TS EAMCET 2014 Physics Question Paper with Answer and Solution

(D) Let the mass of the shell be $2m$. At the highest point,the vertical component of velocity is zero,and the horizontal velocity is $u \cos \theta$.
Thus,the momentum of the shell just before the explosion is $P = (2m)(u \cos \theta)$.
After the explosion,the shell breaks into two equal parts of mass $m$ each.
One part retraces its path,meaning its velocity is $v_1 = -u \cos \theta$.
Let the velocity of the second part be $v_2$.
By the law of conservation of linear momentum:
$2m u \cos \theta = m v_1 + m v_2$
$2m u \cos \theta = m(-u \cos \theta) + m v_2$
$2m u \cos \theta = -m u \cos \theta + m v_2$
$m v_2 = 3m u \cos \theta \Rightarrow v_2 = 3u \cos \theta$.
The kinetic energy of the first part is $E_1 = \frac{1}{2} m v_1^2 = \frac{1}{2} m (-u \cos \theta)^2 = \frac{1}{2} m u^2 \cos^2 \theta$.
The kinetic energy of the second part is $E_2 = \frac{1}{2} m v_2^2 = \frac{1}{2} m (3u \cos \theta)^2 = \frac{1}{2} m (9 u^2 \cos^2 \theta) = 9 \left( \frac{1}{2} m u^2 \cos^2 \theta \right)$.
Therefore,$E_2 = 9 E_1$.

(C) The relative strengths of the four fundamental forces in nature,taking the strongest (strong nuclear force) as $1$,are as follows:
$(A)$ Strong nuclear force: Relative strength $= 1$ (matches $f$)
$(B)$ Weak nuclear force: Relative strength $\approx 10^{-13}$ (matches $h$)
$(C)$ Electromagnetic force: Relative strength $\approx 10^{-2}$ (matches $e$)
$(D)$ Gravitational force: Relative strength $\approx 10^{-39}$ (matches $i$)
Therefore,the correct matching is $A-f, B-h, C-e, D-i$.

(D) For an ideal gas,the internal energy $U$ is given by $U = \frac{n R T}{\gamma-1}$.
Since $n = 1$ mole,$U = \frac{R T}{\gamma-1}$.
From the ideal gas law,$p V = R T$,so $U = \frac{p V}{\gamma-1}$.
The change in internal energy $\Delta U$ is given by $\Delta U = U_f - U_i$.
Initial state: $p V_i = p V$,so $U_i = \frac{p V}{\gamma-1}$.
Final state: $p V_f = p(2V) = 2 p V$,so $U_f = \frac{2 p V}{\gamma-1}$.
Therefore,$\Delta U = \frac{2 p V}{\gamma-1} - \frac{p V}{\gamma-1} = \frac{p V}{\gamma-1}$.

(B) Let $T$ be the tension in the string and $F$ be the horizontal force applied to hold the mass in equilibrium.
At equilibrium,the forces acting on the mass are balanced:
$1$. Vertical direction: $T \cos \theta = M g$ (where $\theta = 60^{\circ}$)
$2$. Horizontal direction: $F = T \sin \theta$
Dividing the horizontal force equation by the vertical force equation:
$\frac{F}{M g} = \frac{T \sin \theta}{T \cos \theta} = \tan \theta$
$F = M g \tan \theta$
Given $\theta = 60^{\circ}$,we have:
$F = M g \tan 60^{\circ} = M g \sqrt{3}$

(A) The elastic potential energy per unit volume $(u)$ is given by the formula:
$u = \frac{1}{2} \times \text{stress} \times \text{strain} = \frac{1}{2} \times \frac{F}{A} \times \frac{F}{AY} = \frac{F^2}{2AY}$
Since the force $(F)$,Young's modulus $(Y)$,and length $(l)$ are the same for both wires,we have:
$u \propto \frac{1}{A} \propto \frac{1}{r^2} \propto \frac{1}{d^2}$
Wait,let us re-evaluate: $u = \frac{1}{2} \times \text{stress} \times \text{strain} = \frac{1}{2} \times \frac{F}{A} \times \frac{F}{AY} = \frac{F^2}{2A^2Y}$.
Since $A = \pi r^2 = \pi (d/2)^2$,we have $A \propto d^2$,so $A^2 \propto d^4$.
Thus,$u \propto \frac{1}{d^4}$.
Given the ratio of diameters $d_1 : d_2 = 1:2$,the ratio of energy per unit volume is:
$\frac{u_1}{u_2} = \left( \frac{d_2}{d_1} \right)^4 = \left( \frac{2}{1} \right)^4 = \frac{16}{1}$.
Therefore,the ratio is $16:1$.

