JEE Main 2021 1st September Evening Shift Question Paper with Solutions

by energy conservation $mgl = {1 \over 2}I{\omega ^2} = {1 \over 2}{{m{l^2}{\omega ^2}} \over 3}$

$\Rightarrow \omega = \sqrt {{{6g} \over l}}$

As we know the relation between the linear speed and angular speed,

$v = \omega r = \omega l = \sqrt {6gl}$

$v = \sqrt {6 \times 10 \times .6}$ = 6 m/s

Hence, the speed of the free end of the rod when it passes through its lowest position is 6 m/s.

For Case I,

The potential difference of the uniform wire, V = 240 V

The resistance of the uniform wire, R₁ = 36 $\Omega$

The power dissipation in the first case,

${P_1} = {{{V^2}} \over {{R_1}}} = {{{{(240)}^2}} \over {36}}$

For Case II,

The resistance of each half, ${R_2} = {{{R_1}} \over 2} = {{36} \over 2} = 18\Omega$

${P_2} = {{{V^2}} \over {{R_2}}} + {{{V^2}} \over {{R_2}}} = {{{{(240)}^2}} \over {18}} + {{{{(240)}^2}} \over {18}} = {{{{(240)}^2}} \over 9}$

Thus, the ratio of the total power dissipation in the first case to the second case

${{{P_1}} \over {{P_2}}} = {{{{(240)}^2}/36} \over {{{(240)}^2}/9}} \Rightarrow {{{P_1}} \over {{P_2}}} = {1 \over 4}$

Comparing with, ${{{P_1}} \over {{P_2}}} = {1 \over x}$

The value of the x = 4.

According to given circuit diagram, equivalent resistance between point P and Q.

${R_{PQ}} = (4 + 4)||(4 + 4)$

$= {{8 \times 8} \over {8 + 8}} = 4\,\Omega$

The equivalent circuit can be drawn as,


Equivalent resistance, R_eq = 4 + 1 = 5 $\Omega$

Magnetic field, B = 5T

The side of the square loop, I = 20 cm = 0.20 m

The steady value of the current, I = 2 mA = 2 $\times$ 10^-3 A

Induced emf, e = Bv₀I

Induced current, I = e / R_eq

Substituting the values in the above equation, we get

$2 \times {10^{ - 3}} = {{5 \times {v_0} \times 0.2} \over 5}$

$\Rightarrow$ v₀ = 10^-2 m/s = 1 cm/s

$\therefore$ The value of v₀ = 1 cm/s, so that a steady current of 2 mA flows in the loop.

During upward motion

F = 2N = (+ve) constant


During downward motion

$\Rightarrow$ F = 2N = ($-$ve) constant

$\Rightarrow$ Best possible answer is option (c)

Consider the actual height of the tumbler be H.

The refractive index of the water, $\mu$ = 4/3

The refractive index of the air, $\mu$ = 1

As we know that,

${\mu _{water}} = {{{H_{real}}} \over {{H_{apparent}}}} \Rightarrow {4 \over 3} = {H \over {{H_{apparent}}}}$

$\Rightarrow {H_{apparent}} = {{3H} \over 4}$

Height of air observed by observer = 17.5 $-$ H

According to question, both height observed by observer is same.

${{3H} \over 4}$ = 17.5 $-$ H

$\Rightarrow$ H = 10 cm

Option (b)

On smooth incline

a = g sin30$^\circ$

by S = ut + ${1 \over 2}$at²

S = ${1 \over 2}$${g \over 2}$T² = ${g \over 4}$T² ........ (i)


For rough surface,

$ma = mg\sin 30^\circ - \mu mg\cos 30^\circ$

$a = g\sin 30^\circ - \mu g\cos 30^\circ$

Distance covered by the block on the rough surface in time $\alpha$T,

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

$s = 0 + {1 \over 2}(g\sin 30^\circ - \mu g\sin 30^\circ ){t^2}$

$s = {g \over 4}(1 - \sqrt 3 \mu ){(\alpha T)^2}$ .... (ii)

