JEE Advanced 2023 Paper 1 Question Paper with Solutions

A bar, having a mass of $M = 1$ kg and a length of $L = 0.2$ m, is hinged at a point P. A particle with a mass of $m = 0.1$ kg moves in a direction perpendicular to the bar at a velocity of $u = 5$ m/s. This particle collides elastically with the bar at a point that is $\frac{L}{2}$ (0.1 m) away from the pivot point P. As a result of this collision, the bar begins to rotate with an angular velocity $\omega$, and the particle recoils with a new velocity $v$.

We can analyze the bar and the particle as a combined system. Initially, before the impact occurs, the system's angular momentum around the pivot point P is in a clockwise direction. The magnitude of this initial angular momentum can be expressed as

$L_i = m u \left( \frac{L}{2} \right)$

Following the collision, the angular momentum of the system about the pivot point P changes. This final angular momentum can be represented as

$L_f = I_{\text{bar}} \omega - m v \left( \frac{L}{2} \right)$

Here, $I_{\text{bar}}$, the moment of inertia of the bar, is given by $\frac{1}{3} M L^2$, and the term $-\frac{1}{2} m v L$ accounts for the particle's contribution post-collision.

During the collision, the angular momentum of the system around the pivot point P is conserved. This means $L_i = L_f$. Applying this principle, we derive the following relationship :

$2 M L \omega = 3 m (u + v)$ ..........(i)

This equation represents the conservation of angular momentum in the system, correlating the initial and final states of the bar and the particle.

The kinetic energy of the system is conserved in an elastic collision. The kinetic energy of the system before the collision is

$K_i=\frac{1}{2} m u^2 .$

In an elastic collision, the total kinetic energy of the system is preserved. Prior to the collision, the kinetic energy of the system can be quantified as

$K_i = \frac{1}{2} m u^2$

This equation denotes the initial kinetic energy, primarily attributed to the particle's motion.

After the collision takes place, the kinetic energy of the system can be expressed as :

$K_f = \frac{1}{2} m v^2 + \frac{1}{2} I_{\text{rot}} \omega^2 = \frac{1}{2} m v^2 + \frac{1}{2} \times \frac{1}{3} M L^2 \omega^2$

Here, the first term represents the kinetic energy due to the particle's velocity post-collision, and the second term represents the rotational kinetic energy of the bar.

By applying the principle of conservation of kinetic energy, which states $K_i = K_f$, we obtain the following relation :

$M L^2 \omega^2 = 3 m(u^2 - v^2)$ ............(ii)

This equation links the initial and final kinetic energies of the system, comprising the particle and the rotating bar.

By resolving equations (i) and (ii), we determine that the velocity $v$ of the particle post-collision is $4.3$ m/s and the angular velocity $\omega$ of the bar is $6.98$ rad/s.

In the given nuclear reaction :

${}_{{Z_1}}{Z^{{A_1}}} \to {}_{{Z_2}}{Y^{{A_2}}} + {N_1}\,\,\left( {{}_2{H^4}} \right) + {N_2}\,\,\left( {{}_1{e^0}} \right) + {N_3}\,\,\left( {{}_{ - 1}{e^0}} \right)$

we have the conservation of charge expressed as :

$ Z_1 = Z_2 + 2N_{1} + N_{2} - N_{3} $ ...............(i)

and the conservation of nucleons (mass number) is given by :

$ \begin{aligned} A_1 &= A_2 + 4N_{1} \\\\ N_{1} &= \frac{A_1 - A_2}{4} ..........(ii) \end{aligned} $

From (i) and (ii), we get

$\mathrm{N}_2-\mathrm{N}_3=\mathrm{Z}_1-\mathrm{Z}_2-\left(\frac{\mathrm{A}_1-\mathrm{A}_2}{2}\right)$

$\begin{aligned} & \text { (P) }{ }_{92} \mathrm{U}^{238} \rightarrow{ }_{91} \mathrm{~Pa}^{234} \\ & \mathrm{~N}_1=\frac{238-234}{4}=1 \rightarrow 1 \alpha \\ & \mathrm{N}_2-\mathrm{N}_3=(92-91)-\left(\frac{4}{2}\right)=-1 \rightarrow 1 \beta^{-}\end{aligned}$