(B) Let the mass $M$ be expressed as $M = C^a h^b G^c$.
The dimensions are:
$C = [LT^{-1}]$
$h = [ML^2T^{-1}]$
$G = [M^{-1}L^3T^{-2}]$
Substituting these into the equation:
$[M^1L^0T^0] = [LT^{-1}]^a [ML^2T^{-1}]^b [M^{-1}L^3T^{-2}]^c$
$[M^1L^0T^0] = M^{b-c} L^{a+2b+3c} T^{-a-b-2c}$
Comparing the powers of $M, L, T$ on both sides:
$b - c = 1$ $(i)$
$a + 2b + 3c = 0$ (ii)
$-a - b - 2c = 0$ (iii)
Adding (ii) and (iii): $b + c = 0$,so $b = -c$.
Substituting $b = -c$ into $(i)$: $-c - c = 1 \implies -2c = 1 \implies c = -1/2$.
Then $b = 1/2$.
Substituting $b = 1/2$ and $c = -1/2$ into (iii): $-a - 1/2 - 2(-1/2) = 0 \implies -a - 1/2 + 1 = 0 \implies a = 1/2$.
Thus,$M = C^{1/2} h^{1/2} G^{-1/2}$.

(C) The period of revolution $T$ of a satellite at a distance $r = R_p + h$ from the center of a planet of mass $M_p$ is given by $T = 2 \pi \sqrt{\frac{r^3}{G M_p}}$.
Since the satellite is revolving very close to the planet,$h \approx 0$,so $r \approx R_p$.
The mass of the planet in terms of its density $\rho$ and radius $R_p$ is $M_p = \frac{4}{3} \pi R_p^3 \rho$.
Substituting $M_p$ into the period formula:
$T = 2 \pi \sqrt{\frac{R_p^3}{G (\frac{4}{3} \pi R_p^3 \rho)}}$.
Simplifying the expression:
$T = 2 \pi \sqrt{\frac{3}{4 \pi G \rho}} = \sqrt{\frac{4 \pi^2 \cdot 3}{4 \pi G \rho}} = \sqrt{\frac{3 \pi}{G \rho}}$.

(C) For upward motion,the force required is $F_{\text{up}} = mg(\sin \theta + \mu \cos \theta)$.
For downward motion,the force required to prevent sliding is $F_{\text{down}} = mg(\sin \theta - \mu \cos \theta)$.
According to the problem,$F_{\text{up}} = 2 F_{\text{down}}$.
Substituting the expressions,we get $mg(\sin \theta + \mu \cos \theta) = 2mg(\sin \theta - \mu \cos \theta)$.
Dividing both sides by $mg$,we have $\sin \theta + \mu \cos \theta = 2 \sin \theta - 2 \mu \cos \theta$.
Rearranging the terms,$3 \mu \cos \theta = \sin \theta$,which simplifies to $\mu = \frac{1}{3} \tan \theta$.
Given $\theta = 60^{\circ}$,we have $\mu = \frac{1}{3} \tan 60^{\circ} = \frac{1}{3} \times \sqrt{3} = \frac{1}{\sqrt{3}}$.

(D) The energy of a drop is given by $U = T \times A$,where $T$ is surface tension and $A$ is surface area.
Energy of $n$ small drops: $U_i = n \times (4\pi r^2 T)$.
Energy of the big drop: $E = 4\pi R^2 T$.
Energy loss is given as $\Delta U = U_i - E = 3E$.
Substituting the values: $n(4\pi r^2 T) - 4\pi R^2 T = 3(4\pi R^2 T)$.
$n(4\pi r^2 T) = 4\pi R^2 T + 12\pi R^2 T$.
$n(4\pi r^2 T) = 16\pi R^2 T$.
$n = \frac{16\pi R^2 T}{4\pi r^2 T} = 4\frac{R^2}{r^2}$.

(B) Time taken to reach the market: $t_1 = \frac{\text{distance}}{\text{speed}} = \frac{2.5 \,km}{5 \,km/h} = 0.5 \,h = 30 \,min$.
Time taken to return to the house: $t_2 = \frac{\text{distance}}{\text{speed}} = \frac{2.5 \,km}{7.5 \,km/h} = \frac{1}{3} \,h = 20 \,min$.
Total time taken for the round trip is $30 \,min + 20 \,min = 50 \,min$.
Total distance covered = $2.5 \,km + 2.5 \,km = 5 \,km = 5000 \,m$.
Total time in seconds = $50 \,min \times 60 \,s/min = 3000 \,s$.
Average speed = $\frac{\text{Total distance}}{\text{Total time}} = \frac{5000 \,m}{3000 \,s} = \frac{5}{3} \,m/s$.

(A) According to the work-energy theorem,the work done by the retarding force is equal to the change in kinetic energy of the bus.
$W = \Delta K$
$F \cdot s = \frac{1}{2} m v^2$
Since the retarding force $F$ and the initial velocity $v$ are constant,we have:
$s = \frac{m v^2}{2 F} \implies s \propto m$
Let the initial mass be $m_1 = m$ and the initial stopping distance be $s_1 = x$.
The new mass after increasing by $25\%$ is $m_2 = m + 0.25m = 1.25m$.
Using the proportionality $s \propto m$:
$\frac{s_2}{s_1} = \frac{m_2}{m_1}$
$s_2 = s_1 \cdot \frac{1.25m}{m}$
$s_2 = 1.25 x$.

