TS EAMCET 2011 Physics Question Paper with Answer and Solution

(B) Let the mass of the entire disc be $M$.
The mass per unit area is $\sigma = \frac{M}{\pi(2R)^2} = \frac{M}{4\pi R^2}$.
The mass of the removed circular disc of radius $R$ is $M_1 = \sigma \cdot \pi R^2 = \frac{M}{4\pi R^2} \cdot \pi R^2 = \frac{M}{4}$.
The mass of the remaining disc is $M_2 = M - M_1 = M - \frac{M}{4} = \frac{3M}{4}$.
Let the centre of the bigger disc be the origin $(0,0)$. The centre of mass of the removed disc is at $x_1 = R$ and the centre of mass of the remaining disc is at $x_2 = -\alpha R$.
Since the centre of mass of the original disc was at the origin,we have:
$M_1 x_1 + M_2 x_2 = 0$
$\frac{M}{4} \cdot R + \frac{3M}{4} \cdot (-\alpha R) = 0$
$\frac{M}{4} \cdot R = \frac{3M}{4} \cdot \alpha R$
$\alpha = \frac{1}{3}$.

(B) Let $\lambda$ be the linear mass density of the chain. The total length of the chain is $L$. Let $l^{\prime}$ be the length of the chain hanging over the edge.
Then,the length of the chain remaining on the table is $(L - l^{\prime})$.
The mass of the hanging part is $m_h = \lambda l^{\prime}$,and the mass of the part on the table is $m_t = \lambda (L - l^{\prime})$.
The force pulling the chain down is the weight of the hanging part: $F_g = m_h g = \lambda l^{\prime} g$.
The maximum static frictional force acting on the part on the table is $f_{max} = \mu N = \mu m_t g = \mu \lambda (L - l^{\prime}) g$.
For the chain to be in equilibrium,the pulling force must be equal to the maximum frictional force:
$\lambda l^{\prime} g = \mu \lambda (L - l^{\prime}) g$
$l^{\prime} = \mu (L - l^{\prime})$
$l^{\prime} = \mu L - \mu l^{\prime}$
$l^{\prime} (1 + \mu) = \mu L$
$l^{\prime} = \frac{\mu L}{(1 + \mu)}$

(B) Let the maximum accelerations of the two boys be $a_1$ and $a_2$ respectively.
The total tension $T$ in the rope must not exceed $60 \,kg$-wt.
The equation of motion for the two boys climbing up is:
$T = m_1(g + a_1) + m_2(g + a_2)$
Given $T = 60 \,kg$-wt,$m_1 = 20 \,kg$,and $m_2 = 30 \,kg$:
$60g = 20(g + a_1) + 30(g + a_2)$
$60g = 20g + 20a_1 + 30g + 30a_2$
$60g = 50g + 20a_1 + 30a_2$
$10g = 20a_1 + 30a_2$
Dividing by $10$:
$g = 2a_1 + 3a_2$
To find the ratio of maximum accelerations,we consider the individual limits for each boy. If only one boy were climbing,the maximum acceleration would be $a = (T/m) - g$. For boy $1$: $a_{1,max} = (60/20)g - g = 2g$. For boy $2$: $a_{2,max} = (60/30)g - g = g$.
However,the question asks for the ratio of accelerations when they climb together. Assuming they climb with the same acceleration $a$ to find the limit,$g = 2a + 3a = 5a \Rightarrow a = g/5$. If we look for the ratio of their individual maximum possible accelerations,it is $2g : g = 2 : 1$. Given the options,the intended ratio is $a_1 : a_2 = 2 : 1$.

(B) The initial vector is $\vec{A} = 4\hat{i} + 10\hat{j}$.
The magnitude of the vector is $|\vec{A}| = \sqrt{(4)^2 + (10)^2} = \sqrt{16 + 100} = \sqrt{116} \,m$.
When the vector is rotated, its magnitude remains constant.
Let the new vector be $\vec{A}' = 8\hat{i} + n\hat{j}$, where the $x$-component is doubled $(4 \times 2 = 8)$.
Since the magnitude is conserved, $|\vec{A}'| = |\vec{A}|$.
$\sqrt{8^2 + n^2} = \sqrt{116}$.
Squaring both sides, we get $64 + n^2 = 116$.
$n^2 = 116 - 64 = 52$.
$n = \sqrt{52} \approx 7.21 \,m$.
Thus, the new $y$-component is approximately $7.2 \,m$.

