TS EAMCET 2008 Physics Question Paper with Answer and Solution

(B) $1$. The ball is dropped from height $h$. The distance covered during the first fall is $h$.
$2$. After the first collision with the floor,the ball rebounds to a height $h_1 = e^2h$.
$3$. The ball then travels upward to height $h_1$ and then falls back down to the floor to make the second hit.
$4$. The distance covered during the rebound is $h_1$ (upward) + $h_1$ (downward) = $2h_1 = 2e^2h$.
$5$. The total distance covered by the ball just before the second hit is the sum of the initial fall and the rebound distance: $D = h + 2e^2h = h(1 + 2e^2)$.

(A) The velocity of the centre of mass $(CM)$ is given by the formula $V_{CM} = \frac{m_A v_A + m_B v_B}{m_A + m_B}$.
Since the particles are initially at rest,the initial velocity of the centre of mass is $V_{CM, initial} = 0$.
According to the law of conservation of momentum,if the net external force on a system is zero,the velocity of the centre of mass remains constant.
In this system,the particles move under mutual forces of attraction,which are internal forces.
Therefore,the net external force on the system is zero.
Since the initial velocity of the centre of mass was zero,it must remain zero at all times,regardless of the individual speeds of the particles.

(D) The time period $T$ of a satellite in a circular orbit is given by the formula $T = 2\pi \sqrt{\frac{r^3}{GM}}$,where $r$ is the orbital radius,$G$ is the gravitational constant,and $M$ is the mass of the earth.
$1$. The orbital radius $r$ is equal to $R + h$,where $R$ is the radius of the earth and $h$ is the height of the satellite from the earth's surface.
$2$. From the formula,it is clear that $T$ depends on the mass of the earth $(M)$,the radius of the orbit $(r)$,and the height $(h)$ because $r = R + h$.
$3$. The time period $T$ is independent of the mass of the satellite $(m)$.
Therefore,the time period depends on (ii),(iii),and (iv).

(B) Let the mass be $m = 150 \ kg$ and the maximum tension the wire can withstand be $T_{max} = 2940 \ N$.
When the load is at the lowest point (equilibrium position),the tension $T$ in the wire is given by $T = mg + \frac{mv^2}{r}$.
However,the question asks for the angle $\theta$ such that the wire does not break when the load passes through the equilibrium position.
At the equilibrium position,the tension $T$ must not exceed $T_{max} = 2940 \ N$.
Using the conservation of energy from the release point at angle $\theta$ to the equilibrium position:
$mgL(1 - \cos \theta) = \frac{1}{2}mv^2 \Rightarrow mv^2 = 2mgL(1 - \cos \theta)$.
The tension at the equilibrium position is $T = mg + \frac{mv^2}{L} = mg + 2mg(1 - \cos \theta) = mg(3 - 2 \cos \theta)$.
Setting $T = T_{max}$:
$2940 = 150 \times 9.8 \times (3 - 2 \cos \theta)$.
$2940 = 1470 \times (3 - 2 \cos \theta)$.
$2 = 3 - 2 \cos \theta$.
$2 \cos \theta = 1 \Rightarrow \cos \theta = 0.5$.
Therefore,$\theta = 60^{\circ}$.

(D) For a smooth inclined plane,the acceleration is $a_s = g \sin \theta$. The time taken to cover distance $s$ is $t_s = \sqrt{\frac{2s}{g \sin \theta}}$.
For a rough inclined plane,the acceleration is $a_r = g(\sin \theta - \mu \cos \theta)$. The time taken is $t_r = \sqrt{\frac{2s}{g(\sin \theta - \mu \cos \theta)}}$.
Given $t_r = n t_s$,where $n = 2$.
Squaring both sides: $\frac{2s}{g(\sin \theta - \mu \cos \theta)} = n^2 \frac{2s}{g \sin \theta}$.
This simplifies to $\sin \theta = n^2(\sin \theta - \mu \cos \theta)$.
Rearranging for $\mu$: $\mu = \tan \theta \left(1 - \frac{1}{n^2}\right)$.
Substituting $\theta = 45^{\circ}$ and $n = 2$: $\mu = \tan 45^{\circ} \left(1 - \frac{1}{2^2}\right) = 1 \times (1 - 0.25) = 0.75$.