Distance covered by the block is same for both the case,

$\Rightarrow {g \over 4}(1 - \sqrt 3 \mu ){(\alpha T)^2} = {g \over 4}{T^2}$ [from Eq. (i) and Eq. (ii)]

$\Rightarrow 1 - \sqrt 3 \mu = {1 \over {{\alpha ^2}}} \Rightarrow \mu = \left( {{{{\alpha ^2} - 1} \over {{\alpha ^2}}}} \right){1 \over {\sqrt 3 }}$

Comparing with $\mu = \left( {{{{\alpha ^2} - 1} \over {{\alpha ^2}}}} \right){1 \over {\sqrt x }}$

The value of the x = 3.

From the potential energy curve,

${U_{\max }} = {1 \over 2}k{A^2}$

$10 = {1 \over 2}k{(2)^2}$

$\Rightarrow$ k = 5 N/m

The length of the simple pendulum, L = 4 m

Time period of spring,

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

Time period of simple pendulum,

$T = 2\pi \sqrt {{l \over g}}$

The time period of simple pendulum is same as the time period of the spring oscillation.

$\Rightarrow$ $2\pi \sqrt {{l \over g}}$ = $2\pi \sqrt {{k \over m}}$

Substituting the values in the above equations, we get

$2\pi \sqrt {{4 \over g}}$ = $2\pi \sqrt {{5 \over 5}}$

g = 4m/s²

$\therefore$ The acceleration due to the gravity on the planet is 4 m/s².

Given, ($\tau$_1/2)_x = ($\tau$)_y

Here, $\tau$_1/2 = half-life period of radioactive element and $\tau$ = mean life period of radioactive element.

As we know the expression,

Half-life of the radioactive element x,

${\tau _{1/2}} = {{\ln (2)} \over {{\lambda _x}}}$

Mean life of the radioactive element y,

$\tau = {1 \over {{\lambda _y}}}$

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

${{\ln 2} \over {{\lambda _x}}} = {1 \over {{\lambda _y}}} \Rightarrow {\lambda _x} = 0.693{\lambda _y}$

Initially they have same number of atoms,

${N_x} = {N_y} = {N_0}$

As we know that,

Activity, $A = \lambda N$

As, ${\lambda _x} < {\lambda _y} \Rightarrow {A_x} < {A_y}$

Therefore, y will decay faster than x.

Here, two projectiles are projected at angles 42$^\circ$ and 48$^\circ$ with same initial velocity.

As we know the expression of range of projectile,

Range $= {{{u^2}\sin 2\theta } \over g}$

AT $\theta$₁ = 42$^\circ$,

Range, ${R_1} = {{{u^2}\sin 2(42)^\circ } \over g} = {{0.99{u^2}} \over g}$

At $\theta$₂ = 48$^\circ$

Range, ${R_2} = {{{u^2}\sin 2(48)^\circ } \over g} = {{0.99{u^2}} \over g}$

The range of the projectile is same for the two projectiles.

Therefore, R₁ = R₂

Now, as we know the expression of height of the projectile,

${H_{\max }} = {{{u^2}\sin \theta } \over {2g}}$

At 42$^\circ$, ${H_{\max }} = {H_1} = {{{u^2}\sin 42^\circ } \over {2g}} = {{0.669{u^2}} \over {2g}}$

At 48$^\circ$, ${H_{\max }} = {H_2} = {{{u^2}\sin 48^\circ } \over {2g}} = {{0.743{u^2}} \over {2g}}$

Higher the value of $\theta$ higher the value of maximum height. Therefore, H₁ < H₂.