(A) Let $V$ be the volume of the body and $\rho_b$ be its density. Let $\rho_1$ and $\rho_2$ be the densities of the liquid at $20^{\circ} C$ and $100^{\circ} C$ respectively.
By the law of flotation,the weight of the body equals the weight of the displaced liquid.
At $20^{\circ} C$: $V \rho_b g = (\frac{2}{3} V) \rho_1 g \implies \rho_b = \frac{2}{3} \rho_1$.
At $100^{\circ} C$: $V \rho_b g = (\frac{3}{4} V) \rho_2 g \implies \rho_b = \frac{3}{4} \rho_2$.
Equating the two expressions for $\rho_b$: $\frac{2}{3} \rho_1 = \frac{3}{4} \rho_2 \implies \frac{\rho_1}{\rho_2} = \frac{9}{8}$.
Since density $\rho = \frac{\rho_0}{1 + \gamma \Delta t}$,we have $\frac{\rho_1}{\rho_2} = \frac{1 + \gamma \Delta t_2}{1 + \gamma \Delta t_1} = 1 + \gamma \Delta T$ (where $\Delta T = 80^{\circ} C$).
$1 + \gamma (80) = \frac{9}{8} \implies 80 \gamma = \frac{1}{8} \implies \gamma = \frac{1}{640} = 0.0015625 {}^{\circ} C^{-1}$.
$\gamma = 15.6 \times 10^{-4} {}^{\circ} C^{-1}$.

(C) The excess pressure inside a soap bubble is given by $P = \frac{4T}{r}$,where $T$ is the surface tension and $r$ is the radius of the bubble.
Since the excess pressure is inversely proportional to the radius $(P \propto \frac{1}{r})$,the smaller bubble has a higher internal pressure compared to the larger bubble.
When connected by a tube,air flows from the region of higher pressure to the region of lower pressure.
Therefore,air flows from the smaller bubble to the larger bubble.

(A) The velocity of the ball just before the first impact is $v_0 = \sqrt{2gh_0}$.
After the first collision,the velocity is $v_1 = e v_0$.
After the second collision,the velocity is $v_2 = e v_1 = e^2 v_0$.
Following this pattern,after $n$ collisions,the velocity of the ball is $v_n = e^n v_0$.
The height $h_n$ reached after the $n$th rebound is given by $h_n = \frac{v_n^2}{2g}$.
Substituting $v_n = e^n v_0$,we get $h_n = \frac{(e^n v_0)^2}{2g} = e^{2n} \frac{v_0^2}{2g}$.
Since $h_0 = \frac{v_0^2}{2g}$,we have $h_n = e^{2n} h_0$.
Rearranging for $e$,we get $e^{2n} = \frac{h_n}{h_0}$,which implies $e = \left[\frac{h_n}{h_0}\right]^{1/2n}$.

(D) The gravitational force acting on the astronaut is $F = G \frac{Mm}{r^2}$,which is finite and non-zero. However,the astronaut is in a state of free fall along with the spaceship. In the frame of reference of the spaceship,the gravitational force is balanced by the pseudo-force (centrifugal force),leading to a state of weightlessness. Therefore,the astronaut does not 'experience' any gravitational force (the effective weight is zero). Thus,Assertion $(A)$ is false.
Reason $(R)$ is a standard physical fact describing the orbital motion of satellites,which is true.

(C) The total mass of the system is $M = m + m = 2m$.
Since the sphere is rolling,the total kinetic energy is the sum of translational kinetic energy and rotational kinetic energy.
Translational kinetic energy $K_t = \frac{1}{2} M v^2 = \frac{1}{2} (2m) v^2 = m v^2$.
For a hollow sphere filled with liquid,the liquid does not rotate with the sphere (assuming non-viscous liquid). Thus,only the shell rotates. The moment of inertia of a thin hollow sphere is $I = \frac{2}{3} m r^2$.
Rotational kinetic energy $K_r = \frac{1}{2} I \omega^2 = \frac{1}{2} (\frac{2}{3} m r^2) (\frac{v}{r})^2 = \frac{1}{3} m v^2$.
Total kinetic energy $K = K_t + K_r = m v^2 + \frac{1}{3} m v^2 = \frac{4}{3} m v^2$.

(C) The pendulum bob is in equilibrium under the action of three forces: the tension $T$ in the string,the horizontal force $F = 2 \text{ N}$,and the weight $W = 1 \text{ N}$ acting vertically downwards.
Resolving the tension $T$ into horizontal and vertical components:
Horizontal component: $T \sin \theta = F = 2 \text{ N}$
Vertical component: $T \cos \theta = W = 1 \text{ N}$
Squaring and adding both equations:
$(T \sin \theta)^2 + (T \cos \theta)^2 = F^2 + W^2$
$T^2 (\sin^2 \theta + \cos^2 \theta) = F^2 + W^2$
$T^2 = F^2 + W^2$
$T = \sqrt{F^2 + W^2} = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5} \text{ N}$

(A) The volume flow rate (quantity of water per second) is given by $Q = A v$,where $A$ is the area of the hole and $v$ is the velocity of efflux.
According to Torricelli's law,the velocity of efflux at a depth $h$ is $v = \sqrt{2gh}$.
For the square hole: Area $A_1 = L^2$ and depth $h_1 = y$. So,$v_1 = \sqrt{2gy}$.
The flow rate $Q_1 = A_1 v_1 = L^2 \sqrt{2gy}$.
For the circular hole: Area $A_2 = \pi R^2$ and depth $h_2 = 4y$. So,$v_2 = \sqrt{2g(4y)} = 2\sqrt{2gy}$.
The flow rate $Q_2 = A_2 v_2 = \pi R^2 (2\sqrt{2gy})$.
Given that $Q_1 = Q_2$,we have:
$L^2 \sqrt{2gy} = 2\pi R^2 \sqrt{2gy}$.
Canceling $\sqrt{2gy}$ from both sides,we get $L^2 = 2\pi R^2$.
Therefore,$R^2 = \frac{L^2}{2\pi}$,which gives $R = \frac{L}{\sqrt{2\pi}}$.