(C) Let $\overrightarrow{B} = \hat{i} - \hat{j}$.
To find the component of vector $\overrightarrow{A}$ along the direction of $\overrightarrow{B}$,we calculate the scalar projection of $\overrightarrow{A}$ onto the unit vector of $\overrightarrow{B}$.
The unit vector along $\overrightarrow{B}$ is $\hat{b} = \frac{\overrightarrow{B}}{|\overrightarrow{B}|} = \frac{\hat{i} - \hat{j}}{\sqrt{1^2 + (-1)^2}} = \frac{\hat{i} - \hat{j}}{\sqrt{2}}$.
The component of $\overrightarrow{A}$ along $\overrightarrow{B}$ is given by $\overrightarrow{A} \cdot \hat{b}$.
$\overrightarrow{A} \cdot \hat{b} = (a_x \hat{i} + a_y \hat{j} + a_z \hat{k}) \cdot \left( \frac{\hat{i} - \hat{j}}{\sqrt{2}} \right)$.
$= \frac{a_x(1) + a_y(-1) + a_z(0)}{\sqrt{2}} = \frac{a_x - a_y}{\sqrt{2}}$.

(B) soap bubble has two surfaces (inner and outer) in contact with air. Therefore,the change in surface area is $2 \times \Delta A$.
Given:
Surface tension $T = 0.03 \,N/m$
Surface area $A = 40 \,cm^2 = 40 \times 10^{-4} \,m^2$
The work done $W$ is given by the formula:
$W = T \times \Delta A_{total} = T \times 2 \times A$
Substituting the values:
$W = 0.03 \times 2 \times 40 \times 10^{-4}$
$W = 0.06 \times 40 \times 10^{-4}$
$W = 2.4 \times 10^{-4} \,J$

(D) The terminal velocity $v_T$ of a spherical drop is given by $v_T = \frac{2}{9} \frac{r^2 g (\rho - \sigma)}{\eta}$,where $r$ is the radius of the drop.
Thus,$v_T \propto r^2$.
Given the ratio of terminal velocities is $\frac{v_{T_1}}{v_{T_2}} = \frac{9}{4}$.
Since $\frac{v_{T_1}}{v_{T_2}} = \frac{r_1^2}{r_2^2}$,we have $\frac{r_1^2}{r_2^2} = \frac{9}{4}$.
Taking the square root on both sides,we get $\frac{r_1}{r_2} = \sqrt{\frac{9}{4}} = \frac{3}{2}$.
The volume $V$ of a spherical drop is $V = \frac{4}{3} \pi r^3$,so $V \propto r^3$.
The ratio of their volumes is $\frac{V_1}{V_2} = \frac{r_1^3}{r_2^3} = \left(\frac{r_1}{r_2}\right)^3 = \left(\frac{3}{2}\right)^3 = \frac{27}{8}$.

(B) The surface tension of the soap solution is given as $T = 0.03 \,N/m$.
The surface area of the soap bubble is $A = 40 \,cm^2 = 40 \times 10^{-4} \,m^2$.
$A$ soap bubble has two surfaces (inner and outer), so the change in surface area is $\Delta A = 2 \times A$.
The work done $W$ in blowing the soap bubble is given by the formula:
$W = T \times \Delta A_{total} = T \times 2 \times A$.
Substituting the values:
$W = 0.03 \times 2 \times 40 \times 10^{-4} \,J$.
$W = 0.06 \times 40 \times 10^{-4} \,J$.
$W = 2.4 \times 10^{-4} \,J$.

(C) The elongation due to a tensile force is given by the formula: $\frac{\Delta l}{l} = \frac{F}{A \cdot Y}$.
Given: $F = 33000 \,N$, $A = 10^{-3} \,m^2$, $Y = 3 \times 10^{11} \,N/m^2$.
Substituting the values: $\frac{\Delta l}{l} = \frac{33000}{10^{-3} \times 3 \times 10^{11}} = \frac{33000}{3 \times 10^8} = 11 \times 10^{-5}$.
The elongation due to thermal expansion is given by: $\frac{\Delta l}{l} = \alpha \Delta T$.
Given: $\alpha = 1.1 \times 10^{-5} /{ }^{\circ}C$.
Equating the two elongations: $11 \times 10^{-5} = 1.1 \times 10^{-5} \times \Delta T$.
Solving for $\Delta T$: $\Delta T = \frac{11 \times 10^{-5}}{1.1 \times 10^{-5}} = 10^{\circ}C$.