T₁ = 1 hour

$\Rightarrow$ $\omega$₁ = 2$\pi$ rad/hour

T₂ = 8 hours

$\Rightarrow$ $\omega$₂ = ${\pi \over 4}$ rad/hour

R₁ = 2 $\times$ 10³ km

As T² $\propto$ R³

$\Rightarrow {\left( {{{{R_2}} \over {{R_1}}}} \right)^3} = {\left( {{{{T_2}} \over {{T_1}}}} \right)^2}$

$\Rightarrow {{{R_2}} \over {{R_1}}} = {\left( {{8 \over 1}} \right)^{2/3}} = 4 \Rightarrow {R_2} = 8 \times {10^3}$ km


V₁ = $\omega$₁R₁ = 4$\pi$ $\times$ 10³ km/h

V₂ = $\omega$₂R₂ = 2$\pi$ $\times$ 10³ km/h

Relative $\omega$ = ${{{V_1} - {V_2}} \over {{R_2} - {R_1}}} = {{2\pi \times {{10}^3}} \over {6 \times {{10}^3}}}$

$= {\pi \over 3}$ rad/hour

x = 3

As per diagram,

Diode D₁ & D₂ are in forward bias i.e. R = 30$\Omega$ whereas diode D₃ is in reverse bias i.e. R = infinite $\Rightarrow$ Equivalent circuit will be

Applying KVL starting from point A


$- \left( {{{{I_1}} \over 2}} \right) \times 30 - \left( {{{{I_1}} \over 2}} \right) \times 130 - {I_1} \times 20 + 200 = 0$

$\Rightarrow$ $-$100 I₁ + 200 = 0

I₁ = 2

Option (c)

Given, Planck's constant, h = 6.6 $\times$ 10^$-$34 Js

Boltzmann constant, k_B = 1.38 $\times$ 10^$-$23 J/K

Mass of an electron, m_e = 9 $\times$ 10^$-$31 kg

Temperature of an ideal gas, T = 300 K

As we know that, de-Broglie wavelength,

$\lambda = {h \over {mv}} = {h \over {\sqrt {2mE} }}$ .... (i)

Here, E is the kinetic energy,

$E = {{3{K_B}T} \over 2}$

Substituting value of E in Eq. (i), we get

$\lambda = {h \over {\sqrt {3m{K_B}T} }}$

Substituting the given values in the above equation, we get

$\lambda = {{6.6 \times {{10}^{ - 34}}} \over {\sqrt {3 \times 9 \times {{10}^{ - 31}} \times 1.38 \times {{10}^{ - 23}} \times 300} }}$

= 6.26 nm

$\therefore$ The corresponding de-Broglie wavelength of an electron approximately at 300 K is 6.26 nm.

Given, the length of the water pipe, L = 1 m

The cross-sectional area of the water pipe, A = 1 cm² = 10^$-$4 m²

The temperature of the ice = $-$ 10$^\circ$C

Current passing in the conductor, I = 0.5 A

Resistance of the conductor, R = 4 k$\Omega$

The latent heat of fusion for ice, L_f = 3.33 $\times$ 10⁵ J/kg

The density of the ice, d = 1000 kg/m³

The specific heat of the ice, c_{p, ice} = 2 $\times$ 10³ J/kg

Heat required to melt the ice at 10$^\circ$C to 0$^\circ$C

Q = mc_p$\Delta$T + mL_f $\Rightarrow$ Q = dVc_p$\Delta$T + dVL_f

= 1000 $\times$ 10^$-$4 $\times$ 2 $\times$ 10³ $\times$ (10) + 1000 $\times$ 10^$-$4 $\times$ 3.33 $\times$ 10⁵ ($\because$ V = A $\times$ L)

= 35300 J

According to the Joule's law of heating,

H = I²Rt

$\Rightarrow$ 35300 = (0.5)²(4000) (t)

$\Rightarrow$ t = 35.3 s

Thus, the minimum time required to melt the ice is 35.3 s.

