TS EAMCET 2007 Physics Question Paper with Answer and Solution

(D) The coefficient of restitution $e$ is defined as the ratio of the relative velocity of separation to the relative velocity of approach:
$e = \frac{v_{\text{sep}}}{v_{\text{app}}}$
For the first collision,we are given that the relative velocity of approach is twice the relative velocity of separation:
$v_{\text{app}, 1} = 2 v_{\text{sep}, 1} \implies e_1 = \frac{v_{\text{sep}, 1}}{v_{\text{app}, 1}} = \frac{1}{2}$
We are given the ratio of the coefficients of restitution as $e_1 : e_2 = 3 : 1$,which means:
$e_2 = \frac{e_1}{3} = \frac{1/2}{3} = \frac{1}{6}$
Since $e_2 = \frac{v_{\text{sep}, 2}}{v_{\text{app}, 2}} = \frac{1}{6}$,the ratio of the relative velocity of approach to the relative velocity of separation in the second collision is:
$\frac{v_{\text{app}, 2}}{v_{\text{sep}, 2}} = \frac{1}{e_2} = 6 = 6: 1$

(D) The force exerted by the bullets on the rifle is equal to the rate of change of momentum of the bullets.
Given:
Mass of rifle $M = 20 \,kg$
Number of bullets per second $n = 4$
Mass of each bullet $m = 35 \times 10^{-3} \,kg$
Velocity of each bullet $v = 400 \,ms^{-1}$
The force $F$ required to keep the rifle stationary is equal to the recoil force exerted by the bullets.
The recoil force is given by the rate of change of momentum of the bullets:
$F = n \times (m \times v)$
$F = 4 \times (35 \times 10^{-3} \,kg) \times (400 \,ms^{-1})$
$F = 4 \times 35 \times 0.4$
$F = 140 \times 0.4 = 56 \,N$
To prevent the rifle from moving backwards,an external force of $56 \,N$ must be applied in the direction of the bullet's motion.

(C) The formula for escape velocity is $v_e = \sqrt{\frac{2GM}{R}}$.
Let $M_e$ and $R_e$ be the mass and radius of the Earth,and $M_p$ and $R_p$ be the mass and radius of the planet.
Given: $M_p = \frac{M_e}{2}$ and $R_p = \frac{R_e}{4}$.
The ratio of escape velocities is $\frac{v_p}{v_e} = \sqrt{\frac{M_p}{M_e} \times \frac{R_e}{R_p}}$.
Substituting the values: $\frac{v_p}{v_e} = \sqrt{\frac{M_e/2}{M_e} \times \frac{R_e}{R_e/4}} = \sqrt{\frac{1}{2} \times 4} = \sqrt{2}$.
Therefore,$v_p = v_e \times \sqrt{2} = 11 \times 1.414 = 15.55 \ km \ s^{-1}$.

(A) According to Dalton's Law of Partial Pressures,the total pressure exerted by a mixture of non-reacting gases is the sum of the partial pressures of the individual gases.
For gas $A$,the pressure is $P$ when it occupies volume $V$ at temperature $T$.
For gas $B$,the pressure is $P$ when it occupies volume $V$ at temperature $T$.
When both gases are mixed into the same volume $V$ at the same temperature $T$,the total pressure $P_{mix}$ is the sum of the individual pressures.
$P_{mix} = P_A + P_B = P + P = 2 P$.

(D) The coefficient of restitution $e$ is defined as the ratio of the relative velocity of separation to the relative velocity of approach:
$e = \frac{v_{sep}}{v_{app}}$
For the first collision,we are given that the relative velocity of approach is twice the relative velocity of separation:
$v_{app,1} = 2 v_{sep,1} \implies e_1 = \frac{v_{sep,1}}{v_{app,1}} = \frac{1}{2}$
We are given the ratio of the coefficients of restitution as $e_1 : e_2 = 3 : 1$,which means $e_2 = \frac{e_1}{3}$.
Substituting the value of $e_1$:
$e_2 = \frac{1/2}{3} = \frac{1}{6}$
Since $e_2 = \frac{v_{sep,2}}{v_{app,2}} = \frac{1}{6}$,the ratio of the relative velocity of approach to the relative velocity of separation in the second collision is:
$\frac{v_{app,2}}{v_{sep,2}} = \frac{1}{e_2} = \frac{6}{1}$

(C) Given,$m_1 = 6 \,kg, m_2 = 4 \,kg$ and $\overrightarrow{v}_1 = 5 \hat{i}-2 \hat{j}+10 \hat{k}, \overrightarrow{v}_2 = 10 \hat{i}-2 \hat{j}+5 \hat{k}$.
The velocity of the centre of mass $\overrightarrow{v}_{cm}$ is given by the formula:
$\overrightarrow{v}_{cm} = \frac{m_1 \overrightarrow{v}_1 + m_2 \overrightarrow{v}_2}{m_1 + m_2}$
Substituting the values:
$\overrightarrow{v}_{cm} = \frac{6(5 \hat{i}-2 \hat{j}+10 \hat{k}) + 4(10 \hat{i}-2 \hat{j}+5 \hat{k})}{6 + 4}$
$\overrightarrow{v}_{cm} = \frac{(30 \hat{i}-12 \hat{j}+60 \hat{k}) + (40 \hat{i}-8 \hat{j}+20 \hat{k})}{10}$
$\overrightarrow{v}_{cm} = \frac{70 \hat{i}-20 \hat{j}+80 \hat{k}}{10}$
$\overrightarrow{v}_{cm} = 7 \hat{i}-2 \hat{j}+8 \hat{k}$