(B) Let the total length of the cylinder be $L$. Initially, the piston is at the midpoint, so the length of the gas column is $L_1 = L/2$. The initial temperature is $T_1 = 0^{\circ} C = 273 \, K$.
When the gas is heated to $T_2 = 100^{\circ} C = 373 \, K$, the piston moves by $5 \, cm$. The new length of the gas column is $L_2 = (L/2) + 5$.
Assuming the pressure remains constant (as the piston is weightless and moves freely), we use Charles's Law: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$.
Since the cross-sectional area $A$ is constant, $V = A \times \text{length}$, so $\frac{L_1}{T_1} = \frac{L_2}{T_2}$.
Substituting the values: $\frac{L/2}{273} = \frac{(L/2) + 5}{373}$.
$373(L/2) = 273(L/2 + 5)$.
$373(L/2) = 273(L/2) + 273 \times 5$.
$(373 - 273)(L/2) = 273 \times 5$.
$100(L/2) = 1365$.
$L/2 = 13.65$.
$L = 27.3 \, cm$.

(B) First, calculate the tension $T$ in the wire. For a system of two blocks connected by a string over a pulley, the acceleration $a$ is given by $a = \frac{(m_2 - m_1)g}{m_1 + m_2} = \frac{(2 - 1)10}{2 + 1} = \frac{10}{3} \,m s^{-2}$.
The tension $T$ in the wire is $T = m_1(g + a) = 1(10 + \frac{10}{3}) = \frac{40}{3} \,N$.
The breaking stress is defined as $\sigma = \frac{T}{A}$, where $A = \pi r^2$ is the cross-sectional area.
Given $\sigma = \frac{40}{3 \pi} \times 10^6 \,N m^{-2}$.
Equating the two: $\frac{40}{3 \pi} \times 10^6 = \frac{40/3}{\pi r^2}$.
Simplifying: $\frac{40}{3 \pi} \times 10^6 = \frac{40}{3 \pi r^2}$.
Thus, $10^6 = \frac{1}{r^2}$, which implies $r^2 = 10^{-6} \,m^2$.
Therefore, $r = 10^{-3} \,m = 1 \,mm$.

(C) Let the police party catch the thief after time $t$.
Distance traveled by the police party in time $t$ is $d_p = v t$.
Distance traveled by the thief in time $t$ starting from rest with acceleration $a$ is $d_t = x + \frac{1}{2} a t^2$.
For the police to catch the thief,the distance traveled by the police must be greater than or equal to the distance traveled by the thief:
$v t \geq x + \frac{1}{2} a t^2$
Rearranging the terms,we get a quadratic inequality in $t$:
$\frac{1}{2} a t^2 - v t + x \leq 0$
For a real solution for $t$ to exist,the discriminant of the corresponding quadratic equation $\frac{1}{2} a t^2 - v t + x = 0$ must be greater than or equal to zero.
The discriminant $D = b^2 - 4ac$ is given by:
$D = (-v)^2 - 4(\frac{1}{2} a)(x) \geq 0$
$v^2 - 2 a x \geq 0$
$v^2 \geq 2 a x$

(C) The initial time period of the simple harmonic oscillator is given by $T = 2 \pi \sqrt{\frac{m}{k}}$.
When a spring of length $l$ and spring constant $k$ is cut into two equal parts,the length of each part becomes $l' = \frac{l}{2}$.
Since the spring constant $k$ is inversely proportional to the length of the spring $(k \propto \frac{1}{l})$,the new spring constant $k'$ for each part will be $k' = 2k$.
The new time period $T'$ with the same mass $m$ and the new spring constant $k'$ is given by $T' = 2 \pi \sqrt{\frac{m}{k'}} = 2 \pi \sqrt{\frac{m}{2k}}$.
Substituting $T = 2 \pi \sqrt{\frac{m}{k}}$,we get $T' = \frac{1}{\sqrt{2}} \times (2 \pi \sqrt{\frac{m}{k}}) = \frac{T}{\sqrt{2}}$.

(A) The moment of inertia $(I)$ of a uniform circular disc is given by $I = \frac{1}{2} M R^2$.
Since the mass $(M)$ and thickness $(t)$ are constant,we express the radius $(R)$ in terms of density $(\rho)$:
$M = \pi R^2 t \rho \Rightarrow R^2 = \frac{M}{\pi t \rho}$.
Substituting this into the formula for $I$:
$I = \frac{1}{2} M \left( \frac{M}{\pi t \rho} \right) = \frac{M^2}{2 \pi t \rho}$.
Since $M$,$t$,and $\pi$ are constants,we have $I \propto \frac{1}{\rho}$.
Therefore,the disc made from the more dense material (higher $\rho$) will have a smaller rotational inertia.