(A) The Young's modulus $Y$ is defined as $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta l/l}$.
Rearranging for the fractional change in length (strain), we get $\frac{\Delta l}{l} = \frac{F}{AY}$.
At the lowest position of the pendulum, the tension $T$ in the wire provides the centripetal force and balances the weight. However, assuming the load is released from a horizontal position, the tension at the lowest point is $T = 3mg$ (from energy conservation and circular motion dynamics).
Given $m = 1 \,kg$, $g = 10 \,m/s^2$, $A = 3 \,mm^2 = 3 \times 10^{-6} \,m^2$, and $Y = 10^{11} \,N/m^2$.
Using $F = T = 3mg = 3 \times 1 \times 10 = 30 \,N$.
$\frac{\Delta l}{l} = \frac{30}{3 \times 10^{-6} \times 10^{11}} = \frac{30}{3 \times 10^5} = 10 \times 10^{-5} = 10^{-4}$.
Wait, re-evaluating the standard interpretation of this specific textbook problem where the tension is often taken as $mg$ for static equilibrium or small oscillations: If $F = mg = 10 \,N$, then $\frac{\Delta l}{l} = \frac{10}{3 \times 10^5} = 0.33 \times 10^{-4} \approx 0.3 \times 10^{-4}$.
Thus, the correct option is $A$.

(B) Let the initial velocity be $u$. At maximum height $h$,the final velocity is $0$. Using $v = u - gt$,we get $0 = u - gt$,so $u = gt$.
The maximum height is $h = \frac{u^2}{2g} = \frac{(gt)^2}{2g} = \frac{gt^2}{2}$.
The body reaches the maximum height in time $t$.
While returning,the body falls from height $h$ to $h/2$. The distance covered is $h/2$.
Using $s = ut + \frac{1}{2}at^2$ for the downward motion starting from rest at the maximum height:
$\frac{h}{2} = 0 + \frac{1}{2}g(t')^2$,where $t'$ is the time taken to fall from the maximum height to half the maximum height.
Substituting $h = \frac{gt^2}{2}$,we get $\frac{gt^2}{4} = \frac{1}{2}g(t')^2$.
Solving for $t'$,we get $(t')^2 = \frac{t^2}{2}$,so $t' = \frac{t}{\sqrt{2}}$.
The total time from projection is $T = t + t' = t + \frac{t}{\sqrt{2}} = \left(1 + \frac{1}{\sqrt{2}}\right)t$.

(C) The velocity of a projectile is always tangent to its path. At the maximum height,the vertical component of velocity becomes zero,while the horizontal component remains constant. Therefore,the velocity vector is purely horizontal at the maximum height,making an angle of $0^{\circ}$ with the horizontal.

(D) Simple harmonic motion is a periodic motion.
By definition,the time period $T$ is the smallest interval of time after which the motion of the particle repeats itself.
This means that at any time $t = nT$ (where $n$ is an integer),the particle returns to its initial position.
Since the particle starts from its initial position,after $2T$ time,it will have completed two full cycles and returned to the starting point.
Therefore,the displacement from the initial position is $0$.

(D) The moment of inertia of a circular disc about an axis passing through its centre and perpendicular to its plane is given by $I = \frac{1}{2} M R^2$.
From this,we have $M R^2 = 2 I$.
The moment of inertia of the disc about its diameter is $I_d = \frac{1}{4} M R^2$.
Using the parallel axis theorem,the moment of inertia about an axis parallel to the diameter and touching the edge of the rim is $I' = I_d + M R^2$.
Substituting the values,$I' = \frac{1}{4} M R^2 + M R^2 = \frac{5}{4} M R^2$.
Since $M R^2 = 2 I$,we get $I' = \frac{5}{4} (2 I) = \frac{5}{2} I$.