(A) According to Dalton's Law of Partial Pressures,the total pressure exerted by a mixture of non-reacting gases is the sum of the partial pressures of the individual gases.
For gas $A$,the pressure is $P$ at volume $V$ and temperature $T$.
For gas $B$,the pressure is $P$ at volume $V$ and temperature $T$.
When mixed in a container of volume $V$ at temperature $T$,the total pressure $P_{mix}$ is given by the sum of the individual pressures:
$P_{mix} = P_A + P_B = P + P = 2 P$.

(C) Let the mass of the man be $m$. The weight of the man is $w = mg$. The forces acting on the man are his weight $mg$ acting downwards and the frictional force $F$ acting upwards.
Since the man is sliding down with an acceleration $a = g/4$,according to Newton's second law of motion:
$mg - F = ma$
Substituting $a = g/4$:
$mg - F = m(g/4)$
$mg - F = mg/4$
$F = mg - mg/4$
$F = 3mg/4$
Since $w = mg$,we get:
$F = 3w/4$

(C) According to Bernoulli's equation for a horizontal pipe (where height $h_1 = h_2$):
$P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2$
Given:
$P_1 = 50 \ kPa = 50 \times 10^3 \ Pa$
$v_1 = 1 \ m/s$
$v_2 = 2 \ m/s$
Density of water $\rho = 1000 \ kg/m^3 = 10^3 \ kg/m^3$
Rearranging the equation to solve for $P_2$:
$P_2 = P_1 + \frac{1}{2} \rho (v_1^2 - v_2^2)$
$P_2 = 50 \times 10^3 + \frac{1}{2} \times 10^3 \times (1^2 - 2^2)$
$P_2 = 50 \times 10^3 + 0.5 \times 10^3 \times (1 - 4)$
$P_2 = 50 \times 10^3 + 0.5 \times 10^3 \times (-3)$
$P_2 = 50 \times 10^3 - 1.5 \times 10^3$
$P_2 = 48.5 \times 10^3 \ Pa = 48.5 \ kPa$

(D) liquid does not wet a solid surface if the angle of contact is obtuse,i.e.,greater than $90^{\circ}$.
In this condition,the cohesive forces between the liquid molecules are stronger than the adhesive forces between the liquid and the solid surface.
Consequently,the liquid tends to minimize contact with the surface,resulting in the liquid not wetting the solid.

(A) Poisson's ratio $(\sigma) = \frac{\text{lateral strain}}{\text{longitudinal strain}}$.
Given lateral strain $= 0.01 \times 10^{-3} \ m$ (Note: Strain is dimensionless,assuming the value represents the change in diameter $\Delta D$ or similar,but here we use it directly as the numerator for the ratio).
$\sigma = \frac{\text{lateral strain}}{\Delta L / L} = 0.4$.
$\frac{\Delta L}{L} = \frac{0.01 \times 10^{-3}}{0.4} = 0.025 \times 10^{-3} = 2.5 \times 10^{-5}$.
Young's modulus $Y = \frac{F/A}{\Delta L/L} = \frac{F \cdot L}{A \cdot \Delta L}$.
$Y = \frac{100 \ N}{0.025 \ m^2 \times (2.5 \times 10^{-5})} = \frac{100}{0.0625 \times 10^{-5}} = \frac{100}{6.25 \times 10^{-7}} = 16 \times 10^7 = 1.6 \times 10^8 \ N/m^2$.

(C) The initial horizontal velocity of the object is $u_x = u \cos \theta$. Given $\sin \theta = 3/5$,then $\cos \theta = 4/5$. So,$u_x = 100 \times (4/5) = 80 \ m/s$.
At the highest point,the vertical velocity is $0$,and the horizontal velocity is $80 \ m/s$. The momentum of the system at the highest point is $P = (2m) \times 80 = 160m$.
After the explosion,the first piece $(m)$ comes to rest,so its velocity is $0$. Let the velocity of the second piece $(m)$ be $v$. By conservation of linear momentum: $160m = m(0) + m(v)$,which gives $v = 160 \ m/s$.
The time taken to reach the highest point is $t = (u \sin \theta) / g = (100 \times 3/5) / 10 = 6 \ s$. The horizontal distance covered to reach the highest point is $x_1 = u_x \times t = 80 \times 6 = 480 \ m$.
The second piece travels from the highest point to the ground in time $t = 6 \ s$ with a constant horizontal velocity of $160 \ m/s$. The additional horizontal distance covered is $x_2 = v \times t = 160 \times 6 = 960 \ m$.
The total distance from the point of projection is $x_1 + x_2 = 480 + 960 = 1440 \ m$.