$\begin{aligned} & \text { (Q) }{ }_{82} \mathrm{~Pb}^{214} \rightarrow { }_{82} \mathrm{~Pb}^{210} \\ & \quad \mathrm{~N}_1=\frac{214-210}{4}=1 \rightarrow 1 \alpha \\ & \mathrm{N}_2-\mathrm{N}_3=(82-82)-\left(\frac{4}{2}\right)=-2 \rightarrow 2 \beta^{-}\end{aligned}$

$\begin{aligned} & (\mathrm{R}){ }_{81} \mathrm{~T} \ell^{210} \rightarrow { }_{82} \mathrm{~Pb}^{206} \\ & \mathrm{~N}_1=\frac{210-206}{4}=1 \rightarrow 1 \alpha \\ & \mathrm{N}_2-\mathrm{N}_3=(81-83)-\frac{4}{2}=-3 \rightarrow 3 \beta^{-}\end{aligned}$

$\begin{aligned} & \text { (S) }{ }_{91} \mathrm{~Pa}^{228} \rightarrow_{88} \mathrm{Ra}^{224} \\ & \mathrm{~N}_1=\frac{228-224}{4}=1 \rightarrow 1 \alpha \\ & \mathrm{N}_2-\mathrm{N}_3=(91-88)-\frac{4}{2}=1 \beta^{+}\end{aligned}$

$\begin{aligned} & \omega_0=10^5 \mathrm{rad} \mathrm{s}^{-1}, L=50 \times 10^{-3} \mathrm{H} \\ & \Rightarrow X_L=5000 \Omega=X_C \text { (at resonance) } \\ & \frac{1}{\sqrt{L C}}=\omega_0 \\ & \Rightarrow L_C=\frac{1}{\omega_0^2} \\ & \Rightarrow C=2 \times 10^{-9} \mathrm{~F} \\ & I_0=\frac{\varepsilon_0}{\sqrt{R^2+\left(X_L-X_C\right)^2}} \\ & I_0=\frac{\varepsilon_0}{R} ..........(i) \end{aligned}$

$$\begin{aligned} 0.05 I_0 & =\frac{\varepsilon_0}{\sqrt{R^2+\left(X_L-X_C\right)^2}} ..........(ii) \\ \text { At } \omega & =8 \times 10^4 \mathrm{rad} \mathrm{s}^{-1} \\ X_L & =8 \times 10^4 \times 50 \times 10^{-3}=4000 \Omega \\ X_C & =\frac{1}{\omega_C}=6250 \Omega \end{aligned}$$

From (i) and (ii), we get

$$\begin{aligned} & \sqrt{R^2+\left(X_L-X_C\right)^2} =\frac{R}{0.05} \\ & R^2+(6250-4000)^2 =400 R^2 \\ & \Rightarrow R^2 \simeq \frac{(2250)^2}{400} \\ & R =\frac{2250}{20}=112.5 \Omega \\ &I_0 =\frac{\varepsilon_0}{R}=\frac{45}{112.5} \\ & =400 \mathrm{~m} \mathrm{~A} \end{aligned}$$

$(\mathrm{P} \rightarrow 3)$

$\begin{aligned} \rightarrow Q & =\frac{1}{R} \sqrt{\frac{L}{C}} \\ \Rightarrow Q & =\frac{2}{2250} \times \sqrt{\frac{50 \times 10^{-3}}{2 \times 10^{-7}}}=\frac{10^4}{225} \\ & =44.4\\ (\mathrm{Q} \rightarrow 1) \\ Q & =\frac{\omega_0}{B W}\end{aligned}$

$\Rightarrow B W=\frac{\omega_0}{Q}=\frac{10^5}{\frac{10^4}{225}}=2250$

$(\mathrm{R} \rightarrow 4)$

$\begin{array}{rrr}P =\frac{\varepsilon_0^2}{R} \text { (peak power) } \\\\ =\frac{45^2}{112.5}=18 \mathrm{~W} \end{array}$

$(S \rightarrow 2)$

When light strikes a surface at Brewster's angle, it becomes polarized perpendicular to the plane of incidence.

Therefore, there is no reflection of blue light observed.

This implies that the light must be polarized within the plane of incidence.



