AP EAMCET 2016 Physics Question Paper with Answer and Solution

(D) Let $m$ be the mass of the ball, $h = 40 \,m$ be the height, and $g = 10 \,m/s^2$ be the acceleration due to gravity.
Total mechanical energy just before hitting the ground is $E_i = \frac{1}{2}mv^2 + mgh$.
After losing $\frac{1}{3}$ of its energy, the remaining energy is $E_f = \frac{2}{3}E_i$.
This remaining energy is used to rebound to the same height $h$, so the potential energy at the peak of the rebound is $mgh$.
Equating the energy: $\frac{2}{3}(\frac{1}{2}mv^2 + mgh) = mgh$.
Dividing by $m$: $\frac{2}{3}(\frac{1}{2}v^2 + gh) = gh$.
$\frac{1}{3}v^2 + \frac{2}{3}gh = gh$.
$\frac{1}{3}v^2 = gh - \frac{2}{3}gh = \frac{1}{3}gh$.
$v^2 = gh$.
$v = \sqrt{10 \times 40} = \sqrt{400} = 20 \,m/s$.

(C) The energy lost in an inelastic collision is given by the formula: $\Delta KE = \frac{m_1 m_2}{2(m_1 + m_2)} (1 - e^2) (u_1 + u_2)^2$.
Given: $m_1 = 1 \ kg$,$m_2 = 0.5 \ kg$,$u_1 = 6 \ ms^{-1}$,$u_2 = 9 \ ms^{-1}$ (since they move in opposite directions,the relative velocity is $u_1 + u_2$),and $e = \frac{1}{3}$.
Substituting the values:
$\Delta KE = \frac{1 \times 0.5}{2(1 + 0.5)} \left(1 - (\frac{1}{3})^2\right) (6 + 9)^2$
$\Delta KE = \frac{0.5}{3} \times (1 - \frac{1}{9}) \times (15)^2$
$\Delta KE = \frac{1}{6} \times \frac{8}{9} \times 225$
$\Delta KE = \frac{8}{54} \times 225 = \frac{4}{27} \times 225 = 33.33 \ J$.

(D) Let the mass of the upper prism be $m$ and the mass of the lower prism be $3m$.
Since the horizontal plane is smooth and there are no external horizontal forces acting on the system of the two prisms,the horizontal position of the centre of mass of the system remains unchanged.
Let the lower prism move a distance $k$ to the left. Then the upper prism must move a distance $(a-b-k)$ to the right relative to the ground to maintain the centre of mass position.
Applying the principle of conservation of the centre of mass:
$3m \cdot k = m \cdot (a - b - k)$
Dividing both sides by $m$:
$3k = a - b - k$
$4k = a - b$
$k = \frac{a - b}{4}$
Thus,the distance moved by the lower prism is $\frac{a - b}{4}$.

(B) Let the point where the gravitational field is zero be at a distance $x$ from the mass $m$. The gravitational field due to both masses must be equal in magnitude and opposite in direction.
$E_1 = E_2 \Rightarrow \frac{Gm}{x^2} = \frac{G(9m)}{(r-x)^2}$
Taking the square root on both sides: $\frac{1}{x} = \frac{3}{r-x}$
$r-x = 3x \Rightarrow 4x = r \Rightarrow x = \frac{r}{4}$.
The distance from mass $9m$ is $r-x = r - \frac{r}{4} = \frac{3r}{4}$.
The gravitational potential $V$ at this point is the sum of the potentials due to both masses:
$V = V_1 + V_2 = -\frac{Gm}{x} - \frac{G(9m)}{r-x}$
$V = -\frac{Gm}{r/4} - \frac{9Gm}{3r/4} = -\frac{4Gm}{r} - \frac{12Gm}{r}$
$V = -\frac{16Gm}{r}$.

(B) Let the particle be at an angle $\theta$ with the vertical. The forces acting on the particle are its weight $mg$ downwards,the normal force $N$ radially outwards,and the frictional force $f$ tangential to the surface.
For the particle to remain stationary,the component of weight along the tangent must be balanced by the limiting friction: $mg \sin \theta = f = \mu N$.
The component of weight perpendicular to the surface is balanced by the normal force: $N = mg \cos \theta$.
Substituting $N$ into the friction equation: $mg \sin \theta = \mu (mg \cos \theta) \implies \tan \theta = \mu = \frac{1}{\sqrt{3}}$.
Thus,$\theta = 30^\circ$.
The height $h$ of the particle from the bottom of the hemisphere is $h = R - R \cos \theta = R(1 - \cos 30^\circ)$.
$h = R(1 - \frac{\sqrt{3}}{2})$.