(D) The given equation of the trajectory is $y = ax - bx^2$.
The standard equation of a projectile trajectory is $y = x \tan \theta - \frac{gx^2}{2u^2 \cos^2 \theta}$.
Comparing the coefficients of $x$ and $x^2$ in both equations:
$\tan \theta = a \implies \theta = \tan^{-1}(a)$.
Also,$b = \frac{g}{2u^2 \cos^2 \theta}$.
The maximum height $H$ is given by $H = \frac{u^2 \sin^2 \theta}{2g}$.
From $\tan \theta = a$,we have $\sin \theta = \frac{a}{\sqrt{1+a^2}}$ and $\cos \theta = \frac{1}{\sqrt{1+a^2}}$.
Substituting $u^2 = \frac{g}{2b \cos^2 \theta}$ into the height formula:
$H = \frac{g}{2b \cos^2 \theta} \cdot \frac{\sin^2 \theta}{2g} = \frac{\tan^2 \theta}{4b} = \frac{a^2}{4b}$.
Thus,the maximum height is $\frac{a^2}{4b}$ and the angle of projection is $\tan^{-1}(a)$.

(C) The range of a projectile is given by $R = \frac{u^2 \sin 2\theta}{g}$.
For maximum range,the angle of projection must be $\theta = 45^{\circ}$.
Thus,$R_{\max} = \frac{u^2 \sin(90^{\circ})}{g} = \frac{u^2}{g} \quad \dots (i)$.
The time of flight $T$ is given by $T = \frac{2u \sin \theta}{g}$.
Substituting $\theta = 45^{\circ}$,we get $T = \frac{2u \sin 45^{\circ}}{g} = \frac{2u}{g \sqrt{2}} = \frac{u \sqrt{2}}{g}$.
From this,we find $u = \frac{Tg}{\sqrt{2}}$.
Substituting the value of $u$ into equation $(i)$:
$R_{\max} = \frac{1}{g} \left( \frac{Tg}{\sqrt{2}} \right)^2 = \frac{1}{g} \cdot \frac{T^2 g^2}{2} = \frac{g T^2}{2}$.

(D) Let the amplitude of the simple harmonic motion be $a$. The displacement of the body from the mean position is $x = \frac{a}{N}$.
The kinetic energy $(KE)$ of the body is given by:
$KE = \frac{1}{2} m \omega^2 (a^2 - x^2) = \frac{1}{2} m \omega^2 \left(a^2 - \frac{a^2}{N^2}\right) = \frac{1}{2} m \omega^2 a^2 \left(1 - \frac{1}{N^2}\right) = \frac{1}{2} m \omega^2 a^2 \left(\frac{N^2 - 1}{N^2}\right) \quad (i)$
The potential energy $(PE)$ of the body is given by:
$PE = \frac{1}{2} m \omega^2 x^2 = \frac{1}{2} m \omega^2 \left(\frac{a}{N}\right)^2 = \frac{1}{2} m \omega^2 \frac{a^2}{N^2} \quad (ii)$
Taking the ratio of $KE$ to $PE$:
$\frac{KE}{PE} = \frac{\frac{1}{2} m \omega^2 a^2 \left(\frac{N^2 - 1}{N^2}\right)}{\frac{1}{2} m \omega^2 \frac{a^2}{N^2}} = \frac{N^2 - 1}{1} = N^2 - 1$

(D) The length of the wire is $l$. Let $R$ be the radius of the circular loop. Then,$2 \pi R = l$,which gives $R = \frac{l}{2 \pi}$.
The mass of the wire is $m = \rho l$.
The moment of inertia of a circular loop about its diameter is $I_{diam} = \frac{1}{2} m R^2$.
According to the parallel axis theorem,the moment of inertia about a tangent $AB$ is $I = I_{cm} + m R^2$,where $I_{cm} = I_{diam} = \frac{1}{2} m R^2$.
Therefore,$I = \frac{1}{2} m R^2 + m R^2 = \frac{3}{2} m R^2$.
Substituting $m = \rho l$ and $R = \frac{l}{2 \pi}$:
$I = \frac{3}{2} (\rho l) \left( \frac{l}{2 \pi} \right)^2 = \frac{3}{2} \rho l \left( \frac{l^2}{4 \pi^2} \right) = \frac{3 \rho l^3}{8 \pi^2}$.

(B) According to the principle of calorimetry:
Heat lost by steam = Heat gained by water and calorimeter
Let $m$ be the mass of steam condensed in grams.
Heat lost by steam = $m \times L + m \times C_w \times (T_{steam} - T_{final})$
$= m \times 540 + m \times 1 \times (100 - 90) = 550m \ cal$
Heat gained by water and calorimeter = $(m_{water} + m_{eq}) \times C_w \times (T_{final} - T_{initial})$
$= (1000 \ g + 200 \ g) \times 1 \times (90 - 9) = 1200 \times 81 = 97200 \ cal$
Equating the two:
$550m = 97200$
$m = \frac{97200}{550} \approx 176.7 \ g$
Converting to $kg$,$m \approx 0.1767 \ kg$,which is approximately $0.18 \ kg$.