(B) The maximum acceleration of a simple harmonic oscillator is given by $a_{max} = \omega^2 A$,where $\omega$ is the angular frequency and $A$ is the amplitude.
The maximum velocity is given by $v_{max} = \omega A$.
According to the problem,the magnitude of maximum acceleration is $\pi$ times the maximum velocity:
$a_{max} = \pi \cdot v_{max}$
Substituting the expressions:
$\omega^2 A = \pi \cdot \omega A$
Dividing both sides by $\omega A$ (assuming $\omega, A \neq 0$):
$\omega = \pi$
We know that the angular frequency $\omega$ is related to the time period $T$ by the formula $\omega = \frac{2\pi}{T}$.
Substituting $\omega = \pi$:
$\pi = \frac{2\pi}{T}$
$T = 2 \text{ s}$.

(B) Given that the masses of the two solid spheres are equal,$m_A = m_B$.
Since mass $m = \text{Volume} \times \text{Density} = \frac{4}{3} \pi R^3 \rho$,we have:
$\frac{4}{3} \pi R_A^3 \rho_A = \frac{4}{3} \pi R_B^3 \rho_B$
$\Rightarrow \frac{R_A^3}{R_B^3} = \frac{\rho_B}{\rho_A} \Rightarrow \frac{R_A}{R_B} = \left(\frac{\rho_B}{\rho_A}\right)^{1/3}$.
The moment of inertia of a solid sphere about its diameter is $I = \frac{2}{5} m R^2$.
Since $m_A = m_B$,the ratio of moments of inertia is:
$\frac{I_B}{I_A} = \frac{\frac{2}{5} m_B R_B^2}{\frac{2}{5} m_A R_A^2} = \left(\frac{R_B}{R_A}\right)^2$.
Substituting the ratio of radii:
$\frac{I_B}{I_A} = \left[ \left( \frac{\rho_A}{\rho_B} \right)^{1/3} \right]^2 = \left( \frac{\rho_A}{\rho_B} \right)^{2/3}$.

(D) According to Stefan-Boltzmann law,the radiant energy emitted per unit area per unit time is proportional to the fourth power of the absolute temperature:
$E \propto T^4$
Let $E_1 = E$ at temperature $T_1 = T$.
Let $E_2$ be the new radiant energy at temperature $T_2 = \frac{T}{2}$.
Using the ratio method:
$\frac{E_2}{E_1} = \left(\frac{T_2}{T_1}\right)^4$
Substituting the values:
$\frac{E_2}{E} = \left(\frac{T/2}{T}\right)^4 = \left(\frac{1}{2}\right)^4$
$\frac{E_2}{E} = \frac{1}{16}$
$E_2 = \frac{E}{16}$

(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.
Taking the fractional change,we have $\frac{\Delta T}{T} = \frac{1}{2} \frac{\Delta l}{l} = \frac{1}{2} \alpha \Delta \theta$.
Given $\alpha = 12 \times 10^{-6} /{ }^{\circ} C$ and $\Delta \theta = 40^{\circ} C - 20^{\circ} C = 20^{\circ} C$.
Substituting the values: $\frac{\Delta T}{T} = \frac{1}{2} \times 12 \times 10^{-6} \times 20 = 120 \times 10^{-6} = 1.2 \times 10^{-4}$.
The time lost or gained in a day $(T = 86400 ~s)$ is $\Delta T = T \times \frac{\Delta T}{T} = 86400 \times 1.2 \times 10^{-4} \approx 10.368 ~s$.
Since the temperature increases,the length of the pendulum increases,the time period increases,and the clock loses time.

(D) For a monoatomic gas, the molar heat capacity at constant volume is $C_v = \frac{3}{2}R$ and at constant pressure is $C_p = \frac{5}{2}R$.
In cylinder $B$, the piston is fixed, so the process is isochoric (constant volume). The heat supplied is $\Delta Q = n C_v \Delta T_B = n (\frac{3}{2}R) \Delta T_B$.
In cylinder $A$, the piston is free to move, so the process is isobaric (constant pressure). The heat supplied is $\Delta Q = n C_p \Delta T_A = n (\frac{5}{2}R) \Delta T_A$.
Since the heat supplied is the same in both cases, we have $n (\frac{3}{2}R) \Delta T_B = n (\frac{5}{2}R) \Delta T_A$.
Given $\Delta T_A = 42 \,K$, we substitute the values: $\frac{3}{2} \Delta T_B = \frac{5}{2} \times 42$.
$3 \Delta T_B = 5 \times 42 = 210$.
$\Delta T_B = \frac{210}{3} = 70 \,K$.

(B) In an adiabatic process,the system does not exchange heat with the surroundings,so $Q = 0$.
According to the first law of thermodynamics,$\Delta U = Q - W$.
Since $Q = 0$,we have $\Delta U = -W$.
In an adiabatic expansion,the gas does work on the surroundings,so $W > 0$.
This implies $\Delta U < 0$,which means the internal energy of the system decreases.
Since the internal energy of an ideal gas is directly proportional to its temperature $(U \propto T)$,a decrease in internal energy leads to a decrease in the temperature of the system.