$\begin{aligned} \tan i & =\sqrt{3} \\ i & =60^{\circ} \\ \delta & =i+e-A \\ 60 & =60+e-A \\ \Rightarrow A & =e\end{aligned}$

$\begin{aligned} \frac{\sin i}{\sin r_1} & =\sqrt{3} \\ \Rightarrow \frac{\sin 60}{\sin r_1} & =\sqrt{3} \Rightarrow r_1=30^{\circ}\end{aligned}$

$\begin{array}{rlrl}r_2 =A-r_1 \\ r \sin \left(A-r_1\right) =\sin e \\ \Rightarrow \sqrt{3} \sin (A-30) =\sin A \\ r_3\left[\sin A \cos 30^{\circ}-\cos A\right. \sin 30]=\sin A \\ \Rightarrow \frac{\sqrt{3}}{2}(\sqrt{3}-\cot A) =1 \\ \Rightarrow \cot A =\sqrt{3}-\frac{2}{\sqrt{3}}=\frac{1}{\sqrt{3}} \\ \Rightarrow \cot A =\frac{1}{\sqrt{3}} \\ \Rightarrow A =60^{\circ}\end{array}$

So, $\quad e=60^{\circ}$ (for blue)

For red,

$$\begin{aligned} \mu & =\frac{\sin \left(\frac{\delta_m+A}{2}\right)}{\sin \frac{A}{2}} \\ & =\frac{\sin \left(\frac{30+60}{2}\right)}{\sin \frac{60}{2}}=\frac{\sin 45}{\sin 30}=\frac{1 / \sqrt{2}}{1 / 2} \\ \mu_{\text {red }} & =\sqrt{2} \end{aligned}$$

(A), (C), (D) are correct.

Given :

- Object distance $u = 10.0 \pm 0.1 \, \text{cm}$

- Image distance $v = 20.0 \pm 0.2 \, \text{cm}$

According to the lens formula for a thin lens :

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

So, we can calculate the focal length :

$ f = \frac{1}{\left(\frac{1}{v} - \frac{1}{u}\right)} = \frac{1}{\left(\frac{1}{20 \, \text{cm}} - \frac{1}{-10 \, \text{cm}}\right)} = \frac{1}{0.05 \, \text{cm}^{-1} + 0.1 \, \text{cm}^{-1}} = \frac{1}{0.15 \, \text{cm}^{-1}} = \frac{20}{3} \, \text{cm} $

Next, we need to calculate the error in the determination of the focal length. For that, we find the differential of the lens formula :

Then, the derivative of the equation gives us the change in the focal length $(df)$ in terms of the changes in the object distance $(du)$ and the image distance $(dv)$:

$-\frac{1}{v^2} dv + \frac{1}{u^2} du = \frac{-1}{f^2} df$

For maximum error, we get :

$\frac{1}{f^2} df = \frac{1}{v^2} dv + \frac{1}{u^2} du$

This equation tells us how errors in $(u)$ and $v$ propagate to an error in $f$. Now, when you compute the relative error in the focal length, you get :

$\frac{df}{f} \times 100 = \left(\frac{1}{v^2} dv + \frac{1}{u^2} du\right) \times \frac{f}{1} \times 100$

Plugging in your values of $u = 10 \mathrm{~cm}$, $du = 0.1 \mathrm{~cm}$, $v = 20 \mathrm{~cm}$, $dv = 0.2 \mathrm{~cm}$, and $f = 20/3 \mathrm{~cm}$, you indeed get:

$\frac{df}{f} \times 100 = \left(\frac{0.2}{20^2} + \frac{0.1}{10^2}\right) \times \frac{20}{3} \times 100 = 1\%$

So, the error in the focal length of the lens is indeed 1% (i.e., $n = 1$).

First, we identify the heat capacities at constant volume for monatomic and diatomic gases.

For a monatomic ideal gas, the heat capacity at constant volume, $C_{v1}$, is $\left(\frac{3}{2}\right)R$. This comes from the degrees of freedom for a monatomic gas, which are three (x, y, and z motions).

For a diatomic ideal gas, the heat capacity at constant volume, $C_{v2}$, is $\left(\frac{5}{2}\right)R$. This comes from the degrees of freedom for a diatomic gas, which are five (x, y, and z motions and rotation about two axes).