(D) According to Stefan-Boltzmann law,the power $P$ radiated by a black body of surface area $A$ at absolute temperature $T$ is given by $P = \sigma A T^4$.
Given values:
$\sigma = 5.67 \times 10^{-8} \ W m^{-2} K^{-4}$
$T = 727^{\circ} C = 727 + 273 = 1000 \ K$
$A = 200 \ mm^2 = 200 \times 10^{-6} \ m^2$
Substituting these values into the formula:
$P = (5.67 \times 10^{-8}) \times (200 \times 10^{-6}) \times (1000)^4$
$P = 5.67 \times 10^{-8} \times 200 \times 10^{-6} \times 10^{12}$
$P = 5.67 \times 2 \times 10^2 \times 10^{-14} \times 10^{12}$
$P = 11.34 \times 10^2 \times 10^{-2}$
$P = 11.34 \ J/s$.

(C) The change in internal energy $dU$ for an ideal gas is given by the formula:
$dU = n C_v dT$
Since $C_v = \frac{R}{\gamma - 1}$, the formula becomes:
$dU = n \frac{R}{\gamma - 1} (T_2 - T_1)$
Given values:
$n = 5 \, mol$
$T_1 = 273 \, K$ (at $STP$)
$T_2 = 673 \, K$
$R = 8.3 \, J/mol-K$
$\gamma = 1.4$
Substituting these values into the equation:
$dU = 5 \times \frac{8.3}{1.4 - 1} \times (673 - 273)$
$dU = 5 \times \frac{8.3}{0.4} \times 400$
$dU = 5 \times 8.3 \times 1000$
$dU = 41500 \, J$
Converting to kilo joules:
$dU = 41.50 \, kJ$

(B) Let mass $M$ be expressed as $M = C^a h^b G^c$.
The dimensions are:
$C = [LT^{-1}]$
$h = [ML^2T^{-1}]$
$G = [M^{-1}L^3T^{-2}]$
Substituting these into the equation:
$[M^1L^0T^0] = [LT^{-1}]^a [ML^2T^{-1}]^b [M^{-1}L^3T^{-2}]^c$
$[M^1L^0T^0] = [M^{b-c} L^{a+2b+3c} T^{-a-b-2c}]$
Comparing the powers of $M, L, T$ on both sides:
$b - c = 1$ $(i)$
$a + 2b + 3c = 0$ (ii)
$-a - b - 2c = 0$ (iii)
Adding (ii) and (iii): $b + c = 0$,so $b = -c$.
Substituting $b = -c$ into $(i)$: $-c - c = 1 \Rightarrow -2c = 1 \Rightarrow c = -1/2$.
Then $b = 1/2$.
Substituting $b = 1/2$ and $c = -1/2$ into (iii): $-a - 1/2 - 2(-1/2) = 0 \Rightarrow -a - 1/2 + 1 = 0 \Rightarrow a = 1/2$.
Thus,$M = C^{1/2} h^{1/2} G^{-1/2}$.

(C) Let the length of the pipe be $l$ and the speed of sound be $v$.
The fundamental frequency of a closed pipe is $f_c = \frac{v}{4l}$.
The $3^{rd}$ harmonic of the closed pipe is $f_{3,c} = 3 \times f_c = \frac{3v}{4l}$.
The fundamental frequency of the open pipe is $f_o = \frac{v}{2l}$.
According to the problem, the fundamental frequency of the open pipe is less than the $3^{rd}$ harmonic of the closed pipe by $55 \,Hz$:
$f_{3,c} - f_o = 55 \,Hz$
$\frac{3v}{4l} - \frac{v}{2l} = 55$
$\frac{3v - 2v}{4l} = 55$
$\frac{v}{4l} = 55 \,Hz$
Since the fundamental frequency of the closed pipe is $f_c = \frac{v}{4l}$, the value is $55 \,Hz$.

(B) Since the angular acceleration $\alpha$ is constant and the initial angular velocity $\omega_0 = 0$,the angle $\theta$ rotated in time $t$ is given by $\theta = \frac{1}{2} \alpha t^2$.
Given $\theta = 15^{\circ}$ for time $t$,we have $15^{\circ} = \frac{1}{2} \alpha t^2$ (Equation $1$).
Now,we need to find the angle rotated in the next $2t \ s$. The total time elapsed is $t + 2t = 3t \ s$.
The total angle $\theta_{total}$ rotated in time $3t$ is $\theta_{total} = \frac{1}{2} \alpha (3t)^2 = 9 \times (\frac{1}{2} \alpha t^2)$.
Substituting Equation $1$ into this,we get $\theta_{total} = 9 \times 15^{\circ} = 135^{\circ}$.
The increase in angle in the next $2t \ s$ is $\Delta \theta = \theta_{total} - \theta = 135^{\circ} - 15^{\circ} = 120^{\circ}$.