(D) $(1)$ Planck's constant $(h)$: $E = h\nu \implies [h] = [E]/[\nu] = [ML^2 T^{-2}] / [T^{-1}] = [ML^2 T^{-1}]$. This matches (iii).
$(2)$ Gravitational constant $(G)$: $F = G(m_1 m_2)/r^2 \implies [G] = [F r^2] / [M^2] = [MLT^{-2}][L^2] / [M^2] = [M^{-1} L^3 T^{-2}]$. This matches (iv).
$(3)$ Bulk modulus $(B)$: $B = \text{Stress} / \text{Strain} = [ML^{-1} T^{-2}] / [M^0 L^0 T^0] = [ML^{-1} T^{-2}]$. This matches $(i)$.
$(4)$ Coefficient of viscosity $(\eta)$: $F = \eta A (dv/dx) \implies [\eta] = [F] / ([A][dv/dx]) = [MLT^{-2}] / ([L^2][LT^{-1}/L]) = [MLT^{-2}] / [L^2 T^{-1}] = [ML^{-1} T^{-1}]$. This matches (ii).
Therefore,the correct matching is $(1)$-(iii),$(2)$-(iv),$(3)$-$(i)$,$(4)$-(ii). Hence,option $(d)$ is correct.

(B) The velocity of the source $v_s$ is given by $v_s = r \omega$.
Given $r = 2 m$ and $\omega = 15 rad/s$,we have $v_s = 2 \times 15 = 30 m/s$.
The Doppler effect formula for a stationary listener and a moving source is $f' = f \left( \frac{v}{v - v_s \cos \theta} \right)$.
The highest frequency is heard when the source is moving directly towards the listener,which corresponds to $\theta = 0^\circ$,so $\cos \theta = 1$.
Substituting the values: $f' = 540 \left( \frac{330}{330 - 30} \right) = 540 \left( \frac{330}{300} \right) = 540 \times 1.1 = 594 Hz$.

(A) The fundamental frequency $f$ of a vibrating wire is given by the formula: $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$, where $T$ is the tension and $\mu$ is the linear mass density.
Since $L$ and $\mu$ are constant for the same wire, we have $f \propto \sqrt{T}$.
Therefore, the ratio of frequencies is given by: $\frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1}}$.
Given $f_1 = 450 \,Hz$, $T_1 = 9 \,kg-wt$, and $f_2 = 900 \,Hz$.
Substituting the values: $\frac{900}{450} = \sqrt{\frac{T_2}{9}}$.
$2 = \sqrt{\frac{T_2}{9}}$.
Squaring both sides: $4 = \frac{T_2}{9}$.
$T_2 = 4 \times 9 = 36 \,kg-wt$.

(C) For water not to spill out of the bucket, the bucket must complete the vertical circular motion.
At the highest point, the minimum velocity required is $v_{top} = \sqrt{gR}$.
Using the principle of conservation of energy between the lowest point (bottom) and the highest point (top):
$\frac{1}{2}mv_{bottom}^2 = \frac{1}{2}mv_{top}^2 + mg(2R)$
$v_{bottom}^2 = v_{top}^2 + 4gR$
$v_{bottom}^2 = gR + 4gR = 5gR$
$v_{bottom} = \sqrt{5gR}$
Given $g = 10 \,m/s^2$ and $R = 0.5 \,m$:
$v_{bottom} = \sqrt{5 \times 10 \times 0.5} = \sqrt{25} = 5 \,m/s$.

(A) When the block strikes the spring,the kinetic energy of the block is converted into the elastic potential energy of the spring. According to the law of conservation of energy:
$\frac{1}{2} mv^2 = \frac{1}{2} kx^2$
where $x$ is the maximum compression in the spring.
Solving for $x$:
$x = \sqrt{\frac{mv^2}{k}} = \sqrt{\frac{25 \times (3)^2}{100}} = \sqrt{\frac{225}{100}} = \sqrt{2.25} = 1.5 \ m$
When the block returns to its original position,the potential energy stored in the spring is fully converted back into the kinetic energy of the block. Since the surface is smooth (no friction),energy is conserved. The magnitude of the velocity remains the same,but the direction is reversed as the block moves away from the spring.
Therefore,the velocity of the block is $v = -3 \ ms^{-1}$.