Given,

R₁ = (4 $\pm$ 0.8) $\Omega$

R₂ = (4 $\pm$ 0.4) $\Omega$

Equivalent resistance when the resistors are connected in parallel is given by

${1 \over {{R_{eq}}}} = {1 \over {{R_1}}} + {1 \over {{R_2}}} \Rightarrow {1 \over {{R_{eq}}}} = {1 \over 4} + {1 \over 4}$

${R_{eq}} = 2\,\Omega$

Now, ${{\Delta {R_{eq}}} \over {R_{eq}^2}} = {{\Delta {R_1}} \over {R_1^2}} + {{\Delta {R_2}} \over {R_2^2}}$

Substituting the values in the above equation, we get

${{\Delta {R_{eq}}} \over 4} = {{0.8} \over {16}} + {{0.4} \over {16}} \Rightarrow \Delta {R_{eq}} = 0.3\,\Omega$

$\therefore$ The equivalent resistance in parallel combination is ${R_{eq}} = (2 \pm 0.3)\Omega$.

We know in study state potential difference across inductor = 0

So, equivalent circuit becomes


${1 \over {{R_{eq}}}} = {1 \over 3} + {1 \over 3} + {1 \over 3} = 1$

$\Rightarrow$ R_eq = 1$\Omega$

$\Rightarrow$ Circuit becomes


$\Rightarrow i = {{30} \over 3} = 10$ A

Let us consider the gravitational force acting on each mass M by adjacent particles be F.

and the gravitational force acting on each mass M diagonally be F₁.

${F_{net}} = {{M{V^2}} \over R}$

Along the centre of circle,

$\sqrt 2 F + {F_1} = {{M{V^2}} \over R}$

$\sqrt 2 {{GMM} \over {{{\left( {\sqrt 2 R} \right)}^2}}} + {{GMM} \over {{{(2R)}^2}}} = {{M{V^2}} \over R}$

${{GM} \over R}\left( {{1 \over {\sqrt 2 }} + {1 \over 4}} \right) = {V^2}$

${{GM} \over R}\left( {{{4 + \sqrt 2 } \over {4\sqrt 2 }}} \right) = {V^2}$

$V = \sqrt {{{GM\left( {4 + \sqrt 2 } \right)} \over {R4\sqrt 2 }}}$

$V = {1 \over 2}\sqrt {{{GM\left( {2\sqrt 2 + 1} \right)} \over R}}$

Option (b)

Given, the mass of the body = m

The height from which the body dropped = h

The speed of the body when reached the ground, ${v_f} = 0.8\sqrt {gh}$

Initial velocity of the body, v = 0 m/s

Using the work-energy theorem,

Work done by gravity + Work done by air-friction = Final kinetic energy $-$ Initial kinetic energy.

${W_{mg}} + {W_{air - friction}} = {1 \over 2}mv_f^2 - {1 \over 2}mv_i^2$

Here, work done by gravity = mgh

$\Rightarrow mgh + {W_{air - friction}} = {1 \over 2}m{(0.8\sqrt {gh} )^2} - {1 \over 2}m{(0)^2}$

$\Rightarrow {W_{air - friction}} = {{0.64mgh} \over 2} - mgh$

$\Rightarrow 0.32mgh - mgh = - 0.68mgh$

The value of the work done by the air friction is $-$ 0.68 mgh.

Given, the length of the shock absorber, l = 1.5 m

The total mass of the system, M = 40000 kg

The speed of the wagon, v = 72 km/h

When brakes are applied, the final velocity, v_f = 0

The compressed spring of the shock absorber, x = 1 m

Applying the work-energy theorem,

Work done by the system = Change in kinetic energy

$W = \Delta KE$

${W_{friction}} + {W_{spring}} = {1 \over 2}mv_f^2 + {1 \over 2}mv_i^2$

$- {{90} \over {100}}\left( {{1 \over 2}m{v^2}} \right) + {W_{spring}} = 0 - {1 \over 2}mv_i^2$ ($\because$ 90% energy lost due to friction)

${W_{spring}} = - {{10} \over {100}} \times {1 \over 2}m{v^2}$

$- {1 \over 2}k{x^2} = {1 \over {20}}m{v^2}$

$k = {{m{v^2}} \over {10 \times {x^2}}}$

Substituting the values in the above equation, we get

$k = {{40000 \times {{\left( {72 \times {5 \over {18}}} \right)}^2}} \over {10{{(1)}^2}}}$

= 16 $\times$ 10⁵ N/m

Comparing the spring constant, k = x $\times$ 10⁵

The value of the x = 16.