(A) Let the thickness of slab $B$ be $d$ and its thermal conductivity be $K$. Then,for slab $A$,the thickness is $2d$ and thermal conductivity is $2K$.
Let the temperature at the contact surface be $T$.
Since the slabs are in series,the rate of heat flow $H$ through both slabs must be the same:
$H = \frac{K_A A (T_1 - T)}{d_A} = \frac{K_B A (T - T_2)}{d_B}$
Substituting the given values:
$\frac{2K \cdot A (100 - T)}{2d} = \frac{K \cdot A (T - 25)}{d}$
$(100 - T) = (T - 25)$
$2T = 125$
$T = 62.5^{\circ} C$

(A) For an adiabatic process,the equation of state is given by $p V^\gamma = \text{constant}$.
Using the ideal gas law $pV = nRT$,we can substitute $V = \frac{nRT}{p}$ into the adiabatic equation:
$p \left( \frac{nRT}{p} \right)^\gamma = \text{constant}$
$p \cdot p^{-\gamma} \cdot T^\gamma = \text{constant}$
$p^{1-\gamma} T^\gamma = \text{constant}$.
Thus,option $A$ is correct.

(D) For an adiabatic process,the relation between pressure $P$ and volume $V$ is given by $PV^{\gamma} = \text{constant}$.
As the volume $V$ decreases during compression,the pressure $P$ must increase to keep the product constant.
According to the ideal gas law $PV = nRT$,since $P$ increases and $V$ decreases,the temperature $T$ must increase because the work done on the gas increases its internal energy in an adiabatic process ($Q = 0$,$\Delta U = -W$).
Therefore,in adiabatic compression,both temperature and pressure increase.

(C) The ideal gas equation is given by $pV = nRT$.
For oxygen: $p_1 = 1 \,atm$,$V_1 = 1 \,L$. The number of moles is $n_{O_2} = \frac{p_1 V_1}{RT} = \frac{1 \times 1}{RT} = \frac{1}{RT}$.
For nitrogen: $p_2 = 0.5 \,atm$,$V_2 = 2 \,L$. The number of moles is $n_{N_2} = \frac{p_2 V_2}{RT} = \frac{0.5 \times 2}{RT} = \frac{1}{RT}$.
When these gases are introduced into a vessel of volume $V_{mix} = 1 \,L$ at the same temperature $T$,the total number of moles is $n_{mix} = n_{O_2} + n_{N_2} = \frac{1}{RT} + \frac{1}{RT} = \frac{2}{RT}$.
The final pressure $p_{mix}$ is given by $p_{mix} = \frac{n_{mix} RT}{V_{mix}} = \frac{(2/RT) \times RT}{1} = 2 \,atm$.

(A) Let the Planck's constant $h$ be expressed as $h = k G^x L^y E^z$,where $k$ is a dimensionless constant.
Dimensional formulas are:
$[h] = [M^1 L^2 T^{-1}]$
$[G] = [M^{-1} L^3 T^{-2}]$
$[L] = [M^1 L^2 T^{-1}]$
$[E] = [M^1 L^2 T^{-2}]$
Substituting these into the equation:
$[M^1 L^2 T^{-1}] = [M^{-1} L^3 T^{-2}]^x [M^1 L^2 T^{-1}]^y [M^1 L^2 T^{-2}]^z$
$[M^1 L^2 T^{-1}] = [M^{-x+y+z} L^{3x+2y+2z} T^{-2x-y-2z}]$
Comparing the powers of $M, L,$ and $T$ on both sides:
$(i)$ $-x + y + z = 1$
(ii) $3x + 2y + 2z = 2$
(iii) $-2x - y - 2z = -1$
Adding $(i)$ and (iii):
$(-x + y + z) + (-2x - y - 2z) = 1 - 1$
$-3x - z = 0 \implies z = -3x$
Substitute $z = -3x$ into $(i)$:
$-x + y - 3x = 1 \implies y - 4x = 1 \implies y = 1 + 4x$
Substitute $y$ and $z$ into (ii):
$3x + 2(1 + 4x) + 2(-3x) = 2$
$3x + 2 + 8x - 6x = 2$
$5x + 2 = 2 \implies 5x = 0 \implies x = 0$
Thus,the dimension of $G$ in the formula for $h$ is $0$.