(A) Let the tension in the string be $T$ and the angle the string makes with the vertical be $\theta$.
For the particle moving in a horizontal circle,the forces acting on it are the tension $T$ and the weight $mg$.
The vertical component of tension balances the weight: $T \cos \theta = mg$.
The horizontal component of tension provides the necessary centripetal force: $T \sin \theta = \frac{mv^2}{r}$.
Dividing the two equations,we get: $\tan \theta = \frac{v^2}{rg} \Rightarrow v = \sqrt{rg \tan \theta}$.
From the geometry of the figure,$\sin \theta = \frac{r}{L}$,so $\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \frac{r^2}{L^2}} = \frac{\sqrt{L^2 - r^2}}{L}$.
Thus,$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{r/L}{\sqrt{L^2 - r^2}/L} = \frac{r}{\sqrt{L^2 - r^2}}$.
Substituting this into the expression for $v$:
$v = \sqrt{rg \cdot \frac{r}{\sqrt{L^2 - r^2}}} = r \sqrt{\frac{g}{\sqrt{L^2 - r^2}}}$.

(D) Let $R$ be the radius of the sphere and $r$ be the radius of the cavity.
Given: $R = 2 \ cm$,relative density of material $\rho_s = 8$,density of water $\rho_w = 1 \ g/cm^3$.
The sphere just sinks in water,which means its weight equals the buoyant force.
Weight of sphere = $\text{Volume of material} \times \rho_s \times g = \frac{4}{3} \pi (R^3 - r^3) \times 8 \times g$.
Buoyant force = $\text{Total volume of sphere} \times \rho_w \times g = \frac{4}{3} \pi R^3 \times 1 \times g$.
Equating the two: $\frac{4}{3} \pi (R^3 - r^3) \times 8 = \frac{4}{3} \pi R^3$.
$8(R^3 - r^3) = R^3$.
$8R^3 - 8r^3 = R^3 \Rightarrow 7R^3 = 8r^3$.
$r^3 = \frac{7}{8} R^3 = \frac{7}{8} \times (2)^3 = \frac{7}{8} \times 8 = 7 \ cm^3$.
Volume of cavity $V_c = \frac{4}{3} \pi r^3 = \frac{4}{3} \times \frac{22}{7} \times 7 = \frac{88}{3} \ cm^3$.

(D) According to Hooke's Law,the extension $\Delta l$ is proportional to the applied force $T$. Let $l_0$ be the natural length and $k$ be the force constant of the string.
The length $l$ under tension $T$ is given by $l = l_0 + \frac{T}{k}$.
For $T_1 = 80 \,N$,$l_1 = 101 \,mm = l_0 + \frac{80}{k}$ --- $(1)$
For $T_2 = 100 \,N$,$l_2 = 102 \,mm = l_0 + \frac{100}{k}$ --- $(2)$
Subtracting $(1)$ from $(2)$: $102 - 101 = \frac{100-80}{k} \implies 1 = \frac{20}{k} \implies k = 20 \,N/mm$.
Substituting $k$ into $(1)$: $101 = l_0 + \frac{80}{20} \implies 101 = l_0 + 4 \implies l_0 = 97 \,mm$.
Now,for $T_3 = 160 \,N$,the length $l_3 = l_0 + \frac{T_3}{k} = 97 + \frac{160}{20} = 97 + 8 = 105 \,mm$.
Converting to cm: $105 \,mm = 10.5 \,cm$.

(B) The initial distance between the trains is $d_0 = 300 \, m$.
The distance covered by a train is equal to the area under its velocity-time graph.
For Train $I$, the area under the $v-t$ graph is $A_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \, s \times 40 \, m/s = 200 \, m$.
For Train $II$, the magnitude of the area under the $v-t$ graph is $A_2 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8 \, s \times 20 \, m/s = 80 \, m$.
Since the trains are moving in opposite directions towards each other, the total distance covered by both trains before stopping is $d_{total} = A_1 + A_2 = 200 \, m + 80 \, m = 280 \, m$.
The final separation between the trains when both have stopped is $d_{final} = d_0 - d_{total} = 300 \, m - 280 \, m = 20 \, m$.

(D) Let the initial velocity be $u$. The total time of flight $T$ is the sum of the time to reach point $X$ $(t_1 = 3 \ s)$ and the time from $X$ to the ground $(t_2 = 6 \ s)$.
Total time $T = t_1 + t_2 = 3 + 6 = 9 \ s$.
The formula for total time of flight is $T = \frac{2u}{g}$.
Substituting the values,$9 = \frac{2u}{10}$,which gives $u = 45 \ m/s$.
The vertical distance $h$ of point $X$ from the ground can be calculated using the equation of motion $h = ut_1 - \frac{1}{2}gt_1^2$.
Substituting $u = 45 \ m/s$,$t_1 = 3 \ s$,and $g = 10 \ m/s^2$:
$h = (45 \times 3) - \frac{1}{2} \times 10 \times (3)^2$
$h = 135 - 5 \times 9$
$h = 135 - 45 = 90 \ m$.