(D) Let the mass of the shell be $M$ and its velocity at the highest point be $v_h = u \cos \theta$. The momentum at the highest point is $P = M u \cos \theta$.
After breaking into two equal parts of mass $M/2$,one part retraces its path,meaning its velocity is $-u \cos \theta$. Let the velocity of the second part be $v_2$.
By conservation of linear momentum:
$M u \cos \theta = \frac{M}{2} (-u \cos \theta) + \frac{M}{2} v_2$
$u \cos \theta = -\frac{1}{2} u \cos \theta + \frac{1}{2} v_2$
$\frac{3}{2} u \cos \theta = \frac{1}{2} v_2 \implies v_2 = 3 u \cos \theta$.
The kinetic energy of the first part is $E_1 = \frac{1}{2} (M/2) (u \cos \theta)^2 = \frac{1}{4} M u^2 \cos^2 \theta$.
The kinetic energy of the second part is $E_2 = \frac{1}{2} (M/2) (3 u \cos \theta)^2 = \frac{1}{4} M (9 u^2 \cos^2 \theta) = \frac{9}{4} M u^2 \cos^2 \theta$.
Comparing the two,we get $E_2 = 9 E_1$.

(D) Given: Capacitance $C = 50 \mu F = 50 \times 10^{-6} \text{ F}$ and voltage $V = 220 \sin 50 t \text{ V}$.
Comparing this with the standard equation $V = V_0 \sin \omega t$,we get peak voltage $V_0 = 220 \text{ V}$ and angular frequency $\omega = 50 \text{ rad/s}$.
The capacitive reactance $X_C$ is given by $X_C = \frac{1}{\omega C} = \frac{1}{50 \times 50 \times 10^{-6}} = \frac{1}{2500 \times 10^{-6}} = \frac{10^6}{2500} = 400 \Omega$.
The peak current $i_0$ is $i_0 = \frac{V_0}{X_C} = \frac{220}{400} = 0.55 \text{ A}$.
The $rms$ current $i_{rms}$ is given by $i_{rms} = \frac{i_0}{\sqrt{2}} = \frac{0.55}{\sqrt{2}} \text{ A}$.

(B) In a Young's double slit experiment, the fringe width $\beta$ is given by the formula $\beta = \frac{\lambda D}{d}$, where $\lambda$ is the wavelength of light, $D$ is the distance between the slits and the screen, and $d$ is the distance between the slits.
From the formula, we can see that $\beta \propto \frac{1}{d}$ when $\lambda$ and $D$ are constant.
Given: Initial slit separation $d_1 = 2 \text{ mm}$, initial fringe width $\beta_1 = 1.2 \text{ mm}$.
New slit separation $d_2 = 1.5 \times d_1 = 1.5 \times 2 \text{ mm} = 3 \text{ mm}$.
Using the proportionality $\beta_1 d_1 = \beta_2 d_2$, we get:
$\beta_2 = \beta_1 \times \frac{d_1}{d_2} = 1.2 \text{ mm} \times \frac{2 \text{ mm}}{3 \text{ mm}} = 1.2 \times \frac{2}{3} \text{ mm} = 0.4 \times 2 \text{ mm} = 0.8 \text{ mm}$.
Therefore, the new fringe width is $0.8 \text{ mm}$.

(B) In Young's double-slit interference experiment, the fringe width $\beta$ is given by the formula:
$\beta = \frac{\lambda D}{d}$
where $\lambda$ is the wavelength, $D$ is the distance between the slits and the screen, and $d$ is the separation between the slits.
Given:
$\beta_1 = 1.2 \ \text{mm}$
$d_2 = 1.5 \times d_1$
Since $\beta \propto \frac{1}{d}$ (assuming $\lambda$ and $D$ remain constant),
$\frac{\beta_1}{\beta_2} = \frac{d_2}{d_1} = 1.5$
Substituting the values:
$\frac{1.2}{\beta_2} = 1.5$
$\beta_2 = \frac{1.2}{1.5} = 0.8 \ \text{mm}$
Thus, the new fringe width is $0.8 \ \text{mm}$.

(D) Given: Capacitance $C = 50 \mu F = 50 \times 10^{-6} \text{ F}$ and voltage $V = 220 \sin 50 t \text{ V}$.
Comparing with the standard equation $V = V_0 \sin \omega t$,we get peak voltage $V_0 = 220 \text{ V}$ and angular frequency $\omega = 50 \text{ rad/s}$.
The capacitive reactance $X_C$ is calculated as:
$X_C = \frac{1}{\omega C} = \frac{1}{50 \times 50 \times 10^{-6}} = \frac{1}{2500 \times 10^{-6}} = \frac{10^6}{2500} = 400 \Omega$.
The peak current $i_0$ is given by:
$i_0 = \frac{V_0}{X_C} = \frac{220}{400} = 0.55 \text{ A}$.
The rms current $i_{\text{rms}}$ is given by:
$i_{\text{rms}} = \frac{i_0}{\sqrt{2}} = \frac{0.55}{\sqrt{2}} \text{ A}$.