(C) In a steady state,the rate of heat flow $(H)$ through both slabs connected in series is the same.
$H = \frac{k_1 A (\Delta T_A)}{L} = \frac{k_2 A (\Delta T_B)}{L}$
Since the thickness $(L)$ and cross-sectional area $(A)$ are the same for both slabs,we have:
$k_1 (\Delta T_A) = k_2 (\Delta T_B)$
Given $k_1 = \frac{k_2}{2}$,we can write $k_2 = 2k_1$.
Substituting this into the equation:
$k_1 (\Delta T_A) = 2k_1 (\Delta T_B)$
$\Delta T_A = 2 \Delta T_B$
We know the total temperature difference is $\Delta T_A + \Delta T_B = 12^{\circ} C$.
Substituting $\Delta T_A = 2 \Delta T_B$ into the total temperature equation:
$2 \Delta T_B + \Delta T_B = 12^{\circ} C$
$3 \Delta T_B = 12^{\circ} C \implies \Delta T_B = 4^{\circ} C$
Therefore,the temperature difference across slab $A$ is:
$\Delta T_A = 12^{\circ} C - 4^{\circ} C = 8^{\circ} C$.

(A) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the source temperature and $T_2$ is the sink temperature in Kelvin.
Given $\eta = \frac{1}{6}$,so $1 - \frac{T_2}{T_1} = \frac{1}{6} \implies \frac{T_2}{T_1} = \frac{5}{6} \implies T_2 = \frac{5}{6} T_1$.
When the sink temperature is reduced by $62 \ K$ (a change of $62^{\circ} C$ is equivalent to $62 \ K$),the new efficiency $\eta' = 2\eta = 2 \times \frac{1}{6} = \frac{1}{3}$.
Thus,$1 - \frac{T_2 - 62}{T_1} = \frac{1}{3} \implies \frac{T_2 - 62}{T_1} = \frac{2}{3}$.
Substituting $T_2 = \frac{5}{6} T_1$: $\frac{\frac{5}{6} T_1 - 62}{T_1} = \frac{2}{3} \implies \frac{5}{6} - \frac{62}{T_1} = \frac{4}{6} \implies \frac{62}{T_1} = \frac{1}{6} \implies T_1 = 372 \ K$.
Converting to Celsius: $T_1 = 372 - 273 = 99^{\circ} C$.
Then $T_2 = \frac{5}{6} \times 372 = 310 \ K$. Converting to Celsius: $T_2 = 310 - 273 = 37^{\circ} C$.
Therefore,the temperatures are $99^{\circ} C$ and $37^{\circ} C$.

(D) For an adiabatic process, the relation between pressure $p$ and volume $V$ is given by $pV^\gamma = \text{constant}$, where $\gamma = C_p / C_V$.
Given that $p \propto T^3$, we can write $p = k T^3$.
Using the ideal gas equation $pV = nRT$, we have $T = \frac{pV}{nR}$.
Substituting this into the given relation:
$p = k \left( \frac{pV}{nR} \right)^3$
$p = k \frac{p^3 V^3}{(nR)^3}$
$1 = \left( \frac{k}{(nR)^3} \right) p^2 V^3$
$p^2 V^3 = \text{constant}'$
$p V^{3/2} = \text{constant}''$
Comparing this with the standard adiabatic equation $pV^\gamma = \text{constant}$, we get $\gamma = 3/2$.
Thus, the value of $C_p / C_V$ is $3/2$.

(A) Let the velocity of sound be $v$ and the frequency of the third note be $f_0$. The frequencies of the two given notes are $f_1 = \frac{v}{\lambda_1} = \frac{v}{40/195} = \frac{195v}{40}$ and $f_2 = \frac{v}{\lambda_2} = \frac{v}{40/193} = \frac{193v}{40}$.
Since each note produces $9$ beats per second with the third note,we have:
$|f_1 - f_0| = 9$ and $|f_2 - f_0| = 9$.
This implies $f_1 - f_0 = 9$ and $f_0 - f_2 = 9$ (assuming $f_1 > f_0 > f_2$).
Adding these two equations:
$(f_1 - f_0) + (f_0 - f_2) = 9 + 9$
$f_1 - f_2 = 18$
$\frac{195v}{40} - \frac{193v}{40} = 18$
$\frac{2v}{40} = 18$
$\frac{v}{20} = 18$
$v = 360 \,m/s$.

(C) The frequency of the $n$-th harmonic for a string is given by $f_n = \frac{n}{2l} \sqrt{\frac{T}{\mu}}$,where $\mu = \pi r^2 \rho$.
For the first overtone of $A$ $(n=2)$: $f_{A} = \frac{2}{2l_A} \sqrt{\frac{T}{\pi r_A^2 \rho}} = \frac{1}{l_A r_A} \sqrt{\frac{T}{\pi \rho}}$.
For the second overtone of $B$ $(n=3)$: $f_{B} = \frac{3}{2l_B} \sqrt{\frac{T}{\pi r_B^2 \rho}} = \frac{3}{2l_B r_B} \sqrt{\frac{T}{\pi \rho}}$.
Given $f_A = f_B$ and $r_A = 2r_B$:
$\frac{1}{l_A (2r_B)} = \frac{3}{2l_B r_B}$.
Simplifying,$\frac{1}{2l_A} = \frac{3}{2l_B} \Rightarrow \frac{l_A}{l_B} = \frac{1}{3}$.
Thus,the ratio $l_A : l_B = 1 : 3$.