(D) The radial (centripetal) acceleration is given by $a_r = \frac{v^2}{R}$.
Given $v = K\sqrt{s}$,we have $a_r = \frac{(K\sqrt{s})^2}{R} = \frac{K^2 s}{R}$.
The tangential acceleration is $a_t = \frac{dv}{dt} = \frac{dv}{ds} \cdot \frac{ds}{dt} = v \frac{dv}{ds}$.
Since $v = K s^{1/2}$,then $\frac{dv}{ds} = K \cdot \frac{1}{2} s^{-1/2} = \frac{K}{2\sqrt{s}}$.
Thus,$a_t = (K\sqrt{s}) \cdot \left( \frac{K}{2\sqrt{s}} \right) = \frac{K^2}{2}$.
The angle $\theta$ between the total acceleration and the tangential acceleration is given by $\tan \theta = \frac{a_r}{a_t}$.
Substituting the values,$\tan \theta = \frac{K^2 s / R}{K^2 / 2} = \frac{2s}{R}$.

(C) Given,amplitude $A = 2 \,m$.
Maximum acceleration $a_{max} = A \omega^2$ and maximum velocity $v_{max} = A \omega$.
According to the problem,$|A \omega^2| - |A \omega| = 4$.
Substituting $A = 2$:
$2 \omega^2 - 2 \omega = 4$
$\omega^2 - \omega - 2 = 0$
$(\omega - 2)(\omega + 1) = 0$.
Since $\omega > 0$,we get $\omega = 2 \,rad/s$.
Time period $T = \frac{2 \pi}{\omega} = \frac{2 \pi}{2} = \pi = \frac{22}{7} \,s$.
Velocity at displacement $y = 1 \,m$ is given by $v = \omega \sqrt{A^2 - y^2}$.
$v = 2 \sqrt{2^2 - 1^2} = 2 \sqrt{4 - 1} = 2 \sqrt{3} \,ms^{-1}$.

(A) Let the mass of the first part be $m$ and the mass of the second part be $(M - m)$.
Heat released by $m \ kg$ of water at $t^{\circ} C$ to become ice at $0^{\circ} C$ is:
$Q_1 = m \times c \times (t - 0) + m \times L_f = m \times 1 \times t + m \times 80 = m(t + 80)$.
Heat absorbed by $(M - m) \ kg$ of water at $t^{\circ} C$ to become steam at $100^{\circ} C$ is:
$Q_2 = (M - m) \times c \times (100 - t) + (M - m) \times L_v = (M - m) \times 1 \times (100 - t) + (M - m) \times 540 = (M - m)(640 - t)$.
Equating $Q_1 = Q_2$:
$m(t + 80) = (M - m)(640 - t)$
$mt + 80m = 640M - Mt - 640m + mt$
$80m + 640m = 640M - Mt$
$720m = M(640 - t)$
$\frac{m}{M} = \frac{640 - t}{720}$.

(A) The kinetic energy of the bullet is converted into heat. Since $\frac{1}{4}$ of the heat is absorbed by the obstacle,$\frac{3}{4}$ of the kinetic energy is used to heat and melt the bullet.
The energy balance equation is: $\frac{3}{4} (\frac{1}{2} m v^2) = m s \Delta \theta + m L$.
Given: $s = 0.03 \ cal \ g^{-1} {}^{\circ} C^{-1} = 0.03 \times 4200 \ J \ kg^{-1} K^{-1} = 126 \ J \ kg^{-1} K^{-1}$.
$\Delta \theta = 600 \ K - 300 \ K = 300 \ K$.
$L = 6 \ cal \ g^{-1} = 6 \times 4200 \ J \ kg^{-1} = 25200 \ J \ kg^{-1}$.
Substituting the values: $\frac{3}{8} v^2 = (126 \times 300) + 25200$.
$\frac{3}{8} v^2 = 37800 + 25200 = 63000$.
$v^2 = \frac{63000 \times 8}{3} = 21000 \times 8 = 168000$.
$v = \sqrt{168000} \approx 409.88 \ ms^{-1} \approx 410 \ ms^{-1}$.