Given, the Young's modulus of the steel rod, Y = 2 $\times$ 10¹¹ Pa

Thermal coefficient of the steel rod, $\alpha$ = 10^$-$5$^\circ$C

The length of the steel rod, l = 4 m

The area of the cross-section, A = 10 cm²

The temperature difference, $\Delta$T = 400$^\circ$C

As we know that,

Thermal strain = $\alpha$ $\Delta$T

Using the Hooke's law

Young's modulus (Y) = ${{Thermal\,stress} \over {Thermal\,strain}} = {{F/A} \over {\alpha \,\Delta \,T}}$

Thermal stress, $F = YA\,\alpha \,\Delta \,T$

Substitute the values in the above equation, we get

$F = 2 \times {10^{11}} \times 10 \times {10^{ - 4}} \times {10^{ - 5}} \times (400)$

$= 8 \times {10^5}N$

Comparing with, $F = x \times {10^5}N$

The value of the x = 8.

Given, the average translational kinetic energy of dinitrogen (N₂) = Kinetic energy of an electron .... (i)

Translational kinetic energy of dinitrogen (N₂)

$KE = {3 \over 2}{K_B}T$

Here, T = temperature of the gas,

and K_B = Boltzmann constant.

Kinetic energy of an electron = eV

Given, the potential differential of an electron, V = 0.1 V

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

${3 \over 2}{K_B}T = eV$

$\Rightarrow {3 \over 2} \times 1.38 \times {10^{ - 23}} \times T = 1.6 \times {10^{ - 19}} \times (0.1)$

$T = 773K = 773 - 273^\circ C = 500^\circ C$

Given, the number of diatomic moles, n = 3 mol

The increase in temperature of the diatomic mole,

$\Delta$T = 40$^\circ$C

Now, the degree of freedom

f = linear + rotational + no oscillation

f = 3 + 2 + 0 $\Rightarrow$ f = 5

Change in internal energy,

$\Delta$U = nC_v$\Delta$T .... (i)

where, ${C_v} = {f \over 2}R = {5 \over 2}R$

Substituting the value in Eq. (i), we get

$\Delta U = {{5R} \over 2}nR\Delta T$

Now, work done by the gas for isobaric process,

$W = p\Delta V = nR\Delta T$

The ratio of the change in internal energy to the work done by the gas,

${{\Delta U} \over W} = {{{5 \over 2}nR\Delta T} \over {nR\Delta T}}$

$= {{\Delta U} \over W} = {5 \over 2}$

Multiply and divide the above equation with 5, we get

${{\Delta U} \over W} = {{5 \times 5} \over {2 \times 5}} = {{25} \over {10}}$

Comparing with given equation, ${{\Delta U} \over W} = {x \over {10}}$

The value of the x = 25.

B_{due to wire (1)} = ${{{\mu _0}I} \over {4\pi y}}\left[ {\sin 90 + \sin {\theta _1}} \right]$

$= {{{\mu _0}} \over {4\pi }}{I \over y}\left( {1 + {x \over {\sqrt {{x^2} + {y^2}} }}} \right)$ ....(1)

B_{due to wire (2)} = ${{{\mu _0}} \over {4\pi }}{1 \over x}(\sin 90^\circ + \sin {\theta _2})$

$= {{{\mu _0}} \over {4\pi }}{1 \over x}\left( {1 + {y \over {\sqrt {{x^2} + {y^2}} }}} \right)$ ....(2)

Total magnetic field

$B = {B_1} + {B_2}$

$B = {{{\mu _0}I} \over {4\pi }}\left[ {{1 \over y} + {x \over {y\sqrt {{x^2} + {y^2}} }} + {1 \over x} + {y \over {x\sqrt {{x^2} + {y^2}} }}} \right]$