(D) The thermo-emf $E$ is given by $E = a\theta + b\theta^2$.
The thermoelectric power $P$ is defined as the rate of change of thermo-emf with respect to temperature: $P = \frac{dE}{d\theta} = a + 2b\theta$.
The neutral temperature $T_n$ is the temperature at which the thermoelectric power becomes zero.
Setting $P = 0$,we get $a + 2bT_n = 0$.
Therefore,$T_n = -\frac{a}{2b}$.
Given the ratio $a/b = 700^{\circ}C$,we substitute this into the expression: $T_n = -\frac{700}{2} = -350^{\circ}C$.
Since $-350^{\circ}C$ is not among the given options,the correct choice is $D$.

(A) Given, $Z=31$ and $a=5 \times 10^7 \, \text{Hz}^{1/2}$.
According to Moseley's law for $K_{\alpha}$ lines, $\sqrt{\nu} = a(Z-1)$.
Squaring both sides, we get $\nu = a^2(Z-1)^2$.
Substituting the values: $\nu = (5 \times 10^7)^2 \times (31-1)^2$.
$\nu = 25 \times 10^{14} \times 30^2 = 25 \times 10^{14} \times 900 = 2.25 \times 10^{18} \, \text{Hz}$.
We know that $\lambda = \frac{c}{\nu}$, where $c = 3 \times 10^8 \, \text{m/s}$.
$\lambda = \frac{3 \times 10^8}{2.25 \times 10^{18}} = 1.33 \times 10^{-10} \, \text{m}$.
Since $1 \, \mathring{A} = 10^{-10} \, \text{m}$, the wavelength is $1.33 \, \mathring{A}$.

(D) Given:
Dispersive power of the first lens,$\omega_1 = 0.45$.
Dispersive power of the second lens,$\omega_2 = 0.21$.
Focal length of the second lens (convex),$f_2 = 84 \,cm$.
For an achromatic combination of two thin lenses in contact,the condition is given by:
$\frac{\omega_1}{f_1} + \frac{\omega_2}{f_2} = 0$
$\Rightarrow \frac{\omega_1}{f_1} = -\frac{\omega_2}{f_2}$
$\Rightarrow \frac{f_1}{f_2} = -\frac{\omega_1}{\omega_2}$
Substituting the given values:
$\frac{f_1}{84} = -\frac{0.45}{0.21}$
$f_1 = -\frac{45}{21} \times 84$
$f_1 = -\frac{45}{1} \times 4$
$f_1 = -180 \,cm$
Thus,the focal length of the required lens is $-180 \,cm$.

(B) In a compound microscope,both the objective lens and the eyepiece are convex lenses. Statement $(A)$ is incorrect because the objective lens forms a real and inverted image,while the eyepiece forms a virtual and magnified image.
Statement $(B)$ is correct because the objective lens must have a very short focal length to provide high magnification.
Statement $(C)$ is correct because the eyepiece acts as a simple magnifying glass to view the real image formed by the objective lens.
Statement $(D)$ is incorrect because both lenses are convex,not concave.
Therefore,statements $(B)$ and $(C)$ are true.

(A) In a series $L-R$ circuit, the current $i$ at any time $t$ after closing the switch is given by $i(t) = \frac{V}{R} (1 - e^{-Rt/L})$.
At $t = 0$, the current in the circuit is $i(0) = \frac{V}{R} (1 - e^0) = 0$.
The potential difference across the resistor is $V_R = iR$. Since $i = 0$ at $t = 0$, $V_R = 0 \times R = 0 \, V$.
The potential difference across the inductor is $V_L = L \frac{di}{dt}$. According to Kirchhoff's voltage law, $V = V_R + V_L$. At $t = 0$, $V = 0 + V_L$, which gives $V_L = 25 \, V$.
Therefore, at $t = 0$, the potential difference across the resistor is $0 \, V$ and across the inductor is $25 \, V$.

(A) Let the initial capacitance of each condenser be $C = \frac{A \varepsilon_0}{d}$.
For condenser $M$,the space is filled with a dielectric of constant $K = 8$. The new capacitance is $C_M = K C = 8C$.
For condenser $N$,a copper plate of thickness $t = d/2$ is introduced. The capacitance of a capacitor with a metal plate of thickness $t$ is $C_N = \frac{A \varepsilon_0}{d - t}$.
Substituting $t = d/2$,we get $C_N = \frac{A \varepsilon_0}{d - d/2} = \frac{A \varepsilon_0}{d/2} = 2 \left( \frac{A \varepsilon_0}{d} \right) = 2C$.
Since the capacitors are connected in series,the charge $Q$ on both is the same.
The potential difference $V$ is given by $V = Q/C$,so $V \propto 1/C$.
Therefore,the ratio of potential differences is $V_M : V_N = \frac{1}{C_M} : \frac{1}{C_N} = \frac{1}{8C} : \frac{1}{2C} = \frac{1}{8} : \frac{1}{2} = 2 : 8 = 1 : 4$.