Next, we find the average heat capacity at constant volume for the gas mixture, $C_{v_{\text{mix}}}$, which is a weighted average based on the number of moles of each gas :

$C_{v_{\text{mix}}} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} = \frac{2 \times \left(\frac{3}{2}\right)R + 1 \times \left(\frac{5}{2}\right)R}{2 + 1} = \frac{11R}{6}$

The change in internal energy $\Delta U$ for a given number of moles $n$ and change in temperature $\Delta T$ is given by :

$\Delta U = n C_{v} \Delta T$

Given that the work done by the system at constant pressure is :

$W = n R \Delta T$

We can replace $\Delta T = \frac{W}{nR}$ in the equation for $\Delta U$ to get :

$\Delta U = n C_{v_{\text{mix}}} \times \frac{W}{n R} = C_{v_{\text{mix}}} \times \frac{W}{ R} = \frac{11R}{6} \times \frac{66}{ R} = 121 \, \text{Joule}$

So the change in internal energy is indeed 121 Joule.

To find the atomic number $Z$ for the hydrogen-like atom emitting photons that cause photoelectrons to eject from the metal surface with a maximum kinetic energy of $1.95 \mathrm{eV}$, we need to make use of the concept of energy transitions in atoms, as well as the photoelectric effect equation. We are given the photoelectric threshold wavelength for the metal and the constants $h c$ and $R h c$.

First, let's calculate the energy of the photon emitted during the transition from level $n=4$ to level $n=3$ in the hydrogen-like atom. The energy of a photon emitted when an electron transitions between two levels in a hydrogen-like atom is given by :

$\Delta E = Z^2 R h c \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$

where:

• $Z$ is the atomic number.
• $R h c$ is the ionization energy of hydrogen (13.6 eV).
• $n_1$ and $n_2$ are the principal quantum numbers of the initial and final energy levels, respectively.

For the transition from $n_1 = 4$ to $n_2 = 3$, the energy of the photon is:

$\Delta E = Z^2 \cdot 13.6 \cdot \left( \frac{1}{3^2} - \frac{1}{4^2} \right) \mathrm{eV}$

$\Delta E = Z^2 \cdot 13.6 \cdot \left( \frac{1}{9} - \frac{1}{16} \right) \mathrm{eV}$

$\Delta E = Z^2 \cdot 13.6 \cdot \left( \frac{16 - 9}{144} \right) \mathrm{eV}$

$\Delta E = Z^2 \cdot 13.6 \cdot \frac{7}{144} \mathrm{eV}$

Next, using the photoelectric effect equation, the maximum kinetic energy (K.E.) of the ejected photoelectrons is equal to the energy of the incident photon minus the work function ($\phi$). The work function can be calculated using the given threshold wavelength :

$\phi = \frac{h c}{\lambda}$

Given: $\lambda = 310 \mathrm{nm}$ $h c = 1240 \mathrm{eV} \cdot \mathrm{nm}$

So the work function is:

$\phi = \frac{1240}{310} \mathrm{eV}$

$\phi = 4 \mathrm{eV}$

The maximum kinetic energy is given by :

$K.E. = \Delta E - \phi$

We are given $K.E. = 1.95 \mathrm{eV}$, hence :

$1.95 = Z^2 \cdot 13.6 \cdot \frac{7}{144} - 4$

$\Rightarrow$ $Z^2 \cdot 13.6 \cdot \frac{7}{144} = 1.95 + 4$

$\Rightarrow$ $Z^2 = \frac{1.95 + 4}{13.6 \cdot \frac{7}{144}}$

$\Rightarrow$ $Z^2 = \frac{5.95 \times 144}{13.6 \times 7}$

$\Rightarrow$ $Z^2 = \frac{856.8}{95.2}$

$\Rightarrow$ $Z^2 = 9$

$\Rightarrow$ $Z = \sqrt{9}$

$\Rightarrow$ $Z = 3$

$\lambda_m T=b$

For option $(\mathrm{P})$ temperature is minimum hence $\lambda _\mathrm{m}$ will be maximum.