$B = {{{\mu _0}I} \over {4\pi }}\left[ {{{x + y} \over {xy}} + {{{x^2} + {y^2}} \over {xy\sqrt {{x^2} + {y^2}} }}} \right]$

$B = {{{\mu _0}I} \over {4\pi }}\left[ {{{x + y} \over {xy}} + {{\sqrt {{x^2} + {y^2}} } \over {xy}}} \right]$

$B = {{{\mu _0}I} \over {4\pi xy}}\left[ {\sqrt {{x^2} + {y^2}} + (x + y)} \right]$

Option (1)

$y = {{Mg{L^3}} \over {4b{d^3}\delta }}$

${{\Delta y} \over y} = {{\Delta M} \over M} + {{3\Delta L} \over L} + {{\Delta b} \over b} + {{3\Delta d} \over d} + {{\Delta \delta } \over \delta }$

${{\Delta y} \over y} = {{{{10}^{ - 3}}} \over 2} + {{3 \times {{10}^{ - 3}}} \over 1} + {{{{10}^{ - 2}}} \over 4} + {{3 \times {{10}^{ - 2}}} \over 4} + {{{{10}^{ - 2}}} \over 5}$

$= {10^{ - 3}}[0.5 + 3 + 2.5 + 7.5 + 2] = 0.0155$

Option (b)

For diamagnetic material

$\chi$ is independent of temperature and negative.

Magnetisation (M) is directly proportional to H (The relation between magnetisation and magnetising field, M = –mH)

Option (a)

Given, the peak voltage of the battery, V₀ = 20V

The voltage of the battery, V = 2V

Time, t = 1 $\mu$s = 1 $\times$ 10^$-$6s

Resistance of the capacitor, R = 10 $\Omega$

As we know that,

V = V₀(1 $-$ e^$-$t/RC)

Substituting the values in the above equation, we get

2 = 20(1 $-$ e^$-$t/RC)

$\Rightarrow {t \over {RC}} = \ln \left( {{{10} \over 9}} \right)$

$\Rightarrow C = {t \over {R\ln (10/9)}} = {{{{10}^{ - 6}}} \over {10 \times \ln (10/9)}} = 0.95$ $\mu$F

$\therefore$ The capacitance of the capacitor is 0.95 $\mu$F.

Given, the side of the cube, s = 0.5 m

Electric field, E = 150 y²$\widehat j$

The direction of electric field is as shown in the below figure,


At bottom surface, y = 0

As we know that, the expression of electric flux,

$\phi$ = E . A cos$\theta$

Here, E is the electric field passing through the cube and A is the surface area of the cube.

Substituting the values in the above equations, we get

$\phi$ = 150y² . (0.5 $\times$ 0.5) $\times$ cos180$^\circ$

= 150(0)² . (0.25) $\times$ ($-$1) = 0

Hence, the electric flux is zero at the bottom surface.

At the top surface, y = 0.5 m

Electric field, E = 150 y²$\widehat j$ = 150(0.5)² = 37.5 N/C

Electric flux at the top surface,

$\phi$ = E . A cos$\theta$

= (37.5) . (0.5 $\times$ 0.5) cos0$^\circ$

= 9.375 N / C - m²

By using the Gauss's law, $\phi = {{{Q_{in}}} \over {{\varepsilon _0}}}$

Here, Q_in = net charge enclosed in the cube and ${{\varepsilon _0}}$ = permittivity of the free space.

Substituting the values in the above equation, we get

$9.375 = {{{Q_{in}}} \over {8.85 \times {{10}^{ - 12}}}}$

Q_in = 8.3 $\times$ 10^-11 C

The charge inside the cube is 8.3 $\times$ 10^-11 C.