(B) Let $V$ be the potential across the capacitor when it is fully charged. The energy stored in the capacitor is given by $U = \frac{1}{2} C V^2$.
When the capacitor is fully discharged through the resistance coil,the entire stored energy is dissipated as heat $\Delta H$ in the block.
Since the system is thermally isolated,the heat gained by the block is equal to the energy lost by the capacitor: $\Delta H = \frac{1}{2} C V^2$.
The heat gained by the block is also given by $\Delta H = m s \Delta T$,where $m$ is the mass and $s$ is the specific heat.
Equating the two expressions for $\Delta H$:
$\frac{1}{2} C V^2 = m s \Delta T$
$V^2 = \frac{2 m s \Delta T}{C}$
$V = \left(\frac{2 m s \Delta T}{C}\right)^{1 / 2}$

(A) The sensitivity of a galvanometer is defined as the deflection per unit current,given by $S_g = \frac{\theta}{i_g}$. When a shunt $S$ is connected in parallel with a galvanometer of resistance $G$,the new sensitivity $S'$ is given by the ratio of the current through the galvanometer $i_g$ to the total current $i$.
The current through the galvanometer is $i_g = i \left( \frac{S}{G+S} \right)$.
Therefore,the new sensitivity is $S' = \frac{i_g}{i} = \frac{S}{G+S}$.
Given,initial sensitivity $= 60 \text{ div/A}$ and final sensitivity $= 10 \text{ div/A}$.
The ratio of sensitivities is $\frac{S'}{S_g} = \frac{10}{60} = \frac{1}{6}$.
Substituting the formula: $\frac{1}{6} = \frac{S}{G+S}$.
Cross-multiplying gives $G + S = 6S$,which simplifies to $G = 5S$.
Given $G = 20 \ \Omega$,we have $20 = 5S$.
Thus,$S = \frac{20}{5} = 4 \ \Omega$.

(B) Let $i_1$ be the current flowing through the $500 \Omega$ resistor and the $12 \text{ V}$ battery.
Since the ammeter reading is zero,no current flows through the right branch containing the $2 \text{ V}$ battery and the ammeter.
Therefore,the current $i_1$ flows through the $500 \Omega$ resistor and the resistor $R$ in series.
Applying Kirchhoff's voltage law $(KVL)$ to the left loop:
$12 - 500 i_1 - R i_1 = 0$
$12 = i_1 (500 + R) \quad \dots (i)$
Now,consider the potential difference across the resistor $R$. Since no current flows through the right branch,the potential difference across $R$ must be equal to the electromotive force $(EMF)$ of the battery in the right branch to maintain zero current in that loop.
The potential difference across $R$ is $V_R = i_1 R$.
For the ammeter to read zero,the potential difference across $R$ must balance the $2 \text{ V}$ battery.
Thus,$i_1 R = 2 \text{ V} \Rightarrow i_1 = \frac{2}{R}$.
Substituting $i_1$ into equation $(i)$:
$12 = \frac{2}{R} (500 + R)$
$6 R = 500 + R$
$5 R = 500$
$R = 100 \Omega$

(D) According to Kirchhoff's Current Law $(KCL)$, the sum of currents entering a junction equals the sum of currents leaving it.
At junction $A$:
$i_1 = 3 \text{ A} + 2 \text{ A} = 5 \text{ A}$
At junction $B$:
The current $i_1$ enters the junction, and $2 \text{ A}$ leaves it. Let $i_2$ be the current flowing towards junction $C$.
$i_1 = 2 \text{ A} + i_2$
$5 \text{ A} = 2 \text{ A} + i_2 \implies i_2 = 3 \text{ A}$
At junction $C$:
The currents $i_2$ and $1 \text{ A}$ enter the junction, and $i$ leaves it.
$i = i_2 + 1 \text{ A}$
$i = 3 \text{ A} + 1 \text{ A} = 4 \text{ A}$

(D) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of an emitted photoelectron is given by $K_{max} = E - \phi_0$,where $E$ is the energy of the incident photon and $\phi_0$ is the work function of the metal.
For the first photon with energy $E_1 = 2.5 eV$:
$\frac{1}{2} m v_1^2 = E_1 - \phi_0 = 2.5 eV - 1.5 eV = 1.0 eV$ $(i)$
For the second photon with energy $E_2 = 3.5 eV$:
$\frac{1}{2} m v_2^2 = E_2 - \phi_0 = 3.5 eV - 1.5 eV = 2.0 eV$ (ii)
Dividing equation $(i)$ by equation (ii),we get:
$\frac{\frac{1}{2} m v_1^2}{\frac{1}{2} m v_2^2} = \frac{1.0 eV}{2.0 eV}$
$\frac{v_1^2}{v_2^2} = \frac{1}{2}$
Taking the square root on both sides:
$\frac{v_1}{v_2} = \frac{1}{\sqrt{2}}$
Thus,the ratio of the maximum velocities is $1 : \sqrt{2}$.