(C) The speed of the car is $v_c = 72 \,km/h = 72 \times \frac{5}{18} = 20 \,m/s$.
In $t = 10 \,s$,the distance traveled by the car is $d_c = v_c \times t = 20 \times 10 = 200 \,m$.
Initially,the car is at $1800 \,m$ from the hill. After $10 \,s$,the car has moved $200 \,m$ closer to the hill,so its new distance from the hill is $1800 - 200 = 1600 \,m$.
The sound travels from the car to the hill $(1800 \,m)$ and then reflects back to the car at its new position $(1600 \,m)$.
Total distance traveled by sound $D = 1800 \,m + 1600 \,m = 3400 \,m$.
Since the time taken is $10 \,s$,the speed of sound $v_s = \frac{D}{t} = \frac{3400}{10} = 340 \,m/s$.

(C) The maximum velocity of a particle in a wave is given by $v_{\max} = A \omega$,where $A$ is the amplitude and $\omega$ is the angular frequency.
Given that the maximum particle velocity is equal to the wave velocity $(v)$,we have $v_{\max} = v$.
Substituting the expressions,we get $A \omega = v$.
Since $\omega = 2 \pi f$ and the wave velocity $v = f \lambda$,we can write $\omega = 2 \pi (v / \lambda)$.
Substituting this into the equation: $A \times (2 \pi v / \lambda) = v$.
Dividing both sides by $v$,we get $A \times (2 \pi / \lambda) = 1$.
Therefore,the amplitude $A = \frac{\lambda}{2 \pi}$.

(C) Given, velocity of the river, $v = 2 \,m/s$.
Density of water, $\rho = 1.2 \,g/cc$.
To find the mass of $1 \,m^3$ of water:
$\rho = 1.2 \,g/cm^3 = 1.2 \times \frac{10^{-3} \,kg}{(10^{-2} \,m)^3} = 1.2 \times 10^3 \,kg/m^3$.
Thus, the mass $m$ of $1 \,m^3$ of water is $1.2 \times 10^3 \,kg$.
The kinetic energy $K$ is given by the formula $K = \frac{1}{2}mv^2$.
Substituting the values:
$K = \frac{1}{2} \times (1.2 \times 10^3 \,kg) \times (2 \,m/s)^2$.
$K = 0.5 \times 1.2 \times 10^3 \times 4$.
$K = 2.4 \times 10^3 \,J = 2.4 \,kJ$.

(B) The line joining the two charges $+q$ and $-q$ forms an electric dipole.
Any point on the perpendicular bisector (equatorial line) of the dipole is equidistant from both charges.
Let the distance of a point $P$ on the perpendicular bisector from charge $+q$ be $r_1$ and from charge $-q$ be $r_2$. Since $P$ is on the perpendicular bisector, $r_1 = r_2 = r$.
The electric potential $V$ at point $P$ is given by $V = \frac{1}{4 \pi \varepsilon_0} (\frac{q}{r} + \frac{-q}{r}) = 0$.
However, the electric field $E$ at point $P$ is the vector sum of the fields due to $+q$ and $-q$. Since both fields are non-zero and directed such that they do not cancel out (the resultant field is parallel to the dipole axis), the electric field strength is non-zero.
Therefore, the electric potential is zero, but the electric field strength is non-zero.

(A) In a Huygen's eye-piece,the cross-wires are placed between the field lens and the eye lens.
Because of this,the image of the object is formed by both lenses (field lens and eye lens),resulting in magnification by both.
However,the cross-wires are placed such that they are only magnified by the eye lens.
Since the magnification is not uniform for the object and the cross-wires,the measurements taken using this eye-piece are not accurate.
Therefore,both Statement $(S)$ and Reason $(R)$ are true,and $(R)$ correctly explains $(S)$.

