i_1} \times 4 = {{1.5} \over {4 + r}} \times 4$

From equation 4 and 5, we get

${{4R} \over {(R + {R_h})l}} \times 80 = 1.5 - {{1.5r} \over {4 + r}}$

$\Rightarrow {{1.5} \over {120}} \times 80 = 1.5 - {{1.5r} \over {4 + r}}$ [From equation 3]

$\Rightarrow 1.5 \times {2 \over 3} = 1.5 - {{1.5r} \over {4 + r}}$

$\Rightarrow {{1.5r} \over {4 + r}} = 1.5 - 1.5 \times {2 \over 3}$

$\Rightarrow {{1.5r} \over {4 + r}} = 1.5 \times {1 \over 3}$

$\Rightarrow 3r = 4 + r$

$\Rightarrow 2r = 4$

$\Rightarrow r = 2\,\Omega$

Other Method :

Now this can be converted to single battery and single resistance using combination of battery rule,

We know,

${E_{eq}} = {{{E_1}{r_2} + {E_2}{r_1}} \over {{r_1} + {r_2}}}$

and ${r_{eq}} = {{{r_1}{r_2}} \over {{r_1} + {r_2}}}$

$\therefore$ Here, ${E_{eq}} = {{1.5 \times 4 + 0 \times r} \over {4 + r}} = {6 \over {4 + r}}$

and ${r_{eq}} = {{4r} \over {4 + r}}$

$\therefore$ Circuit becomes

When null point is found at C then no current will flow through the circuit added to the potentiometer wire.

Potential difference between A and C is

(1) From potentiometer wire :

${V_A} - {V_C} = \Delta {V_{AC}} = {{{V_{AB}}} \over l} \times 80 = {{4R} \over {(R + {R_h})l}} \times 80$ .... (4)

(2) From external circuit :

${V_A} - {6 \over {4 + r}} - 0 \times {{4r} \over {4 + r}} = {V_C}$

$\Rightarrow {V_A} - {V_C} = {6 \over {4 + r}}$ ...... (5)

From equation 4 and 5, we get

${{4R} \over {(R + {R_h})l}} \times 80 = {6 \over {4 + r}}$

$\Rightarrow {{1.5} \over {120}} \times 80 = {6 \over {4 + r}}$ [From equation 3]

$\Rightarrow 1.5 \times {2 \over 3} = {6 \over {4 + r}}$

$\Rightarrow 4 + r = 6$

$\Rightarrow r = 2\,\Omega$

Given,

${{{r_A}} \over {{r_B}}} = {2 \over 3}$

${{{\rho _A}} \over {{\rho _B}}} = {3 \over 5}$

We know,

Acceleration due to gravity

$g = {{GM} \over {{r^2}}} = {G \over {{r^2}}} \times {4 \over 3}\pi {r^3} \times \rho$

$= {{4\pi Gr\rho } \over 3}$

$\therefore$ $g \propto \rho r$

$\therefore$ ${{{g_A}} \over {{g_B}}} = {{{\rho _A}} \over {{\rho _B}}} \times {{{r_A}} \over {{r_B}}}$

$= {3 \over 5} \times {2 \over 3}$

$= {2 \over 5}$

Given,

Mass of body A = 5 kg

Mass of body B = 8 kg

Momentum of body B is twice that of body A,

$\therefore$ ${P_B} = 2{P_A}$

We know,

Kinetic Energy $(K) = {{{P^2}} \over {2m}}$

$\therefore$ ${{{K_A}} \over {{K_B}}} = {\left( {{{{P_A}} \over {{P_B}}}} \right)^2} \times {{{m_B}} \over {{m_A}}}$

$= {\left( {{1 \over 2}} \right)^2} \times {8 \over 5}$

$= {1 \over 4} \times {8 \over 5}$

$= {2 \over 5}$

Here, the apparent frequency is given by

$$\begin{aligned} &f^{\prime}=f\left(\frac{V-V_{0}}{V+V_{s}}\right)=720\left(\frac{(340+20)-20}{(340+20)-0}\right) \\ &=\frac{720 \times 340}{360}=680 \mathrm{~Hz} \end{aligned}$$

Refractive index

$$\begin{aligned} & \mu=\frac{\sin \left(\frac{A+\delta \sin }{2}\right)}{\sin \left(\frac{A}{2}\right)} \Rightarrow \sqrt{2}=\sin \frac{\left(\frac{60+\delta \min }{2}\right)}{\sin 30^{\circ}} \\ & \Rightarrow \frac{1}{2}=\sin \left(\frac{60+\delta \min }{2}\right) \Rightarrow 45^{\circ}=\frac{60+\delta \min }{2} \\ & \Rightarrow \delta \min =30^{\circ} \\ & \delta=\mathrm{i}+\mathrm{e}-\mathrm{A} \\ & \text { Here, } e=i \\ & \text { So, } \delta \mathrm{min}=2 \mathrm{e}-\mathrm{A} \Rightarrow 2 \mathrm{e}=\delta \min +\mathrm{A} \\ & e=\frac{\delta \min +A}{2}=\frac{30^{\circ}+60^{\circ}}{2}=45^{\circ} \end{aligned}$$

For monoatomic gas degree of freedom f = 3 and $\gamma$ = ${5 \over 3}$

Here for gas,

Initial pressure (P₁) = 75 kPa

Initial volume (V₁) = 1200 cm³

Final volume (V₂) = 150 cm³

Final pressure (P₂) = ?