(D) The thermo emf is given by $E = a\theta + b\theta^2$.
To find the neutral temperature $(T_n)$,we differentiate $E$ with respect to $\theta$ and set it to zero,as the thermo emf is maximum at the neutral temperature.
$\frac{dE}{d\theta} = a + 2b\theta$.
Setting $\frac{dE}{d\theta} = 0$,we get $a + 2b\theta = 0$.
Therefore,$T_n = -\frac{a}{2b}$.
Given that $a/b = 700^{\circ}C$,we substitute this into the expression:
$T_n = -\frac{1}{2} \times (700^{\circ}C) = -350^{\circ}C$.
Since the calculated value $-350^{\circ}C$ is not among the options,the correct choice is $D$.

(A) Given: Atomic number $Z=31$ and constant $a=5 \times 10^7 \text{ Hz}^{1/2}$.
According to Moseley's Law for the $K_{\alpha}$ line:
$\sqrt{\nu} = a(Z-1)$
Squaring both sides:
$\nu = a^2(Z-1)^2$
Substitute the values:
$\nu = (5 \times 10^7)^2 \times (31-1)^2$
$\nu = 25 \times 10^{14} \times 30^2$
$\nu = 25 \times 10^{14} \times 900 = 2.25 \times 10^{18} \text{ Hz}$
Now, using the relation $\lambda = \frac{c}{\nu}$ where $c = 3 \times 10^8 \text{ m/s}$:
$\lambda = \frac{3 \times 10^8}{2.25 \times 10^{18}}$
$\lambda = 1.33 \times 10^{-10} \text{ m}$
Since $1 \text{ Å} = 10^{-10} \text{ m}$, we get:
$\lambda = 1.33 \text{ Å}$.

(B) The correct matches are as follows:
$(A)$ Fleming's left hand rule is used to find the direction of force on a current-carrying conductor in a magnetic field,which corresponds to (iii).
$(B)$ Right hand thumb rule is used to determine the direction of magnetic field lines around a current-carrying conductor,which corresponds to (iv).
$(C)$ Biot-Savart law is used to calculate the magnitude and direction of magnetic induction due to a small current element,which corresponds to (ii).
$(D)$ Fleming's right hand rule is used to find the direction of induced current in a conductor moving in a magnetic field,which corresponds to $(i)$.
Therefore,the correct matching is $(A)-(iii), (B)-(iv), (C)-(ii), (D)-(i)$.
Thus,the correct option is $(b)$.

(C) Initial pole strength $= m$ and magnetic moment $= M = m \times (2l)$,where $2l$ is the length of the magnet.
When a magnet is cut $n$ times parallel to its axis,the pole strength of each piece becomes $m' = \frac{m}{n+1}$. Here,$n=5$,so $m' = \frac{m}{5+1} = \frac{m}{6}$.
When a magnet is cut $k$ times perpendicular to its axis,the length of each piece becomes $l' = \frac{l}{k+1}$. Here,$k=3$,so the new length is $2l' = \frac{2l}{3+1} = \frac{2l}{4} = \frac{l}{2}$.
The pole strength remains unaffected by cuts perpendicular to the axis,so the final pole strength of each piece is $m' = \frac{m}{6}$.
The new magnetic moment $M'$ is the product of the new pole strength and the new length: $M' = m' \times (2l') = \left(\frac{m}{6}\right) \times \left(\frac{2l}{4}\right) = \frac{m \times 2l}{24} = \frac{M}{24}$.

(A) Energy released in the fission of one nucleus is $E_1 = 200 \text{ MeV}$.
Converting this energy into Joules:
$E_1 = 200 \times 1.6 \times 10^{-13} \text{ J} = 3.2 \times 10^{-11} \text{ J}$.
We need to find the number of nuclei $(n)$ required to release a total energy $E_{total} = 1000 \text{ J}$.
The relationship is $E_{total} = n \times E_1$.
Therefore,$n = \frac{E_{total}}{E_1} = \frac{1000}{3.2 \times 10^{-11}}$.
$n = \frac{10^3}{3.2 \times 10^{-11}} = \frac{1}{3.2} \times 10^{14} = 0.3125 \times 10^{14} = 3.125 \times 10^{13}$ nuclei.