(D) The time period $T$ of a vibrating bar magnet is given by the formula:
$T = 2 \pi \sqrt{\frac{I}{MH}}$
where $I$ is the moment of inertia,$M$ is the magnetic moment,and $H$ is the magnetic field induction.
Squaring both sides,we get:
$T^2 = 4 \pi^2 \frac{I}{MH}$
Rearranging to solve for $M$:
$M = \frac{4 \pi^2 I}{T^2 H}$
Given values:
$I = 49 \times 10^{-2} \,kg-m^2$
$H = 0.5 \times 10^{-4} \,T$
$T = 8.8 \,s$
Substituting these values:
$M = \frac{4 \times (3.14)^2 \times 49 \times 10^{-2}}{(8.8)^2 \times 0.5 \times 10^{-4}}$
$M = \frac{4 \times 9.8596 \times 49 \times 10^{-2}}{77.44 \times 0.5 \times 10^{-4}}$
$M = \frac{1932.48 \times 10^{-2}}{38.72 \times 10^{-4}}$
$M = 49.909 \times 10^2 \approx 5000 \,A-m^2$

(C) In a Ramsden eyepiece,the two planoconvex lenses have the same focal length $f$ and are separated by a distance $d = \frac{2f}{3}$.
Given the separation distance $d = 12 \ cm$,we can find $f$:
$d = \frac{2f}{3} \implies 12 = \frac{2f}{3} \implies f = \frac{12 \times 3}{2} = 18 \ cm$.
The equivalent focal length $f_{eq}$ of a combination of two lenses with focal lengths $f_1$ and $f_2$ separated by distance $d$ is given by:
$\frac{1}{f_{eq}} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}$.
For a Ramsden eyepiece,$f_1 = f_2 = f$ and $d = \frac{2f}{3}$.
Substituting these values:
$\frac{1}{f_{eq}} = \frac{1}{f} + \frac{1}{f} - \frac{2f/3}{f^2} = \frac{2}{f} - \frac{2}{3f} = \frac{6-2}{3f} = \frac{4}{3f}$.
Therefore,$f_{eq} = \frac{3f}{4}$.
Substituting $f = 18 \ cm$:
$f_{eq} = \frac{3 \times 18}{4} = \frac{54}{4} = 13.5 \ cm$.

(C) The natural frequency of an $L-C$ circuit is given by $f = \frac{1}{2 \pi \sqrt{LC}}$.
This implies $f \propto \frac{1}{\sqrt{C}}$.
When the capacitor $C$ is replaced by a capacitor $C'$ with a dielectric medium of constant $K$,the new capacitance becomes $C' = KC$.
The new frequency $f'$ is given as $f' = f - 25 \text{ kHz} = 125 \text{ kHz} - 25 \text{ kHz} = 100 \text{ kHz}$.
Taking the ratio of the frequencies:
$\frac{f'}{f} = \sqrt{\frac{C}{C'}} = \sqrt{\frac{C}{KC}} = \frac{1}{\sqrt{K}}$.
Substituting the values:
$\frac{100}{125} = \frac{1}{\sqrt{K}}$.
$\frac{4}{5} = \frac{1}{\sqrt{K}} \implies \sqrt{K} = \frac{5}{4} = 1.25$.
Therefore,$K = (1.25)^2 = 1.5625 \approx 1.56$.

(D) The correct matches are as follows:
$1$. $A$ burning candle emits light across a wide range of wavelengths,resulting in a continuous spectrum.
$2$. Sodium vapour,when excited,emits light at specific discrete wavelengths,resulting in a line spectrum.
$3$. $A$ Bunsen flame (often containing molecular species) produces a band spectrum.
$4$. Dark lines in the solar spectrum (Fraunhofer lines) are caused by the absorption of specific wavelengths by gases in the solar atmosphere,resulting in an absorption spectrum.
Therefore,the correct matching is $1-(ii), 2-(i), 3-(iii), 4-(iv)$.

(C) The resistance $R$ of a wire is given by the formula $R = \rho \frac{l}{A}$, where $\rho$ is the resistivity, $l$ is the length, and $A$ is the cross-sectional area.
Given: $R = 4 \Omega$, $\rho = 2.2 \times 10^{-8} \Omega \cdot m$, and diameter $d = 1.4 \text{ mm} = 1.4 \times 10^{-3} \text{ m}$.
The radius $r = \frac{d}{2} = 0.7 \times 10^{-3} \text{ m}$.
The cross-sectional area $A = \pi r^2 = \pi (0.7 \times 10^{-3})^2 \text{ m}^2$.
Rearranging the formula for length $l$: $l = \frac{R A}{\rho}$.
Substituting the values: $l = \frac{4 \times \pi \times (0.7 \times 10^{-3})^2}{2.2 \times 10^{-8}}$.
$l = \frac{4 \times 3.14159 \times 0.49 \times 10^{-6}}{2.2 \times 10^{-8}} = \frac{6.1575 \times 10^{-6}}{2.2 \times 10^{-8}} \approx 280 \text{ m}$.

(D) In a meter bridge,the principle is given by the formula $\frac{R}{S} = \frac{l_1}{100 - l_1}$,where $R$ is the resistance in the left gap and $S$ is the resistance in the right gap.
Given the ratio $\frac{R}{S} = \frac{2}{3}$.
Substituting the values into the formula: $\frac{2}{3} = \frac{l_1}{100 - l_1}$.
Cross-multiplying gives: $2(100 - l_1) = 3l_1$.
$200 - 2l_1 = 3l_1$.
$200 = 5l_1$.
$l_1 = \frac{200}{5} = 40 \,cm$.
Therefore,the balance length from the left end is $40 \,cm$.