$\lambda_m=\frac{b}{T}=\frac{2.9 \times 10^{-3}}{2000}=1.45 \times 10^{-6} \mathrm{~m}=1450 \mathrm{~nm}$

$\begin{aligned} & \sin \theta=\frac{\lambda}{d} \\ & 2 \theta=\text { width of central maximum } \\ \Rightarrow & \text { width } \propto \lambda . \\ \Rightarrow & \text {maximum width of }(\mathrm{P}) \\ & \mathrm{P} \rightarrow 3\end{aligned}$

$\begin{aligned} & \Rightarrow \text { For option }(\mathrm{Q}),\mathrm{T}=3000 \\ & \\ & \lambda_\mathrm{m}=\frac{\mathrm{b}}{\mathrm{T}}=\frac{2.9 \times 10^{-3}}{30000} \\ & \lambda_\mathrm{m}=\frac{2.9}{3} \times 10^{-6} \\ & =0.96 \times 10^{-6} \\ & =966.6 \mathrm{~nm} \\ & \mathrm{P}_{3000}=6 \mathrm{~A}(3000)^4 \\ & \mathrm{P}_{6000}=6 \mathrm{~A}(6000)^4 \\ & \frac{\mathrm{P}_{3000}}{\mathrm{P}_{6000}}=\left(\frac{1}{2}\right)^4=\frac{1}{16} \\ & \mathrm{P}_{3000}=\frac{1}{16} \mathrm{P}_{6000} \\ & \mathrm{Q}-4\end{aligned}$

$\begin{aligned} \Rightarrow & \text { For }(\mathrm{R}), \mathrm{T}=5000 \mathrm{~K} \\ & \lambda_\mathrm{m}=\frac{2.9 \times 10^{-3}}{5 \times 10^3}=0.58 \times 10^{-6} \\ & =580 \mathrm{~nm} \\ & \text { Visible to human eyes } \\ & \mathrm{R}-2\end{aligned}$

For $(S), T=10,000 K$

$\lambda_m T=b \Rightarrow \lambda_m=\frac{2.9 \times 10^{-3}}{10,000}=290 \mathrm{~nm}$

$$\begin{aligned} \phi & =4 e \mathrm{~V} \\ \Rightarrow \lambda_{\mathrm{T}} & =\frac{1240}{4}=310 \mathrm{n} \mathrm{m} \end{aligned}$$

To emit photoelectron

$$\begin{aligned} & \lambda_0 < \lambda_{\mathrm{T}} \\ & \Rightarrow \mathrm{S} \rightarrow 1 \end{aligned}$$

Reflect on the scenario of a ball sliding down a frictionless slope. Initially at rest, the ball descends through a height h. Applying the principle of conservation of mechanical energy, we deduce that the initial speed $u_0$ is given by $u_0 = \sqrt{2gh}$. Consequently, the velocity of the ball as it reaches the end of the slide can be expressed as :

$ \vec{u}_0 = \sqrt{2gh} \, \hat{\imath} . $

In this scenario where a ball is launched horizontally with an initial speed $u_0$ from the top of a building that is $3h$ tall, we calculate the time $t$ it takes for the ball to descend the vertical distance of $3h$. This time duration can be determined using the equation of motion under gravity, leading to :

$ t = \sqrt{\frac{2(3h)}{g}} = \sqrt{\frac{6h}{g}}. $

Considering the horizontal distance covered over the time period t, we can express it as :

$ d = u_0 t = \sqrt{2gh} \cdot \sqrt{\frac{6h}{g}} = 2\sqrt{3}h $ ..............(1)

When the ball impacts the ground, its velocity $\vec{u}$ comprises two components : a horizontal component $u_x$ along $\hat{\imath}$ and a vertical component $u_y$ along $\hat{\jmath}$. The horizontal velocity component remains unchanged throughout the motion and is given by $u_x = \sqrt{2gh}$. On the other hand, the vertical velocity component just before the ball hits the ground can be described as :

$ u_y = -\sqrt{2g(3h)} = -\sqrt{6gh} $

Now, let's focus on the ball's impact with the ground. Since the ground can only exert an impulse in the vertical direction, the horizontal momentum of the ball remains unchanged. This means the horizontal component of the ball's velocity after collision (represented as $v \cos \phi )$ is equal to its horizontal velocity component before the collision. Therefore, we can write this relationship as :

$ v \cos \phi = u_x = \sqrt{2gh} $ ..........(2)