(A) For an ideal transformer, the relationship between the number of turns $(N)$ and the current $(I)$ in the primary $(P)$ and secondary $(S)$ coils is given by the inverse ratio:
$\frac{N_P}{N_S} = \frac{I_S}{I_P}$
Given:
$N_P = 50$
$N_S = 200$
$I_P = 4 \,A$
Substituting the values into the formula:
$\frac{50}{200} = \frac{I_S}{4}$
$\frac{1}{4} = \frac{I_S}{4}$
$I_S = 1 \,A$
Therefore, the current in the secondary coil is $1 \,A$.

(B) Let $E$ be the electromotive force $(EMF)$ of the battery and $r$ be its internal resistance.
The terminal voltage $V$ across a resistor $R$ is given by $V = I R$,where $I = \frac{E}{R+r}$.
Thus,$V = E \left( \frac{R}{R+r} \right)$,which can be rewritten as $E = V + I r = V + \left( \frac{V}{R} \right) r$.
For the first case: $R_1 = 16 \ \Omega$,$V_1 = 12 \ V$.
$E = 12 + \left( \frac{12}{16} \right) r = 12 + 0.75 r$ --- $(i)$
For the second case: $R_2 = 10 \ \Omega$,$V_2 = 11 \ V$.
$E = 11 + \left( \frac{11}{10} \right) r = 11 + 1.1 r$ --- (ii)
Equating $(i)$ and (ii):
$12 + 0.75 r = 11 + 1.1 r$
$12 - 11 = 1.1 r - 0.75 r$
$1 = 0.35 r$
$r = \frac{1}{0.35} = \frac{100}{35} = \frac{20}{7} \ \Omega$.

(C) The circuit consists of two parallel branches connected between points $P$ and $Q$.
Branch $PRQ$ has a total resistance $R_1 = 3 \, \Omega + 7 \, \Omega = 10 \, \Omega$.
Branch $PSQ$ has a total resistance $R_2 = 7 \, \Omega + 3 \, \Omega = 10 \, \Omega$.
Since the resistances of both branches are equal, the total current of $2 \, A$ splits equally into $1 \, A$ through each branch.
For branch $PRQ$, the current $I_1 = 1 \, A$. The potential drop across the $3 \, \Omega$ resistor is $V_P - V_R = I_1 \times 3 \, \Omega = 1 \, A \times 3 \, \Omega = 3 \, V$.
For branch $PSQ$, the current $I_2 = 1 \, A$. The potential drop across the $7 \, \Omega$ resistor is $V_P - V_S = I_2 \times 7 \, \Omega = 1 \, A \times 7 \, \Omega = 7 \, V$.
We want to find $V_R - V_S$. From the above equations:
$V_R = V_P - 3$
$V_S = V_P - 7$
Subtracting the two equations:
$V_R - V_S = (V_P - 3) - (V_P - 7) = -3 + 7 = +4 \, V$.

(A) The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$.
Given, $\lambda = 45 \times 10^{-2} \text{ Å} = 45 \times 10^{-12} \text{ m}$.
Substituting the values: $E = \frac{6.62 \times 10^{-34} \times 3 \times 10^8}{45 \times 10^{-12}} \text{ J}$.
$E = \frac{19.86 \times 10^{-26}}{45 \times 10^{-12}} \text{ J} = 0.4413 \times 10^{-14} \text{ J}$.
To convert energy into $eV$, divide by the charge of an electron $(1.6 \times 10^{-19} \text{ C})$:
$E_{eV} = \frac{0.4413 \times 10^{-14}}{1.6 \times 10^{-19}} \text{ eV} \approx 0.2758 \times 10^5 \text{ eV} = 27,580 \text{ eV}$.
Rounding to the nearest given option, the value is $27,500 \text{ eV}$.

(B) The Compton shift formula is given by $\Delta \lambda = \lambda^{\prime} - \lambda = \frac{h}{m_e c} (1 - \cos \phi)$.
Given $\lambda = 0.140 \,nm = 0.140 \times 10^{-9} \,m$ and $\phi = 90^{\circ}$.
Since $\cos 90^{\circ} = 0$,the shift is $\Delta \lambda = \frac{h}{m_e c}$.
The Compton wavelength is $\frac{h}{m_e c} \approx 2.426 \times 10^{-12} \,m = 0.002426 \,nm$.
Thus,$\lambda^{\prime} = \lambda + \Delta \lambda = 0.140 \,nm + 0.002426 \,nm = 0.142426 \,nm$.
Rounding to three decimal places,we get $\lambda^{\prime} \approx 0.142 \,nm$.