(A) The total energy of the $n$-th orbit is given by $E_n = -\frac{m e^4}{8 \varepsilon_0^2 h^2 n^2}$.
Since $E_n \propto m$,the ratio of the energy of the $\mu^{-}$-system to the hydrogen atom is $\frac{E_{\mu}}{E_e} = \frac{m_{\mu}}{m_e} = 207$.
Thus,$E_{\mu} = 207 \times E_e = 207 \times (-13.6 \text{ eV}) = -13.6 \times 207 \text{ eV}$.
The radius of the $n$-th orbit is given by $r_n = \frac{\varepsilon_0 h^2 n^2}{\pi m e^2}$.
Since $r_n \propto \frac{1}{m}$,the ratio of the radius of the $\mu^{-}$-system to the hydrogen atom is $\frac{r_{\mu}}{r_H} = \frac{m_e}{m_{\mu}} = \frac{1}{207}$.
Thus,$r_{\mu} = \frac{r_H}{207}$.

(B) The maximum line-of-sight distance $d$ for a transmitting antenna of height $h_T$ and a receiving antenna at ground level is given by the formula: $d = \sqrt{2 R_e h_T}$.
Given:
$R_e = 6.4 \times 10^6 \ m$
$h_T = 128 \ m$
Substituting the values:
$d = \sqrt{2 \times (6.4 \times 10^6) \times 128}$
$d = \sqrt{12.8 \times 10^6 \times 128}$
$d = \sqrt{128 \times 10^5 \times 128}$
$d = 128 \times \sqrt{10^5} \ m$
$d = 128 \times \sqrt{10 \times 10^4} \ m$
$d = 128 \times 100 \sqrt{10} \ m$
$d = 12800 \sqrt{10} \ m$
Converting to kilometers $(1 \ km = 1000 \ m)$:
$d = \frac{12800 \sqrt{10}}{1000} \ km = 12.8 \sqrt{10} \ km$
Wait,checking the calculation again:
$d = \sqrt{2 \times 6.4 \times 10^6 \times 128} = \sqrt{12.8 \times 10^6 \times 128} = \sqrt{1638.4 \times 10^6} = \sqrt{16.384 \times 10^8} = 40477 \ m \approx 40.5 \ km$.
Re-evaluating the provided options: $128/\sqrt{10} \approx 128/3.16 \approx 40.5 \ km$. Thus,option $B$ is correct.

(A) The heat produced per second is power, $P = I^2 R$. Given $P_2 = 50 \text{ J/s}$ for the $5 \Omega$ resistor.
$I_2^2 \times 5 = 50 \Rightarrow I_2^2 = 10 \Rightarrow I_2 = \sqrt{10} \text{ A}$.
The voltage across the parallel branches is $V = I_2 \times 5 = 5\sqrt{10} \text{ V}$.
The upper branch consists of $2 \Omega$ and $8 \Omega$ resistors in series, so the total resistance of the upper branch is $R_{upper} = 2 + 8 = 10 \Omega$.
The current in the upper branch is $I_1 = \frac{V}{R_{upper}} = \frac{5\sqrt{10}}{10} = \frac{\sqrt{10}}{2} \text{ A}$.
The heat generated per second in the $2 \Omega$ resistor is $P_1 = I_1^2 \times R_1$.
$P_1 = \left( \frac{\sqrt{10}}{2} \right)^2 \times 2 = \frac{10}{4} \times 2 = \frac{20}{4} = 5 \text{ J/s}$.

(A) Given: $\rho_B = 2 \rho_A$ and $R_A = R_B = R$.
Let the radius of wire $A$ be $r_A = r$. Since the diameter of $B$ is twice that of $A$,the radius of wire $B$ is $r_B = 2r$.
The resistance of a wire is given by $R = \rho \frac{l}{A} = \rho \frac{l}{\pi r^2}$.
Since $R_A = R_B$,we have:
$\rho_A \frac{l_A}{\pi r_A^2} = \rho_B \frac{l_B}{\pi r_B^2}$
Substituting the given values:
$\rho_A \frac{l_A}{\pi r^2} = (2 \rho_A) \frac{l_B}{\pi (2r)^2}$
$\rho_A \frac{l_A}{r^2} = 2 \rho_A \frac{l_B}{4r^2}$
$l_A = \frac{2}{4} l_B$
$l_A = \frac{1}{2} l_B$
Therefore,$\frac{l_B}{l_A} = 2$.

(B) Given that the energy of the photon $E$ is equal to the kinetic energy of the proton $K_p = E$.
For a proton,the de-Broglie wavelength $\lambda_1$ is given by $\lambda_1 = \frac{h}{p}$,where $p$ is the momentum of the proton.
Since $E = \frac{p^2}{2m}$,we have $p = \sqrt{2mE}$.
Thus,$\lambda_1 = \frac{h}{\sqrt{2mE}}$.
For a photon,the wavelength $\lambda_2$ is given by $\lambda_2 = \frac{hc}{E}$.
Now,calculating the ratio $\frac{\lambda_1}{\lambda_2}$:
$\frac{\lambda_1}{\lambda_2} = \frac{h / \sqrt{2mE}}{hc / E} = \frac{h}{\sqrt{2mE}} \cdot \frac{E}{hc} = \frac{E}{c \sqrt{2mE}} = \frac{\sqrt{E}}{c \sqrt{2m}}$.
Since $c$ and $m$ are constants,$\frac{\lambda_1}{\lambda_2} \propto \sqrt{E}$ or $E^{1/2}$.