In the vertical axis, the approach velocity, denoted as $v_a$, equals the absolute value of $u_y$, which is $\sqrt{6gh}$. The separation velocity, denoted as $v_s$, is expressed as $v \sin \phi$. Using the coefficient of restitution $e$, which is defined as $\frac{1}{\sqrt{3}}$, we can establish the relationship :

$ e = \frac{1}{\sqrt{3}} = \frac{v_s}{v_a} = \frac{v \sin \phi}{\sqrt{6gh}}. $

This equation leads to the conclusion that :

$ v \sin \phi = \sqrt{2gh} $ ...............(3)

From these two equations (2) and (3), it's clear that $v \cos \phi$ and $v \sin \phi$ are equal. This equality suggests that $\phi = 45^\circ$, as the sine and cosine of 45 degrees are equal.

To find $v$, we can use either of the two equations. Let's use the first one :

$v \cos 45^\circ = \sqrt{2gh}$

Since $\cos 45^\circ = \frac{1}{\sqrt{2}}$, we can rearrange and solve for $v$:

$v = \sqrt{2gh} \times \sqrt{2} = \sqrt{4gh}$

Thus, $\phi = 45^\circ$ and $v = \sqrt{4gh}$.

The ball bounces off with velocity $\vec{v}$ composed of both horizontal and vertical components, as we have established that both $v \cos \phi$ and $v \sin \phi$) are equal to $\sqrt{2gh}$, and the angle $\phi$ is $45^\circ$.

In vector form, the velocity of the ball after bouncing can be represented as the sum of its horizontal and vertical components. Given that both components are equal to $\sqrt{2gh}$ and considering the direction of these components along the unit vectors $\hat{\imath}$ (horizontal) and $\hat{\jmath}$ (vertical), the velocity vector $\vec{v}$ can be expressed as :

$ \vec{v} = \sqrt{2gh}\hat{\imath} + \sqrt{2gh}\hat{\jmath}. $

$\Rightarrow$ $ \vec{v} = \sqrt{2gh}(\hat{\imath} + \hat{\jmath}). $

This equation indicates that after the collision, the ball bounces off with equal horizontal and vertical velocity components, each of magnitude $\sqrt{2gh}$.

The peak height, $h_1$, achieved during the projectile's trajectory from point C to D, can be calculated using the formula :

$ h_1 = \frac{v^2 \sin^2 \phi}{2g}. $

In this scenario, where $v = \sqrt{4gh}$ and $\phi = 45^\circ$, the expression becomes :

$ h_1 = \frac{(\sqrt{4gh})^2 \sin^2 45^\circ}{2g}. $

Using the fact that $\sin 45^\circ$ is $\frac{1}{\sqrt{2}}$ and simplifying, we find that :

$ h_1 = h = \frac{d}{2\sqrt{3}}, $

where the equation (1) has been applied to establish the relationship.

Case-1 : if $\mathrm{I}^{\text {st }}$ refraction on lens :

Given that the object is positioned at the focal point, the resulting light rays will transition to a parallel alignment.

Upon the first reflection at Mirror $M_1$:
As the incoming light rays are parallel, the resultant image will be formed at the focal point.

$\mathrm{II}^{\text {nd }}$ refraction from $\mathrm{L}$ :-

$$\begin{aligned} & \mathrm{u}=-(\mathrm{d}-10) \\ & \mathrm{f}=10 \mathrm{~cm} \\ & \frac{1}{\mathrm{v}}-\frac{1}{\mu}=\frac{1}{\mathrm{f}} \\ & \frac{1}{\mathrm{v}}+\frac{1}{\mathrm{~d}-10}=\frac{1}{10} \\ & \frac{1}{\mathrm{v}}=\frac{1}{10}-\frac{1}{(\mathrm{~d}-10)} .......(i) \end{aligned}$$

This $v$ will be object for $M_2$, and image should be at $10 \mathrm{~cm}$

$$\begin{aligned} & \frac{1}{\mu}+\frac{1}{v_1}=\frac{1}{f} \\ & -\frac{1}{(20-v)}-\frac{1}{10}=-\frac{1}{12} \\ & \frac{1}{12}-\frac{1}{10}=\frac{1}{20-v} \\ & -\frac{2}{120}=\frac{1}{20-v} \\ & 20-v=-60 \\ & v=80 \mathrm{~cm} \end{aligned}$$