(B) Given total charge $Q = 1 \mu C = 10^{-6} \ C$.
The charge is divided in the ratio $2:3$.
Let the two charges be $q_1 = 2x$ and $q_2 = 3x$.
Since $q_1 + q_2 = 1 \mu C$,we have $5x = 1 \mu C$,so $x = 0.2 \mu C$.
Thus,$q_1 = 2 \times 0.2 \mu C = 0.4 \times 10^{-6} \ C$ and $q_2 = 3 \times 0.2 \mu C = 0.6 \times 10^{-6} \ C$.
The distance between them is $r = 1 \ m$.
The electrostatic force $F$ is given by Coulomb's Law:
$F = \frac{k q_1 q_2}{r^2} = \frac{9 \times 10^9 \times (0.4 \times 10^{-6}) \times (0.6 \times 10^{-6})}{1^2}$
$F = 9 \times 10^9 \times 0.24 \times 10^{-12} = 2.16 \times 10^{-3} \ N = 0.00216 \ N$.

(B) The perpendicular bisector of the line joining two equal and opposite charges ($+q$ and $-q$) is known as the equatorial line of the electric dipole.
For any point on the equatorial line at a distance $r$ from the center of the dipole, the electric potential $V$ is given by the sum of potentials due to individual charges:
$V = \frac{1}{4 \pi \varepsilon_0} \left( \frac{q}{r_1} + \frac{-q}{r_2} \right)$
Since any point on the perpendicular bisector is equidistant from both charges, $r_1 = r_2$, which implies $V = 0$.
However, the electric field strength $E$ at this point is not zero. It is directed parallel to the dipole axis (from $+q$ to $-q$) and is given by $E = \frac{1}{4 \pi \varepsilon_0} \frac{p}{r^3}$ (for $r \gg a$).

(B) The magnetic field at the centre of a circular coil is given by $B = \frac{\mu_0 N i}{2r}$.
Since the two coils carry currents in opposite directions,the net magnetic field at the centre will be the difference between the two fields.
Given: $N = 10$,$i_1 = 0.2 \ A$,$r_1 = 0.2 \ m$,$i_2 = 0.3 \ A$,$r_2 = 0.4 \ m$.
$B_{\text{net}} = |B_1 - B_2| = \frac{\mu_0 N}{2} |\frac{i_1}{r_1} - \frac{i_2}{r_2}|$
$B_{\text{net}} = \frac{10 \mu_0}{2} |\frac{0.2}{0.2} - \frac{0.3}{0.4}|$
$B_{\text{net}} = 5 \mu_0 |1 - 0.75|$
$B_{\text{net}} = 5 \mu_0 \times 0.25 = \frac{5}{4} \mu_0$.

(A) The initial magnetic moment of the wire is $M = m \cdot l$,where $m$ is the pole strength.
When the wire of length $l$ is bent into a semicircle,the length $l$ becomes the arc length of the semicircle.
Thus,$l = \pi r$,which implies $r = \frac{l}{\pi}$.
The new magnetic moment $M^{\prime}$ is the product of the pole strength $m$ and the straight-line distance between the poles (the diameter of the semicircle).
$M^{\prime} = m \cdot (2r) = m \cdot \left( \frac{2l}{\pi} \right)$.
Since $M = m \cdot l$,we can substitute $m \cdot l = M$ into the equation:
$M^{\prime} = \frac{2}{\pi} (m \cdot l) = \frac{2 M}{\pi}$.

(C) The time period of a vibration magnetometer is given by $T = 2 \pi \sqrt{\frac{I}{MB}}$, where $I$ is the moment of inertia and $M$ is the magnetic moment.
When the bar magnet is cut parallel to its length into four equal pieces, the mass of each piece becomes $m' = \frac{m}{4}$.
The new moment of inertia of one piece about the axis of rotation is $I' = \frac{I}{4}$.
The new magnetic moment of one piece is $M' = \frac{M}{4}$.
Substituting these into the formula for the new time period $T'$:
$T' = 2 \pi \sqrt{\frac{I'}{M' B}} = 2 \pi \sqrt{\frac{I/4}{(M/4) B}} = 2 \pi \sqrt{\frac{I}{MB}} = T$.
Therefore, the new time period $T' = 4 \,s$.