(C) In a metre bridge,the condition for the balance point is given by the formula: $\frac{P}{Q} = \frac{l}{100-l}$,where $P$ is the resistance in the left gap,$Q$ is the resistance in the right gap,and $l$ is the balance length from the left end.
Given the ratio $\frac{P}{Q} = \frac{2}{3}$.
Substituting the values into the formula:
$\frac{2}{3} = \frac{l}{100-l}$
Cross-multiplying gives:
$2(100-l) = 3l$
$200 - 2l = 3l$
$200 = 5l$
$l = \frac{200}{5} = 40 \ cm$.
Therefore,the balance point from the left is $40 \ cm$.

(A) In the Sun,the primary source of energy is the proton-proton cycle.
In this cycle,hydrogen nuclei undergo a series of nuclear fusion reactions to form helium nuclei.
During this fusion process,a significant amount of mass is converted into energy according to Einstein's mass-energy equivalence principle,$E = mc^2$.

(C) According to Einstein's photoelectric equation: $e V_0 = h \nu - h \nu_0$,where $V_0$ is the stopping potential,$\nu$ is the frequency of incident light,and $\nu_0$ is the threshold frequency.
Rearranging for the threshold frequency $\nu_0$: $\nu_0 = \nu - \frac{e V_0}{h}$.
Since $\nu = \frac{c}{\lambda}$,we have $\nu_0 = \frac{c}{\lambda} - \frac{e V_0}{h}$.
Substituting the given values: $c = 3 \times 10^8 \,m/s$,$\lambda = 2 \times 10^{-7} \,m$,$e = 1.6 \times 10^{-19} \,C$,$V_0 = 2.5 \,V$,and $h = 6.6 \times 10^{-34} \,J-s$.
$\nu_0 = \frac{3 \times 10^8}{2 \times 10^{-7}} - \frac{1.6 \times 10^{-19} \times 2.5}{6.6 \times 10^{-34}}$
$\nu_0 = 1.5 \times 10^{15} - 0.606 \times 10^{15} \approx 0.894 \times 10^{15} \,Hz$.
Rounding to the nearest significant value,$\nu_0 \approx 9.0 \times 10^{14} \,Hz$.

(C) The stopping potential $V_0$ is related to the incident energy and work function by the Einstein's photoelectric equation: $e V_0 = K_{\text{max}} = \frac{hc}{\lambda} - \phi$.
Given:
Work function $\phi = 5 \text{ eV} = 5 \times 1.6 \times 10^{-19} \text{ J} = 8 \times 10^{-19} \text{ J}$.
Wavelength $\lambda = 2000 \text{ Å} = 2 \times 10^{-7} \text{ m}$.
Planck's constant $h = 6.67 \times 10^{-34} \text{ J-s}$.
Speed of light $c = 3 \times 10^8 \text{ m/s}$.
Calculating the energy of incident photon:
$E = \frac{hc}{\lambda} = \frac{6.67 \times 10^{-34} \times 3 \times 10^8}{2 \times 10^{-7}} = 10.005 \times 10^{-19} \text{ J}$.
Now,substituting into the equation:
$e V_0 = 10.005 \times 10^{-19} \text{ J} - 8 \times 10^{-19} \text{ J} = 2.005 \times 10^{-19} \text{ J}$.
$V_0 = \frac{2.005 \times 10^{-19} \text{ J}}{1.6 \times 10^{-19} \text{ C}} \approx 1.25 \text{ V}$.

(A) Let the two unit negative charges be $q_1 = q_2 = -1 \text{ C}$ separated by a distance $d$. The charge $q$ is placed at the midpoint. For the system to be in equilibrium,the net force on each charge must be zero.
Consider the force on charge $q_1$ at point $A$ due to charges $q$ at $B$ and $q_2$ at $C$:
$F_A = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q}{(d/2)^2} + \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{d^2} = 0$
Dividing by $\frac{1}{4 \pi \varepsilon_0} q_1$,we get:
$\frac{q}{(d/2)^2} + \frac{q_2}{d^2} = 0$
$\frac{4q}{d^2} + \frac{q_2}{d^2} = 0$
$4q = -q_2$
Since $q_2 = -1 \text{ C}$,we have $4q = -(-1) = 1$.
Therefore,$q = \frac{1}{4} = 0.25 \text{ C}$.