(A) At $STP$, the molar volume of an ideal gas is $22.4 \text{ litres/mol}$.
Number of moles of helium gas, $n = \frac{67.2 \text{ L}}{22.4 \text{ L/mol}} = 3 \text{ mol}$.
Helium is a monoatomic gas, so its molar heat capacity at constant volume is $C_v = \frac{3}{2} R$.
The heat required to raise the temperature by $dT = 20^{\circ} C$ (or $20 \text{ K}$) at constant volume is given by:
$dQ = n C_v dT$
$dQ = 3 \times \left( \frac{3}{2} \times 8.314 \text{ J/mol K} \right) \times 20 \text{ K}$
$dQ = 3 \times 1.5 \times 8.314 \times 20$
$dQ = 9 \times 83.14 = 748.26 \text{ J} \approx 748 \text{ J}$.
Given the options, the closest value is $748 \text{ J}$ (Note: Using $R = 8.31 \text{ J/mol K}$ gives $747.9 \text{ J}$, and using $R = 8.3 \text{ J/mol K}$ gives $747 \text{ J}$. The provided option $784 \text{ J}$ appears to be a calculation approximation or typo in the source, but we select the closest logical result).

(D) Let the initial temperature of the source be $T_1$ and the sink be $T_2$. The initial efficiency is $\eta = 1 - \frac{T_2}{T_1}$.
When the source temperature is increased by $25 \%$,the new source temperature $T_1' = 1.25 T_1$.
The new efficiency is $\eta' = 1 - \frac{T_2}{1.25 T_1}$.
According to the problem,$\eta' = \eta + 0.80 \eta = 1.8 \eta$.
Substituting the expressions: $1 - \frac{T_2}{1.25 T_1} = 1.8 \left( 1 - \frac{T_2}{T_1} \right)$.
Assuming $T_1 = 100 \text{ K}$ for simplicity,$\eta = 1 - \frac{T_2}{100}$.
Then $1 - \frac{T_2}{125} = 1.8 \left( 1 - \frac{T_2}{100} \right)$.
$1 - \frac{T_2}{125} = 1.8 - 0.018 T_2$.
$0.018 T_2 - 0.008 T_2 = 1.8 - 1 \Rightarrow 0.01 T_2 = 0.8 \Rightarrow T_2 = 80 \text{ K}$.
The new efficiency is $\eta' = 1 - \frac{80}{125} = \frac{125 - 80}{125} = \frac{45}{125} = 0.36 = 36 \%$.

(A) For an isobaric process,the work done is $dW = P dV$ and the heat supplied is $dQ = n C_p dT$.
Using the relation $C_p = \frac{\gamma R}{\gamma - 1}$ and $dW = n R dT$,we have:
$\frac{dW}{dQ} = \frac{nR dT}{n C_p dT} = \frac{R}{C_p} = \frac{R}{\frac{\gamma R}{\gamma - 1}} = \frac{\gamma - 1}{\gamma}$.
Given $\gamma = 1.4$ and $dW = 300 \ J$,we substitute these values:
$\frac{300}{dQ} = \frac{1.4 - 1}{1.4} = \frac{0.4}{1.4} = \frac{4}{14} = \frac{2}{7}$.
Therefore,$dQ = 300 \times \frac{7}{2} = 150 \times 7 = 1050 \ J$.

(B) The Boltzmann constant $k_B$ relates energy to temperature: $E = k_B T$. Thus, $[k_B] = [\text{Energy}] / [\text{Temperature}] = [ML^2T^{-2}] / [K] = [ML^2T^{-2}K^{-1}]$. This matches $III$.
$(B)$ From Stokes' law, $F = 6\pi \eta r v$. The dimensions of viscosity $\eta$ are $[F] / ([L][LT^{-1}]) = [MLT^{-2}] / [L^2T^{-1}] = [ML^{-1}T^{-1}]$. This matches $II$.
$(C)$ Water equivalent is the mass of water that absorbs the same amount of heat as the body for the same temperature change. Its dimension is $[M]$. However, in the given options, $[ML^0T^0]$ is the closest representation of mass. This matches $I$.
$(D)$ From the heat conduction formula $Q/t = KA(\theta_1 - \theta_2)/l$, the coefficient of thermal conductivity $K = (Q \cdot l) / (A \cdot t \cdot \Delta\theta)$. Dimensions: $[ML^2T^{-2} \cdot L] / [L^2 \cdot T \cdot K] = [MLT^{-3}K^{-1}]$. This matches $IV$.
Therefore, the correct sequence is $A-III, B-II, C-I, D-IV$.

(C) The frequency of the tuning fork is $n = 170 \,Hz$. The speed of the student (source and observer) is $v_0 = 2 \,ms^{-1}$ towards the wall.
The wall acts as a stationary source reflecting the sound. The frequency of the echo heard by the student is given by the Doppler effect formula for a moving observer approaching a stationary source:
$n' = n \left( \frac{v + v_0}{v} \right)$
where $v = 340 \,ms^{-1}$ is the speed of sound.
$n' = 170 \left( \frac{340 + 2}{340} \right) = 170 \left( \frac{342}{340} \right) = \frac{342}{2} = 171 \,Hz$.
The beat frequency is the difference between the frequency of the echo and the original frequency of the tuning fork:
$f_{beat} = |n' - n| = |171 \,Hz - 170 \,Hz| = 1 \,Hz$.