(C) The nuclear force is the strong force that acts between nucleons (protons and neutrons) within the nucleus.
Experimental evidence shows that the nuclear force is charge-independent.
This means the force between two protons,two neutrons,or a proton and a neutron is essentially the same,provided the distance between them and their spin states are the same.
Therefore,$F_{pp} = F_{nn} = F_{np}$.

(C) The image of an object in white light formed by a single lens is usually coloured and blurred because the focal length of the lens varies with the wavelength of light. This defect is known as chromatic aberration.
An achromatic combination of lenses is designed to minimize or eliminate this chromatic aberration.
By combining two lenses of different materials (typically a convex lens of crown glass and a concave lens of flint glass),the dispersion caused by one lens is cancelled by the other.
Therefore,the resulting image is free from chromatic aberration,meaning the image quality is not affected by the variation of the refractive index with the wavelength of light.

(A) Huygen's eye-piece consists of two plano-convex lenses separated by a distance.
When used for measurements,the cross-wires are placed between the two lenses.
The final image of the object is formed by the combined effect of both lenses,resulting in a specific magnification.
However,the cross-wires are only magnified by the eye-lens.
Because the object and the cross-wires are magnified by different numbers of lenses,they are not magnified proportionately.
This leads to errors in measurement,making the statement $(S)$ correct and the reason $(R)$ the correct explanation for it.

(A) For a plano-concave lens, one surface is plane $(R_1 = \infty)$ and the other is concave $(R_2 = 0.3 \,m)$. By convention, for a concave surface, $R_2 = -0.3 \,m$.
Using the lens maker's formula:
$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$
Given $\mu = 5/3$, $R_1 = \infty$, and $R_2 = -0.3 \,m$:
$\frac{1}{f} = \left( \frac{5}{3} - 1 \right) \left( \frac{1}{\infty} - \frac{1}{-0.3} \right)$
$\frac{1}{f} = \left( \frac{2}{3} \right) \left( 0 + \frac{1}{0.3} \right)$
$\frac{1}{f} = \frac{2}{3} \times \frac{10}{3} = \frac{20}{9}$
$f = \frac{9}{20} = 0.45 \,m$
Since it is a concave lens, the focal length must be negative by sign convention, so $f = -0.45 \,m$.

(D) In forward biasing,the positive terminal of the battery is connected to the $p$-side and the negative terminal to the $n$-side.
This configuration pushes the majority charge carriers (holes in the $p$-region and electrons in the $n$-region) towards the junction.
As a result,the width of the depletion region decreases,and the potential barrier is lowered.
Therefore,the statement that electrons and holes move away from the junction is incorrect.

(A) The thermoelectric power $S$ is defined as the rate of change of thermo-electromotive force with respect to temperature,given by $S = \frac{dE}{dT}$.
Given the expression $E = k(T-T_r)[T_0 - \frac{1}{2}(T+T_r)]$.
Expanding the expression: $E = k[T_0(T-T_r) - \frac{1}{2}(T^2 - T_r^2)]$.
Differentiating with respect to $T$: $\frac{dE}{dT} = k[T_0 - \frac{1}{2}(2T)] = k(T_0 - T)$.
Substituting $T = \frac{1}{2} T_0$ into the expression for $S$:
$S = k(T_0 - \frac{1}{2} T_0) = k(\frac{1}{2} T_0) = \frac{1}{2} k T_0$.

(C) Fraunhofer diffraction occurs when the source of light and the screen are effectively at an infinite distance from the obstacle or aperture.
In practical terms,this is achieved when the Fresnel distance $z_F = \frac{b^2}{\lambda}$ is much smaller than the distance $L$ between the obstacle and the screen.
Therefore,the condition is $L \gg \frac{b^2}{\lambda}$,which can be rearranged as $\frac{b^2}{L \lambda} \ll 1$.