Given, the amplitude of the carrier wave, A_c = 250 V

The amplitude of the message wave, A_m = 150 V

We know that,

The maximum amplitude, A_max = A_c + A_m

Substituting the values in the above equation, we get

A_max = 250 + 150 = 400 V

We know that,

The minimum amplitude, A_min = A_c $-$ A_m

Substituting the values in the above equation, we get

A_min = 250 $-$ 150 = 100 V

Thus, the ratio of the minimum amplitude to the maximum amplitude of the modulated wave is

${{{A_{\min }}} \over {{A_{\max }}}} = {{100} \over {400}} = {1 \over 4}$

Comparing with, 1 : x

The value of the x = 4.

Let acceleration of wedge is a_w and acceleration of block w.r.t. wedge is a_b

For the wedge w.r.t. ground


N cos60$^\circ$ = Ma_w = 16a_w

$\Rightarrow$ N = 32a_w

F.B.D. of block w.r.t. wedge


Balancing vertical forces

N = 8g cos30$^\circ$ $-$ 8a_w sin30$^\circ$ $\Rightarrow$ 32a_w = 4$\sqrt 3$g $-$ 4a_w

$\Rightarrow$ a_w = ${{\sqrt 3 } \over 9}$g

Along incline plane

8g sin30$^\circ$ + 8a_w cos30$^\circ$ = ma_b = 8a_b

a_b = g $\times$ ${1 \over 2} + {{\sqrt 3 } \over 9}$g . ${{\sqrt 3 } \over 2} = {{2g} \over 3}$

Option (d)

Given, electric field,

E = 20 cos(2 $\times$ 10¹⁰t $-$ 200 x) V/m

Comparing with the standard equation,

E = E₀ cos($\omega$t $-$ kx) V/m, we get

Wave constant, k = 200

Angular frequency, $\omega$ = 2 $\times$ 10¹⁰ rad/s

Speed of the wave, $v = {\omega \over k} = {{2 \times {{10}^{10}}} \over {200}} = {10^8}$ m/s

Refractive index, $\mu = {c \over v} = {{3 \times {{10}^8}} \over {{{10}^8}}} = 3$

As we know the relation between the refractive index and dielectric constant,

$\mu = \sqrt {{\varepsilon _r}{\mu _r}}$

Substituting the value in the above equations, we get

$3 = \sqrt {{\varepsilon _r}(1)}$

${\varepsilon _r} = 9$

Thus, the dielectric constant of the medium is 9.

Given, b₁ = 3b₂

Here, b₁ = width of the one of the two slit and b₂ = width of the other slit.

As we know that,

Intensity, I $\propto$ (Amplitude)²

$\Rightarrow$ ${{{I_1}} \over {{I_2}}} = {\left( {{{{b_1}} \over {{b_2}}}} \right)^2} \Rightarrow {{{I_1}} \over {{I_2}}} = {\left( {{{3{b_2}} \over {{b_2}}}} \right)^2}$

${I_1} = 9{I_2}$

As we know, the ratio of the minimum intensity to the maximum intensity in the interference pattern,

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

Substituting the values in the above equations, we get

${{{I_{\min }}} \over {{I_{\max }}}} = {{{{(\sqrt {9{I_2}} - \sqrt {{I_2}} )}^2}} \over {{{(\sqrt {9{I_2}} + \sqrt {{I_2}} )}^2}}} = {\left( {{{3 - 1} \over {3 + 1}}} \right)^2}$

${{{I_{\min }}} \over {{I_{\max }}}} = {1 \over 4}$

Comparing with, ${{{I_{\min }}} \over {{I_{\max }}}} = {x \over 4}$

The value of x = 1.

Given, mass of Na = 8g

Molar mass of Na = 23 gmol^$-$1

${{Weight\,of\,sodium\,atom} \over {Molecular\,mass\,of\,sodium\,atom}} = {{Number\,of\,atoms} \over {Avogadro's\,number}}$

${{8g} \over {23g}} = {{Number\,of\,atoms} \over {6.022 \times {{10}^{23}}}}$

Number of atoms $= {{8 \times 6.022} \over {23}} \times {10^{23}}$

Number of atoms = 2.09 $\times$ 10²³

x $\approx$ 2

Hence, answer is 2.