(C) The given equation for the electric field is $E = i 30 \cos (k z - 5 \times 10^8 t)$.
Comparing this with the standard wave equation $E = E_0 \cos (k z - \omega t)$,we identify the angular frequency $\omega$ as $5 \times 10^8 \ rad/s$.
The relationship between the speed of light $c$,angular frequency $\omega$,and wave vector $k$ is given by $c = \frac{\omega}{k}$.
Rearranging to solve for $k$,we get $k = \frac{\omega}{c}$.
Substituting the given values: $k = \frac{5 \times 10^8 \ rad/s}{3 \times 10^8 \ m/s}$.
Thus,$k = \frac{5}{3} \approx 1.66 \ rad/m$.

(C) According to Coulomb's law,the force $F$ between two charges $q$ and $Q-q$ separated by a distance $r$ is given by:
$F = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q(Q-q)}{r^2}$
To find the value of $q$ for which the force is maximum,we differentiate $F$ with respect to $q$ and set it to zero:
$\frac{dF}{dq} = \frac{1}{4 \pi \varepsilon_0 r^2} \cdot \frac{d}{dq}(Qq - q^2) = 0$
$Q - 2q = 0$
$2q = Q$
$q = \frac{Q}{2}$
Thus,the force is maximum when the charge $Q$ is divided into two equal parts,$q = \frac{Q}{2}$.

(A) Let the surface charge density be $\sigma$. The total charge $Q$ is distributed on the two shells such that $\sigma_1 = \sigma_2 = \sigma$.
Total charge $Q = \sigma(4 \pi r^2) + \sigma(4 \pi R^2) = 4 \pi \sigma(r^2 + R^2)$.
Thus,$\sigma = \frac{Q}{4 \pi (r^2 + R^2)}$.
The electric potential at the centre of a spherical shell of radius $a$ with charge $q$ is $V = \frac{1}{4 \pi \varepsilon_0} \frac{q}{a}$.
For the two shells,the potential at the centre is the sum of the potentials due to each shell:
$V = \frac{1}{4 \pi \varepsilon_0} \left( \frac{q_1}{r} + \frac{q_2}{R} \right) = \frac{1}{4 \pi \varepsilon_0} \left( \frac{\sigma 4 \pi r^2}{r} + \frac{\sigma 4 \pi R^2}{R} \right)$.
$V = \frac{\sigma}{\varepsilon_0} (r + R)$.
Substituting the value of $\sigma$:
$V = \frac{Q}{4 \pi \varepsilon_0 (r^2 + R^2)} (r + R) = \frac{Q(R+r)}{4 \pi \varepsilon_0 (R^2+r^2)}$.

(D) The magnetic field at the centre of a circular coil with $n$ turns carrying current $i$ and radius $r$ is given by $B = \frac{\mu_0 n i}{2 r}$.
Let the length of the wire be $L$. For a single turn $(n_1 = 1)$,the circumference is $2 \pi r_1 = L$,so $r_1 = \frac{L}{2 \pi}$.
The magnetic field is $B_1 = B = \frac{\mu_0 (1) i}{2 r_1} = \frac{\mu_0 i}{2 (L / 2 \pi)} = \frac{\mu_0 i \pi}{L}$.
When the same wire is bent into $n$ turns $(n_2 = n)$,the new circumference of each turn is $2 \pi r_2 = \frac{L}{n}$,so $r_2 = \frac{L}{2 \pi n}$.
The new magnetic field $B_2$ is $B_2 = \frac{\mu_0 n i}{2 r_2} = \frac{\mu_0 n i}{2 (L / 2 \pi n)} = \frac{\mu_0 n i \pi n}{L} = n^2 \left( \frac{\mu_0 i \pi}{L} \right)$.
Substituting $B$ for the term in the bracket,we get $B_2 = n^2 B$.

(D) When a charged particle with charge $q$ and mass $m$ enters a uniform magnetic field $B$ with velocity $v$ perpendicular to the field,it experiences a magnetic Lorentz force $F = q(v \times B)$.
Since the force is always perpendicular to the velocity,it acts as a centripetal force,causing the particle to move in a circular path.
The radius $r$ of this circular path is given by the formula:
$r = \frac{mv}{qB}$
From this expression,it is clear that the radius $r$ is directly proportional to the velocity $v$ $(r \propto v)$.
Therefore,the particle travels in a circular path with a radius directly proportional to its velocity.

(D) Given:
Horizontal component of earth's magnetic field,$B_H = 0.8 \times 10^{-4} \,T$
Angle of dip,$\theta = 60^{\circ}$
Let the earth's total magnetic field be $B_e$.
We know that the horizontal component of the earth's magnetic field is related to the total magnetic field by the formula:
$B_H = B_e \cos \theta$
Substituting the given values:
$0.8 \times 10^{-4} = B_e \cos 60^{\circ}$
Since $\cos 60^{\circ} = 0.5$ or $\frac{1}{2}$,we have:
$0.8 \times 10^{-4} = B_e \times 0.5$
Solving for $B_e$:
$B_e = \frac{0.8 \times 10^{-4}}{0.5}$
$B_e = 1.6 \times 10^{-4} \,T$
Thus,the earth's overall magnetic field is $1.6 \times 10^{-4} \,T$.