From equation (i), we get

$$\begin{aligned} & \frac{1}{80}=\frac{1}{10}-\frac{1}{d-10} \\ & \frac{1}{d-10}=\frac{1}{10}-\frac{1}{80} \\ & \frac{1}{d-10}=\frac{80-10}{800}=\frac{70}{800} \\ & d-10=\frac{80}{7} \Rightarrow d=10+\frac{80}{7}=\frac{150}{7} \\ & n=150 \end{aligned}$$

Case-2 : If $1^{\text {st }}$ reflection on mirror $m_2$ :



For $\mathrm{m}_2$

$$\begin{aligned} & \frac{1}{\mathrm{~V}_1}+\frac{1}{-10}=\frac{1}{-12} \\ & \mathrm{~V}_1=60 \mathrm{~cm} \end{aligned}$$

Then refraction on lens $\mathrm{L}$

$$\begin{aligned} & \mathrm{u}_2=80 \mathrm{~cm} \\ & \frac{1}{\mathrm{~V}_2}-\frac{1}{-60}=\frac{1}{10} \\ & \mathrm{~V}_2=\frac{80}{7} \end{aligned}$$

Then reflection on $\mathrm{m}_2$

Either $V_2$ is at centre (normal incidence)

$$\begin{aligned} & \mathrm{d}-\frac{80}{7}=20 \\ & \mathrm{~d}=\frac{220}{7} \\ & \frac{\mathrm{n}}{7}=\frac{220}{7} \\ & \mathrm{n}=220 \end{aligned}$$

$V_2$ is at pole of $m_2$

$$\begin{aligned} & \mathrm{d}-\frac{80}{7}=0 \\ & \mathrm{~d}=\frac{80}{7} \\ & \frac{\mathrm{n}}{7}=\frac{80}{7} \\ & \mathrm{n}=80 \end{aligned}$$

$\begin{aligned} & m_1=30 \mathrm{gm} \\ & m_2=20 \mathrm{gm}\end{aligned}$



Moment of inertia about the axis of rotation is

$I=m_1 r_1^2+m_2 r_2^2$

Clearly $r_1=4 \mathrm{~cm}$

And $r_2=6 \mathrm{~cm}$

$$\begin{aligned} & \therefore I=\left(30 \times 10^{-3} \times 16 \times 10^{-4}\right)+\left(20 \times 10^{-3} \times 36 \times 10^{-4}\right) \\ & \Rightarrow I=1200 \times 10^{-7} \mathrm{~kg} \mathrm{~m}^2 \end{aligned}$$

If the system is rotated by small angle ' $\theta$ ', the restoring torque is $\tau_{(R)}=-k \theta$

And $\frac{d^2 \theta}{d t^2}=\frac{-k}{l} \cdot \theta=-\omega^2 \theta=\frac{-1.2 \times 10^{-8}}{1200 \times 10^{-7}} \cdot \theta$

$\therefore \omega^2=10^{-4}$

$\begin{aligned} & \text { So, } \omega=\frac{1}{100} \mathrm{rad} / \mathrm{s} \\ & \Rightarrow \omega=10 \times 10^{-3} \mathrm{rad} / \mathrm{s}\end{aligned}$

Volume of dielectric filled in $10 \mathrm{sec}$

$$\begin{aligned} & =250 \times 10=2500 \mathrm{~cm}^3 \\ h \times 50 \times 5 & =2500 \\ \Rightarrow h & =10 \mathrm{~cm} \end{aligned}$$


$C_1=\frac{\varepsilon_0 A_1}{d}, C_2=\frac{\varepsilon_0 A_2}{d} K$

$\begin{aligned} A_1 & =50 \times(50-h) \\ A_2 & =50 \times h \\ C_{\text {eq }} & =C_1+C_2 \\ & =\frac{\varepsilon_0 50(50-h) \mathrm{cm}^2}{5 \mathrm{~cm}}+\frac{3 \times \varepsilon_0 \times 50 \times h}{5 \mathrm{~cm}} \\ & =\varepsilon_0\left[\frac{50 \times 40}{5}+\frac{3 \times 50 \times 10}{5}\right] \times 10^{-2} \\ & =7 \varepsilon_0=63 \times 10^{-12} \\ & =63 \mathrm{pF}\end{aligned}$