(B) The difference between successive resonance frequencies is $\Delta f = 595 - 425 = 170 \,Hz$ and $765 - 595 = 170 \,Hz$.
Since the difference between successive frequencies is $2f_0$ (where $f_0$ is the fundamental frequency), the pipe must be a closed organ pipe.
Thus, $2f_0 = 170 \,Hz$, which gives $f_0 = 85 \,Hz$.
The fundamental frequency of a closed organ pipe is given by $f_0 = \frac{v}{4l}$.
Substituting the given values: $85 = \frac{340}{4l}$.
Solving for $l$: $l = \frac{340}{4 \times 85} = \frac{340}{340} = 1 \,m$.

(C) The natural frequency of an $L-C$ circuit is given by $f = \frac{1}{2 \pi \sqrt{LC}}$.
When the capacitor is filled with a dielectric material of constant $K$,the new capacitance becomes $C' = KC$.
The new frequency is $f' = \frac{1}{2 \pi \sqrt{L(KC)}} = \frac{1}{\sqrt{K}} f$.
Given $f = 125 \ kHz$ and the frequency decreases by $25 \ kHz$,the new frequency is $f' = 125 \ kHz - 25 \ kHz = 100 \ kHz$.
Substituting these values: $\frac{100}{125} = \frac{1}{\sqrt{K}}$.
$\frac{4}{5} = \frac{1}{\sqrt{K}} \Rightarrow \sqrt{K} = \frac{5}{4} = 1.25$.
Therefore,$K = (1.25)^2 = 1.5625 \approx 1.56$.

(C) In a hydrogen atom,the energy of an electron in the $n^{th}$ orbit is given by $E_n = -13.6/n^2 \ eV$. When an electron transitions from a higher energy level to a lower energy level,$n$ decreases.
As $n$ decreases,the radius of the orbit $r_n \propto n^2$ decreases.
The kinetic energy $K = |E_n| = 13.6/n^2 \ eV$ increases as $n$ decreases.
The velocity $v_n \propto 1/n$ increases as $n$ decreases.
The angular momentum $L = mvr = n(h/2\pi)$ is quantized and depends on the principal quantum number $n$. Since $n$ changes during the transition,the angular momentum does not remain constant.
The de-Broglie wavelength $\lambda = h/p = h/(mv)$. Since $v$ increases,$\lambda$ decreases.
Therefore,the statement that the angular momentum remains constant is incorrect.

(A) The coverage area $A$ of a transmitting antenna is given by the formula $A = \pi d^2$, where $d$ is the range of the antenna.
Given that $d = \sqrt{2Rh}$, where $R$ is the radius of the Earth and $h$ is the height of the antenna.
Substituting $d$ into the area formula: $A = \pi (\sqrt{2Rh})^2 = 2\pi Rh$.
Given values: $h = 105 \,m$ and $R = 6.4 \times 10^6 \,m$.
Substituting these values: $A = 2 \times 3.14 \times (6.4 \times 10^6 \,m) \times (105 \,m)$.
$A = 2 \times 3.14 \times 6.4 \times 105 \times 10^6 \,m^2$.
$A = 4220.16 \times 10^6 \,m^2$.
Since $1 \,km^2 = 10^6 \,m^2$, we have $A = 4220.16 \,km^2$.
Rounding to the nearest given option, $A \approx 4224 \,km^2$.

(A) The drift velocity $v_d$ is given by the formula $v_d = \frac{I}{neA}$.
Since the current $I = \frac{V}{R}$ and the resistance $R = \rho \frac{l}{A}$, we can substitute these into the expression:
$v_d = \frac{V/R}{neA} = \frac{V}{(\rho l/A)neA} = \frac{V}{\rho l ne}$.
Here, $V$ is the potential difference of the cell, $\rho$ is the resistivity of the material, $l$ is the length of the rod, $n$ is the free electron density, and $e$ is the charge of an electron.
As seen from the final expression, $v_d$ is independent of the cross-sectional area $A$.
Therefore, when a hole is made along the axis, the cross-sectional area changes, but the drift velocity remains unchanged.