(C) Given: Refractive indices $\mu_1 = 1.6$,$\mu_2 = 1.8$,$\mu_3 = 1.3$.
At the interface between medium $2$ and $3$,the ray is incident at the critical angle $C$. Therefore,$\sin C = \frac{\mu_3}{\mu_2} = \frac{1.3}{1.8}$.
Let $r$ be the angle of refraction in medium $2$. Since the ray is incident at the critical angle at the interface of $2$ and $3$,the angle of refraction $r$ at the first interface is equal to the critical angle $C$ (i.e.,$r = C$).
Applying Snell's Law at the interface between medium $1$ and $2$:
$\mu_1 \sin \theta = \mu_2 \sin r$
Since $r = C$,we have $\sin r = \sin C = \frac{1.3}{1.8}$.
Substituting the values:
$1.6 \times \sin \theta = 1.8 \times \left(\frac{1.3}{1.8}\right)$
$1.6 \times \sin \theta = 1.3$
$\sin \theta = \frac{1.3}{1.6} = \frac{13}{16}$
$\theta = \sin ^{-1}\left(\frac{13}{16}\right)$

(D) Given:
Dispersive power of the first lens, $\omega_1 = 0.45$.
Dispersive power of the second lens, $\omega_2 = 0.21$.
Focal length of the second lens, $f_2 = 84 \,cm$.
For an achromatic combination of two thin lenses in contact, the condition is given by:
$\frac{\omega_1}{f_1} + \frac{\omega_2}{f_2} = 0$
$\frac{\omega_1}{f_1} = -\frac{\omega_2}{f_2}$
$\frac{f_1}{f_2} = -\frac{\omega_1}{\omega_2}$
Substituting the values:
$f_1 = -f_2 \times \left( \frac{\omega_1}{\omega_2} \right)$
$f_1 = -84 \times \left( \frac{0.45}{0.21} \right)$
$f_1 = -84 \times \left( \frac{45}{21} \right)$
$f_1 = -84 \times \frac{15}{7}$
$f_1 = -12 \times 15 = -180 \,cm$.
Thus, the focal length of the lens is $-180 \,cm$.

(B) In a compound microscope,both the objective and the eyepiece are convex lenses.
An objective lens has a very short focal length,while the eyepiece has a larger focal length.
Statement $(A)$ is false because the objective lens forms a real and inverted image,while the eyepiece forms a virtual and magnified image.
Statement $(B)$ is true as the objective lens has a very short focal length.
Statement $(C)$ is true because the eyepiece acts as a simple magnifying glass to view the intermediate image formed by the objective.
Statement $(D)$ is false because both lenses are convex.
Therefore,statements $(B)$ and $(C)$ are correct.

(C) Given: Barrier potential,$V = 0.3 \,V$.
Thickness of depletion layer,$d = 2 \times 10^{-6} \,m$.
The electric field $E$ is given by the relation $E = \frac{V}{d}$.
Substituting the values,we get:
$E = \frac{0.3}{2 \times 10^{-6}} = 0.15 \times 10^6 = 1.5 \times 10^5 \,V/m$.
In a $p-n$ junction,the electric field is directed from the $n$-region to the $p$-region due to the accumulation of positive ions on the $n$-side and negative ions on the $p$-side of the depletion layer.
Therefore,the correct option is $C$.

(C) The expression $\frac{1}{2} \mu_0 H^2$ represents the energy density of a magnetic field.
Energy density is defined as energy per unit volume.
The dimensional formula for energy is $[ML^2 T^{-2}]$ and for volume is $[L^3]$.
Therefore,the dimensional formula for energy density is $\frac{[ML^2 T^{-2}]}{[L^3]} = [ML^{-1} T^{-2}]$.
Alternatively,using the dimensions of the given quantities:
$[\mu_0] = [MLT^{-2} A^{-2}]$ and $[H] = [AL^{-1}]$.
Substituting these into the expression:
$[\frac{1}{2} \mu_0 H^2] = [MLT^{-2} A^{-2}] \times [AL^{-1}]^2 = [MLT^{-2} A^{-2}] \times [A^2 L^{-2}] = [ML^{-1} T^{-2}]$.

(B) The units for the given physical quantities are as follows:
$1$. Magnetic field intensity $(H)$ is measured in $A \cdot m^{-1}$. Thus,$(A)-(iv)$.
$2$. Magnetic flux $(\phi)$ is measured in Weber $(Wb)$. Thus,$(B)-(i)$.
$3$. Magnetic pole strength $(m)$ is measured in Ampere-meter $(A \cdot m)$. Thus,$(C)-(iii)$.
$4$. Magnetic induction $(B)$ is measured in $Wb \cdot m^{-2}$ (or Tesla). Thus,$(D)-(ii)$.
Therefore,the correct matching is $(A)-(iv), (B)-(i), (C)-(iii), (D)-(ii)$.
The correct option is $(B)$.

(C) The resultant intensity $I_R$ at any point is given by $I_R = I_{\max} \cos^2 \left( \frac{\phi}{2} \right)$,where $\phi$ is the phase difference.
Given that $I_R = 75 \% \text{ of } I_{\max} = 0.75 I_{\max} = \frac{3}{4} I_{\max}$.
Substituting this into the formula:
$\frac{3}{4} I_{\max} = I_{\max} \cos^2 \left( \frac{\phi}{2} \right)$
$\cos^2 \left( \frac{\phi}{2} \right) = \frac{3}{4}$
Taking the square root on both sides:
$\cos \left( \frac{\phi}{2} \right) = \frac{\sqrt{3}}{2}$
Since $\cos \left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2}$,we have:
$\frac{\phi}{2} = \frac{\pi}{6}$
$\phi = \frac{\pi}{3}$