For adiabatic process,

${P_1}V_1^\gamma = {P_2}V_2^\gamma$

$\Rightarrow 75 \times {(1200)^\gamma } = {P_2}{(150)^\gamma }$

$\Rightarrow {P_2} = 75 \times {\left( {{{1200} \over {150}}} \right)^{{5 \over 3}}}$

$= 75 \times {(8)^{{5 \over 3}}}$

$= 75 \times 32$ kPa

= 2400 kPa

Work done in adiabatic process,

$W = {{{P_2}{V_2} - {P_1}{V_1}} \over {1 - \gamma }}$

$= {{2400 \times {{10}^3} \times 150 \times {{10}^{ - 6}} - 75 \times {{10}^3} \times 1200 \times {{10}^{ - 6}}} \over {1 - {5 \over 3}}}$

$= {{(2400 \times 150 - 75 \times 1200) \times {{10}^{ - 3}}} \over { - {2 \over 3}}}$

$= {{ - 810000 \times {{10}^{ - 3}}} \over 2}$

$= - 405$ kJ

We know, equation of wave travelling in positive x-direction is -

$y = A\sin (kx - wt)$

where $k = {{2\pi } \over \lambda }$

and $w = 2\pi f$

Here given $\lambda$ = 4 cm and frequency (f) = 100 Hz

$\therefore$ $k = {{2\pi } \over 4} = 0.5\pi$ cm^$-$1

and $w = 2\pi \times 100 = 200\pi$ s^$-$1

$\therefore$ Equation of travelling wave,

$y = A\sin (0.5\pi x - 200\pi t)$

Let,

Width of first slit = w₁ and width of second slit = w₂

Given, ${{{w_1}} \over {{w_2}}} = {9 \over {16}}$

Given,

Intensity of light $(I) \propto w$

$\therefore$ ${{{I_1}} \over {{I_2}}} = {{{w_1}} \over {{w_2}}} = {9 \over {16}}$

We know,

${{{I_{\max }}} \over {{I_{\min }}}} = {\left( {{{\sqrt {{I_1}} + \sqrt {{I_2}} } \over {\sqrt {{I_1}} - \sqrt {{I_2}} }}} \right)^2}$

$= {\left( {{{\sqrt {{{{I_1}} \over {{I_2}}}} + 1} \over {\sqrt {{{{I_1}} \over {{I_2}}}} - 1}}} \right)^2}$

$= {\left( {{{{3 \over 4} + 1} \over {{3 \over 4} - 1}}} \right)^2}$

$= {\left( {{7 \over 1}} \right)^2}$

$= {{49} \over 1}$

At t = 20 s,

velocity of truck,

v = 0 + 5 $\times$ 20 = 100 m/s

At 20 sec a ball is dropped from the truck, so velocity of ball will be same as truck.

Velocity of truck at x-direction = 100 m/s and in y-direction = 0.

$\therefore$ Velocity of ball v_x = 100 m/s, v_y = 0

Now ball will show projectile motion where vertically downward acceleration g = 10 m/s act on the ball.

As horizontally no acceleration acting on the ball so horizontal velocity 100 m/s will remain unchanged.

Velocity of the ball when it reach the ground along y-direction after 1 sec.

${v_y} = 0 - 10 \times 1$

$\Rightarrow$ ${v_y} = - 10$ m/s

$\therefore$ Velocity of ball $(\overrightarrow v ) = 100\widehat i - 10\widehat j$

Air bubble moves with constant speed v. So net force = 0.

$\therefore$ Buoyant Force = Viscous force

$\Rightarrow {F_b} = {F_v}$

$\Rightarrow \sigma \times {4 \over 3}\pi {r^3}g = 6\pi nrv$

$\Rightarrow n = {{2\sigma {r^2}g} \over {9v}}$

Given,

$f = 10 \times {10^6}$ Hz

$r = 0.6$ m

Charge on proton (q) = e

We know,

Radius $(r) = {{mv} \over {qB}}$

$= {{\sqrt {2mk} } \over {eB}}$ ..... (1)

Also, we know,

Cyclotron oscillation frequency should be equal to the pendulum evolution frequency.

$\therefore$ $f = {{eB} \over {2\pi m}}$

$\Rightarrow eB = 2\pi mf$

Putting this value of eB in equation (1) we get

$r = {{\sqrt {2mk} } \over {2\pi mf}}$

$\Rightarrow {r^2} = {{2mk} \over {4{\pi ^2}{m^2}{f^2}}}$

$\Rightarrow k = 2{\pi ^2}m{f^2}{r^2}$

$= 2 \times {\left( {{{22} \over 7}} \right)^2} \times 1.67 \times {10^{ - 27}} \times {\left( {10 \times {{10}^6}} \right)^2} \times {\left( {0.6} \right)^2}$ J

$= 1.2 \times {10^{ - 12}}$ J

$= {{1.2 \times {{10}^{ - 12}}} \over {1.6 \times {{10}^{ - 19}}}}$ eV

$= {{12} \over {16}} \times {10^7}$ eV

$= 0.75 \times {10^7}$ eV

$= 7.5 \times {10^6}$ eV

= 7.5 MeV