(C) Let $R_L$ be the resistance in the left gap and $R_R$ be the resistance of the nichrome wire in the right gap. The balancing condition is $\frac{R_L}{R_R} = \frac{l}{100-l}$.
Given $l = 60 \ cm$,so $\frac{R_L}{R_R} = \frac{60}{40} = 1.5$.
When the wire is stretched by $20 \%$,its new length $l' = 1.2l_0$. Since volume $V = A \cdot l$ is constant,the new area $A' = \frac{A}{1.2}$.
The new resistance $R_R' = \rho \frac{l'}{A'} = \rho \frac{1.2l_0}{A/1.2} = (1.2)^2 R_R = 1.44 R_R$.
Let the new balancing length be $l_{new}$. Then $\frac{R_L}{R_R'} = \frac{l_{new}}{100-l_{new}}$.
Substituting $R_L = 1.5 R_R$ and $R_R' = 1.44 R_R$:
$\frac{1.5 R_R}{1.44 R_R} = \frac{l_{new}}{100-l_{new}} \Rightarrow \frac{1.5}{1.44} = \frac{l_{new}}{100-l_{new}}$.
$1.04167 = \frac{l_{new}}{100-l_{new}} \Rightarrow 104.167 - 1.04167 l_{new} = l_{new}$.
$2.04167 l_{new} = 104.167 \Rightarrow l_{new} \approx 51 \ cm$.

(B) The de-Broglie wavelength $\lambda_e$ of an electron is given by $\lambda_e = \frac{h}{p}$,where $p$ is the momentum of the electron.
Since the kinetic energy $E_k$ of the emitted electron is related to momentum by $E_k = \frac{p^2}{2m}$,we have $p = \sqrt{2m E_k}$.
Thus,$\lambda_e = \frac{h}{\sqrt{2m E_k}}$.
Given that the work function is neglected,the entire energy of the incident photon is converted into the kinetic energy of the electron: $E_k = E_{photon} = \frac{hc}{\lambda}$.
Substituting $E_k$ into the de-Broglie wavelength formula:
$\lambda_e = \frac{h}{\sqrt{2m \left(\frac{hc}{\lambda}\right)}}$.
Simplifying the expression:
$\lambda_e = \sqrt{\frac{h^2}{2m \frac{hc}{\lambda}}} = \sqrt{\frac{h^2 \lambda}{2mhc}} = \sqrt{\frac{h \lambda}{2mc}}$.

(A) Given: Current $I = 2 \ A$, Resistance $R = 7 \ \Omega$, $EMF$ $E = 4 \ V$, Inductance $L = 9 \ mH = 9 \times 10^{-3} \ H$, Rate of change of current $\frac{dI}{dt} = -10^3 \ A \ s^{-1}$ (since current is decreasing).
Applying Kirchhoff's voltage law from $A$ to $B$:
$V_A - I R - E - L \frac{dI}{dt} = V_B$
$V_A - V_B = I R + E + L \frac{dI}{dt}$
Substituting the values:
$V_{AB} = (2 \ A)(7 \ \Omega) + 4 \ V + (9 \times 10^{-3} \ H)(-10^3 \ A \ s^{-1})$
$V_{AB} = 14 \ V + 4 \ V - 9 \ V$
$V_{AB} = 18 \ V - 9 \ V = 9 \ V$
Wait, let's re-evaluate the polarity of the inductor. The induced $EMF$ in the inductor is $\varepsilon = -L \frac{dI}{dt}$. Since current is decreasing, $\frac{dI}{dt} = -10^3 \ A \ s^{-1}$, so $\varepsilon = - (9 \times 10^{-3})(-10^3) = +9 \ V$. The potential drop across the inductor is $V_L = L \frac{dI}{dt} = -9 \ V$.
Path $A \to B$: $V_A - I R + E - V_L = V_B$ (considering the battery polarity in the image: $A \to$ resistor $ o$ negative terminal $ o$ positive terminal $ o$ inductor $ o B$).
$V_A - (2)(7) + 4 - (9 \times 10^{-3})(-10^3) = V_B$
$V_A - 14 + 4 + 9 = V_B$
$V_A - 1 = V_B \Rightarrow V_A - V_B = 1 \ V$.

(B) The electromagnetic spectrum in increasing order of wavelength is: Gamma rays < $X$-rays < Ultraviolet < Visible < Infrared < Microwaves < Radio waves.
$1$. Radioactive source produces Gamma rays $(\lambda \approx 10^{-12} \text{ m})$.
$2$. $X$-ray tube produces $X$-rays $(\lambda \approx 10^{-10} \text{ m})$.
$3$. Sodium lamp produces Visible light $(\lambda \approx 10^{-7} \text{ m})$.
$4$. Magnetron valve produces Microwaves $(\lambda \approx 10^{-3} \text{ m})$.
Therefore, the correct sequence in increasing order of wavelength is: Radioactive source $\rightarrow$ $X$-ray tube $\rightarrow$ Sodium lamp $\rightarrow$ Magnetron valve.