(D) The radius of a nucleus is given by the formula $R = R_0 A^{1/3}$, where $R_0$ is a constant and $A$ is the mass number.
Given:
$R_1 = 1.5 \text{ fermi}$, $A_1 = 125$
$A_2 = 64$
We need to find $R_2$.
Taking the ratio of the radii:
$\frac{R_1}{R_2} = \left( \frac{A_1}{A_2} \right)^{1/3}$
$\frac{1.5}{R_2} = \left( \frac{125}{64} \right)^{1/3}$
$\frac{1.5}{R_2} = \frac{5}{4}$
$R_2 = \frac{1.5 \times 4}{5} = 0.3 \times 4 = 1.2 \text{ fermi}$.

(C) According to the first condition,for a converging lens,the lens formula is $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Given $f = 25 \,cm$ and $v = 75 \,cm$,we have:
$\frac{1}{25} = \frac{1}{75} - \frac{1}{u}$
$\frac{1}{u} = \frac{1}{75} - \frac{1}{25} = \frac{1-3}{75} = -\frac{2}{75}$
$u = -37.5 \,cm$.
So,the object is placed at a distance of $37.5 \,cm$ from the lens.
According to the second condition,the screen is moved closer to the lens by $25 \,cm$,so the new image distance is $v_1 = 75 \,cm - 25 \,cm = 50 \,cm$.
Using the lens formula again:
$\frac{1}{25} = \frac{1}{50} - \frac{1}{u_1}$
$\frac{1}{u_1} = \frac{1}{50} - \frac{1}{25} = \frac{1-2}{50} = -\frac{1}{50}$
$u_1 = -50 \,cm$.
So,the new object distance is $50 \,cm$.
The distance through which the object has to be shifted is $\Delta u = |u_1| - |u| = 50 \,cm - 37.5 \,cm = 12.5 \,cm$.

(C) For a biconvex lens with equal radii of curvature $R_1 = R$ and $R_2 = -R$,the Lens Maker's formula is given by:
$\frac{1}{f} = (\mu - 1) \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$
Substituting the values:
$\frac{1}{f} = (\mu - 1) \left[ \frac{1}{R} - \frac{1}{-R} \right] = (\mu - 1) \left[ \frac{2}{R} \right]$
Thus,$R = 2f(\mu - 1)$.
When the lens is cut vertically into two identical plano-convex lenses,each new lens has one surface with radius $R$ and the other surface as a plane (radius $\infty$).
Using the Lens Maker's formula for the new lens with focal length $f'$:
$\frac{1}{f'} = (\mu - 1) \left[ \frac{1}{R} - \frac{1}{\infty} \right]$
$\frac{1}{f'} = (\mu - 1) \left[ \frac{1}{R} \right]$
$f' = \frac{R}{\mu - 1}$
Substituting $R = 2f(\mu - 1)$:
$f' = \frac{2f(\mu - 1)}{\mu - 1} = 2f$.

(B) The output characteristics of a transistor are defined as the variation of the collector current $(I_C)$ with the collector-emitter voltage $(V_{CE})$ while keeping the base current $(I_B)$ constant.
This relationship is graphically represented by plotting $I_C$ on the y-axis and $V_{CE}$ on the x-axis for different fixed values of $I_B$.

(D) Given:
Intrinsic carrier concentration $n_i = 1.6 \times 10^{16} / m^3$
Donor concentration $N_D = 4.8 \times 10^{20} / m^3$
In an $n$-type semiconductor,the electron concentration $n_e \approx N_D = 4.8 \times 10^{20} / m^3$.
According to the law of mass action for semiconductors,$n_i^2 = n_e \times n_h$,where $n_h$ is the hole concentration.
Substituting the values:
$(1.6 \times 10^{16})^2 = (4.8 \times 10^{20}) \times n_h$
$2.56 \times 10^{32} = 4.8 \times 10^{20} \times n_h$
$n_h = \frac{2.56 \times 10^{32}}{4.8 \times 10^{20}}$
$n_h = 0.533 \times 10^{12} / m^3 = 5.33 \times 10^{11} / m^3$.
Thus,the concentration of holes is approximately $5.3 \times 10^{11} / m^3$.

(B) The formula for fringe width in Young's double slit experiment is given by $\beta = \frac{\lambda D}{d}$.
Given:
Initial fringe width $\beta_1 = 1.2 \text{ mm}$.
Initial slit separation $d_1 = 2 \text{ mm}$.
The slit separation is increased by one-and-half times the previous value,so $d_2 = 1.5 d_1$.
Since $\beta \propto \frac{1}{d}$,we have $\frac{\beta_1}{\beta_2} = \frac{d_2}{d_1}$.
Substituting the values:
$\frac{1.2}{\beta_2} = 1.5$.
$\beta_2 = \frac{1.2}{1.5} = 0.8 \text{ mm}$.