(B) The charges are located at $x=0$ $(q/2)$,$x=a$ $(-q)$,and $x=2a$ $(q/2)$. Point $P$ is at distance $r$ from the charge $-q$ at $x=a$. Since $P$ is on the $x$-axis at distance $r$ from $x=a$,its coordinate is $x = a + r$.
The potential $V$ at point $P$ is the sum of potentials due to individual charges:
$V = \frac{1}{4 \pi \varepsilon_0} \left[ \frac{q/2}{r+a} - \frac{q}{r} + \frac{q/2}{r-a} \right]$
Factoring out $q/2$:
$V = \frac{q}{8 \pi \varepsilon_0} \left[ \frac{1}{r+a} - \frac{2}{r} + \frac{1}{r-a} \right]$
Combining the terms:
$V = \frac{q}{8 \pi \varepsilon_0} \left[ \frac{r(r-a) - 2(r^2-a^2) + r(r+a)}{r(r^2-a^2)} \right]$
Simplifying the numerator:
$V = \frac{q}{8 \pi \varepsilon_0} \left[ \frac{r^2 - ar - 2r^2 + 2a^2 + r^2 + ar}{r(r^2-a^2)} \right] = \frac{q}{8 \pi \varepsilon_0} \left[ \frac{2a^2}{r(r^2-a^2)} \right]$
Since $r \gg a$,we can approximate $r^2 - a^2 \approx r^2$:
$V \approx \frac{q}{8 \pi \varepsilon_0} \cdot \frac{2a^2}{r^3} = \frac{q a^2}{4 \pi \varepsilon_0 r^3}$

(B) The electric potential $V$ at a point due to a system of charges is the algebraic sum of the potentials due to individual charges.
Let the point $P$ be at $x = a + r$. The distances of the charges from $P$ are:
For charge $\frac{q}{2}$ at $x=0$: distance $d_1 = (a+r) - 0 = r+a$
For charge $-q$ at $x=a$: distance $d_2 = (a+r) - a = r$
For charge $\frac{q}{2}$ at $x=2a$: distance $d_3 = |(a+r) - 2a| = |r-a| = r-a$ (since $r >> a$)
The total potential $V_P$ is:
$V_P = \frac{1}{4 \pi \varepsilon_0} \left[ \frac{q/2}{r+a} - \frac{q}{r} + \frac{q/2}{r-a} \right]$
$V_P = \frac{q}{4 \pi \varepsilon_0} \left[ \frac{1}{2(r+a)} - \frac{1}{r} + \frac{1}{2(r-a)} \right]$
$V_P = \frac{q}{4 \pi \varepsilon_0} \left[ \frac{r(r-a) - 2(r^2-a^2) + r(r+a)}{2r(r^2-a^2)} \right]$
$V_P = \frac{q}{4 \pi \varepsilon_0} \left[ \frac{r^2 - ar - 2r^2 + 2a^2 + r^2 + ar}{2r(r^2-a^2)} \right]$
$V_P = \frac{q}{4 \pi \varepsilon_0} \left[ \frac{2a^2}{2r(r^2-a^2)} \right] = \frac{q a^2}{4 \pi \varepsilon_0 r(r^2-a^2)}$
Since $r >> a$,we have $r^2 - a^2 \approx r^2$.
Therefore,$V_P = \frac{q a^2}{4 \pi \varepsilon_0 r^3}$.

(A) When a magnet is displaced by a very small angle $\theta$,the restoring torque acting on the magnet is given by $\tau = -M B_H \sin \theta$.
The negative sign indicates the restoring nature of the torque.
Since $\tau = I \alpha$,where $I$ is the moment of inertia and $\alpha$ is the angular acceleration,we have $I \alpha = -M B_H \sin \theta$.
For small angular displacements,$\sin \theta \approx \theta$.
Therefore,$I \alpha = -M B_H \theta$.
The magnitude of the angular acceleration is $\alpha = \frac{M B_H \theta}{I}$.

(A) For wire $A$,the magnetic field at the center of a circular loop is $B_1 = \frac{\mu_0 i_1}{2r}$.
Given the length of wire $A$ is $40 \ cm$,the circumference is $2\pi r = 40 \ cm$,so $r = \frac{40}{2\pi}$.
For wire $B$,the magnetic field at the center of an arc of radius $r$ subtending an angle $\theta$ is $B_2 = \frac{\mu_0 i_2 \theta}{4\pi r}$.
The length of wire $B$ is $30 \ cm$,so the arc length $s = r\theta = 30 \ cm$.
Thus,$\theta = \frac{30}{r} = \frac{30}{40/2\pi} = \frac{3}{2}\pi$.
Setting $B_1 = B_2$:
$\frac{\mu_0 i_1}{2r} = \frac{\mu_0 i_2 \theta}{4\pi r}$
$\frac{i_1}{2} = \frac{i_2 \theta}{4\pi} = \frac{i_2 (3\pi/2)}{4\pi} = \frac{3i_2}{8}$
$\frac{i_1}{i_2} = \frac{3}{8} \times 2 = \frac{3}{4}$.
Therefore,the ratio $i_1: i_2$ is $3: 4$.