(C) From the figure,the net force $F_{\text{net}}$ due to $F_1$ and $F_2$ makes an angle $\theta$ with force $F_2$.
Since $F_1$ and $F_2$ are perpendicular to each other,the angle $\theta$ is given by $\tan \theta = \frac{F_1}{F_2}$.
Using Coulomb's Law,the force $F_1$ on charge at $B$ due to charge at $A$ is $F_1 = k \cdot \frac{q_A q_B}{(AB)^2} = k \cdot \frac{q^2}{(3)^2}$.
The force $F_2$ on charge at $B$ due to charge at $C$ is $F_2 = k \cdot \frac{q_C q_B}{(BC)^2} = k \cdot \frac{q^2}{(4)^2}$.
Therefore,$\tan \theta = \frac{F_1}{F_2} = \frac{k \cdot q^2 / 9}{k \cdot q^2 / 16} = \frac{16}{9}$.
Thus,$\theta = \tan ^{-1}\left(\frac{16}{9}\right)$.

(C) Let the two charges of $+10 \mu C$ be at $A(0, a)$ and $C(0, -a)$. The charge $-20 \mu C$ is displaced to $B(x, 0)$.
The distance between the charge at $A$ and the charge at $B$ is $r = \sqrt{a^2 + x^2}$.
The electrostatic force $F$ exerted by each charge on the charge at $B$ is given by Coulomb's law:
$F = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2} = (9 \times 10^9) \frac{(10 \times 10^{-6})(20 \times 10^{-6})}{a^2 + x^2} = \frac{1.8}{a^2 + x^2} \text{ N}$.
The force $F$ is attractive, directed towards $A$ and $C$. The vertical components of these forces cancel out, while the horizontal components add up.
The net force $F_{\text{net}}$ along the $X$-axis is:
$F_{\text{net}} = 2F \cos \theta$, where $\cos \theta = \frac{x}{r} = \frac{x}{\sqrt{a^2 + x^2}}$.
Substituting the values:
$F_{\text{net}} = 2 \left( \frac{1.8}{a^2 + x^2} \right) \left( \frac{x}{\sqrt{a^2 + x^2}} \right) = \frac{3.6 x}{(a^2 + x^2)^{3/2}}$.
Since $x \ll a$, we can neglect $x^2$ in the denominator:
$F_{\text{net}} \approx \frac{3.6 x}{(a^2)^{3/2}} = \frac{3.6 x}{a^3} \text{ N}$.

(D) The work done by an electric field is independent of the path taken because the electric field is a conservative field. The work done $W$ is given by $W = q \Delta V = -q \int_{A}^{B} \vec{E} \cdot d\vec{r} = -q \vec{E} \cdot \vec{d}$.
First,calculate the displacement vector $\vec{d} = \vec{r}_B - \vec{r}_A = (2-1)\hat{i} + (1-2)\hat{j} + (3-0)\hat{k} = (1\hat{i} - 1\hat{j} + 3\hat{k}) \ m$.
The electric field is $\vec{E} = (10\hat{i} + 30\hat{j}) \ Vm^{-1}$.
The potential difference $\Delta V = V_B - V_A = -\vec{E} \cdot \vec{d} = -[(10)(1) + (30)(-1) + (0)(3)] = -[10 - 30] = 20 \ V$.
The work done is $W = q \Delta V = (0.8 \ C)(20 \ V) = 16 \ J$.

(D) The electric potential $V$ at a distance $r$ from a point charge $q$ is given by $V = \frac{kq}{r}$,where $k = 9 \times 10^9 \text{ Nm}^2\text{C}^{-2}$.
From the figure,for the equipotential surface with potential $V = 60 \text{ V}$,the radius is $r = 10 \text{ cm} = 0.1 \text{ m}$.
Substituting these values into the potential formula:
$60 = \frac{kq}{0.1}$
$kq = 60 \times 0.1 = 6 \text{ Vm}$.
The magnitude of the electric field $E$ at a distance $r$ from a point charge is given by $E = \frac{kq}{r^2}$.
Substituting the value of $kq = 6$ into the electric field formula,we get:
$E = \frac{6}{r^2} \text{ Vm}^{-1}$.

(C) Let the tension developed in the wire be $T$ and the radius of the circular loop be $r$.
Consider a small element of the wire of length $dl$ subtending an angle $d\theta$ at the center.
The magnetic force acting on this element is $dF = B I dl = B I (r d\theta)$.
This force is directed radially outward.
The tension $T$ at the ends of this element provides a restoring force directed radially inward.
The component of tension providing the inward force is $2 T \sin(d\theta / 2) \approx 2 T (d\theta / 2) = T d\theta$.
Equating the forces for equilibrium: $T d\theta = B I r d\theta$, which gives $T = B I r$.
Since the total length of the wire is $l = 2 \pi r$, we have $r = l / (2 \pi)$.
Substituting $r$ in the tension formula: $T = B I (l / 2 \pi)$.
Given $B = 2.0 \,T$, $I = 1.1 \,A$, and $l = 1 \,m$:
$T = (2.0 \times 1.1 \times 1) / (2 \times 3.14) = 2.2 / 6.28 \approx 0.35 \,N$.

