AP 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 frequency of a stretched string is given by $f = \frac{n}{2l} \sqrt{\frac{T}{m}}$,where $n$ is the harmonic number,$l$ is the length,$T$ is the tension,and $m$ is the mass per unit length.
Since $m = \pi r^2 \rho$ (where $\rho$ is the density),the frequency formula becomes $f = \frac{n}{2l r} \sqrt{\frac{T}{\pi \rho}}$.
For the first overtone of $A$,$n = 2$. Thus,$f_A = \frac{2}{2 l_A r_A} \sqrt{\frac{T}{\pi \rho}} = \frac{1}{l_A r_A} \sqrt{\frac{T}{\pi \rho}}$.
For the second overtone of $B$,$n = 3$. Thus,$f_B = \frac{3}{2 l_B r_B} \sqrt{\frac{T}{\pi \rho}}$.
Given $f_A = f_B$,we have $\frac{1}{l_A r_A} = \frac{3}{2 l_B r_B}$.
Rearranging for the ratio of lengths: $\frac{l_A}{l_B} = \frac{2 r_B}{3 r_A}$.
Given $r_A = 2 r_B$,we substitute to get $\frac{l_A}{l_B} = \frac{2 r_B}{3 (2 r_B)} = \frac{1}{3}$.

(A) The frequency of the $n$-th harmonic for a stretched 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 $r_A = 2r_B$,we equate the expressions:
$\frac{1}{l_A r_A} = \frac{3}{2l_B r_B} \Rightarrow \frac{1}{l_A (2r_B)} = \frac{3}{2l_B r_B}$.
Simplifying,we get $\frac{1}{2l_A} = \frac{3}{2l_B} \Rightarrow \frac{l_A}{l_B} = \frac{1}{3}$.
Thus,the ratio of the lengths $l_A : l_B = 1:3$.

(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 $T' = \frac{1}{\sqrt{2}} \times (2 \pi \sqrt{\frac{m}{k}}) = \frac{T}{\sqrt{2}}$.

(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:
$(f_1 - f_0) + (f_0 - f_2) = 9 + 9$
$f_1 - f_2 = 18$
$\frac{195v}{40} - \frac{193v}{40} = 18$
$\frac{2v}{40} = 18$
$\frac{v}{20} = 18$
$v = 360 \,m/s$.

(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:
$\frac{\phi}{2} = \frac{\pi}{6}$
$\phi = \frac{\pi}{3}$