(C) When a charged particle moves perpendicularly to a uniform magnetic field,it follows a circular path.
The radius $r$ of the circular path is given by the formula:
$r = \frac{mv}{qB}$
Given values:
Mass of electron $m = 9 \times 10^{-31} \ kg$
Velocity $v = 1.6 \times 10^7 \ m/s$
Charge of electron $e = 1.6 \times 10^{-19} \ C$
Magnetic field $B = 0.1 \ T$
Substituting these values into the formula:
$r = \frac{9 \times 10^{-31} \times 1.6 \times 10^7}{1.6 \times 10^{-19} \times 0.1}$
$r = \frac{9 \times 1.6 \times 10^{-24}}{1.6 \times 10^{-20}}$
$r = 9 \times 10^{-4} \ m$

(C) When a charged particle moves perpendicular to a uniform magnetic field,it follows a circular path. On a circular path,the velocity vector is always tangent to the path,and the acceleration vector (centripetal acceleration) is directed towards the center of the circle. Therefore,the velocity and acceleration vectors are always perpendicular to each other at any instant.
Given,$\overrightarrow{v}=2 \hat{i}+c \hat{j}$ and $\overrightarrow{a}=3 \hat{i}+4 \hat{j}$.
Since $\overrightarrow{v} \perp \overrightarrow{a}$,their dot product must be zero:
$\overrightarrow{v} \cdot \overrightarrow{a} = 0$
$(2 \hat{i} + c \hat{j}) \cdot (3 \hat{i} + 4 \hat{j}) = 0$
$(2)(3) + (c)(4) = 0$
$6 + 4c = 0$
$4c = -6$
$c = -1.5$

(D) The time period $T$ of a vibrating bar magnet in a magnetic field is given by the formula:
$T = 2 \pi \sqrt{\frac{I}{MH}}$
Where $I$ is the moment of inertia,$M$ is the magnetic moment,and $H$ is the magnetic field induction.
Squaring both sides,we get:
$T^2 = 4 \pi^2 \frac{I}{MH}$
Rearranging for $M$:
$M = \frac{4 \pi^2 I}{T^2 H}$
Substituting the given values:
$I = 49 \times 10^{-2} \,kg-m^2$,$H = 0.5 \times 10^{-4} \,T$,$T = 8.8 \,s$,and $\pi \approx 3.14$:
$M = \frac{4 \times (3.14)^2 \times 49 \times 10^{-2}}{(8.8)^2 \times 0.5 \times 10^{-4}}$
$M = \frac{4 \times 9.8596 \times 49 \times 10^{-2}}{77.44 \times 0.5 \times 10^{-4}}$
$M = \frac{1932.48 \times 10^{-2}}{38.72 \times 10^{-4}}$
$M = 49.909 \times 10^2 \approx 5000 \,A-m^2$.

(C) For a Ramsden eyepiece,the separation distance $d$ between the two plano-convex lenses is given by $d = \frac{2f}{3}$.
Given that $d = 12 \ cm$,we can find the focal length $f$ of each lens:
$12 = \frac{2f}{3} \implies f = \frac{12 \times 3}{2} = 18 \ cm$.
The equivalent focal length $f_{eq}$ of a Ramsden eyepiece is given by the formula:
$f_{eq} = \frac{f_1 f_2}{f_1 + f_2 - d}$.
Substituting $f_1 = f_2 = f$ and $d = \frac{2f}{3}$:
$f_{eq} = \frac{f^2}{2f - \frac{2f}{3}} = \frac{f^2}{\frac{4f}{3}} = \frac{3f}{4}$.
Substituting $f = 18 \ cm$:
$f_{eq} = \frac{3 \times 18}{4} = \frac{54}{4} = 13.5 \ cm$.

(A) The lens maker's formula is given by $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
For a double convex lens, the radii of curvature are $R_1 = R$ and $R_2 = -R$.
Given $\mu = 1.5$ and $f = 5 \,cm$.
Substituting these values into the formula:
$\frac{1}{5} = (1.5 - 1) \left( \frac{1}{R} - \frac{1}{-R} \right)$
$\frac{1}{5} = 0.5 \times \left( \frac{1}{R} + \frac{1}{R} \right)$
$\frac{1}{5} = 0.5 \times \frac{2}{R}$
$\frac{1}{5} = \frac{1}{R}$
Therefore, $R = 5 \,cm$.

(A) In an $n$-type semiconductor,the addition of donor impurity atoms creates discrete energy levels just below the conduction band.
As the concentration of electrons in the conduction band increases due to these donor levels,the Fermi energy level shifts upward from the middle of the forbidden gap.
Therefore,in an $n$-type semiconductor,the Fermi energy level lies in the forbidden energy gap,closer to the conduction band.

(B) The relationship between the temperature of inversion $(T_i)$,neutral temperature $(T_n)$,and the temperature of the cold junction $(T_0)$ is given by the formula:
$T_n = \frac{T_i + T_0}{2}$
Rearranging the formula to solve for $T_i$:
$T_i = 2T_n - T_0$
Given values are $T_0 = 10^{\circ} C$ and $T_n = 270^{\circ} C$.
Substituting these values into the equation:
$T_i = 2 \times 270^{\circ} C - 10^{\circ} C$
$T_i = 540^{\circ} C - 10^{\circ} C$
$T_i = 530^{\circ} C$
Therefore,the temperature of inversion is $530^{\circ} C$.

(B) The Huygen eyepiece is designed such that the distance between the two plano-convex lenses is equal to half the sum of their focal lengths. This specific configuration satisfies the condition for achromatism (elimination of chromatic aberration) and also minimizes spherical aberration.