(D) The potential energy of a magnetic dipole in a magnetic field is given by $U = -MB \cos \theta$.
Initially, the dipole is in the East-West direction $(\theta_1 = 90^{\circ})$, so $U_i = -MB \cos 90^{\circ} = 0$.
Finally, the dipole is in the North-South position $(\theta_2 = 0^{\circ})$, so $U_f = -MB \cos 0^{\circ} = -MB$.
By the law of conservation of energy, the loss in potential energy is equal to the gain in kinetic energy:
$KE = U_i - U_f = 0 - (-MB) = MB$.
Given $M = 2.5 \text{ A m}^2$ and $B_H = 3 \times 10^{-5} \text{ T}$.
$KE = 2.5 \times 3 \times 10^{-5} = 7.5 \times 10^{-5} \text{ J}$.
Converting to microjoules: $7.5 \times 10^{-5} \text{ J} = 75 \times 10^{-6} \text{ J} = 75 \mu\text{J}$.

(A) The magnetic dipole moment $M$ is given by $M = i A$,where $i$ is the current and $A$ is the area of the ring.
Since the charge $q$ is rotating with angular velocity $\omega$,the time period of one revolution is $T = \frac{2 \pi}{\omega}$.
The equivalent current $i$ is given by $i = \frac{q}{T} = \frac{q \omega}{2 \pi}$.
The area of the ring is $A = \pi R^2$.
Substituting these values into the formula for $M$:
$M = \left( \frac{q \omega}{2 \pi} \right) (\pi R^2) = \frac{1}{2} q \omega R^2$.

(A) The radius of a nucleus is given by the relation $R = R_0 A^{1/3}$,where $A$ is the number of nucleons.
Therefore,the ratio of the radii of two nuclei is given by $\frac{R_1}{R_2} = \left(\frac{A_1}{A_2}\right)^{1/3}$.
Given that $R_{Ge} = 2 R_{Be}$ and $A_{Be} = 9$.
Substituting these values into the formula:
$2 = \left(\frac{A_{Ge}}{9}\right)^{1/3}$.
Cubing both sides,we get $2^3 = \frac{A_{Ge}}{9}$.
$8 = \frac{A_{Ge}}{9}$.
$A_{Ge} = 8 \times 9 = 72$.
Thus,the number of nucleons in germanium is $72$.

(D) For a concave mirror,the mirror formula is $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
Using sign convention,$u$ is replaced by $-u$ and $f$ by $-f$.
So,$\frac{1}{v} - \frac{1}{u} = -\frac{1}{f}$.
Rearranging for $v$,we get $\frac{1}{v} = \frac{1}{u} - \frac{1}{f} = \frac{f-u}{uf}$.
Thus,$v = \frac{uf}{f-u}$.
The magnitude of the image position is $|v| = \left| \frac{uf}{f-u} \right| = \frac{uf}{u-f}$.
The rod extends from $u$ to $\infty$. The image of the nearer end is at $v_1 = \frac{uf}{u-f}$.
The image of the farther end (at $\infty$) is at the focus $v_2 = f$.
The length of the image is $L = |v_1 - v_2| = \left| \frac{uf}{u-f} - f \right|$.
$L = \left| \frac{uf - f(u-f)}{u-f} \right| = \left| \frac{uf - uf + f^2}{u-f} \right| = \frac{f^2}{u-f}$.

(B) In a common-emitter configuration, the current gain $\beta$ is defined as $\beta = \frac{I_C}{I_B}$.
Given $\beta = 60$, we have $I_C = 60 I_B$.
We know that the emitter current $I_E$ is the sum of the base current $I_B$ and the collector current $I_C$:
$I_E = I_B + I_C$
Substituting $I_C = 60 I_B$ into the equation:
$I_E = I_B + 60 I_B = 61 I_B$
Given $I_E = 6.6 \,mA$, we can solve for $I_B$:
$6.6 \,mA = 61 I_B$
$I_B = \frac{6.6}{61} \,mA \approx 0.108 \,mA$.

(B) The position of the $n$th bright fringe from the central maximum is given by $y_n = n \frac{\lambda D}{d}$.
For red light,the position of the $n$th bright fringe is $y_P = n \frac{\lambda_R D}{d}$.
For green light,the position of the $(n+1)$th bright fringe is $y_P = (n+1) \frac{\lambda_G D}{d}$.
Since the point $P$ is the same in both cases,we equate the two expressions:
$n \lambda_R \frac{D}{d} = (n+1) \lambda_G \frac{D}{d}$.
Canceling the common terms $\frac{D}{d}$,we get:
$n \lambda_R = (n+1) \lambda_G$.
Substituting the given wavelengths $\lambda_R = 6000 \ Å$ and $\lambda_G = 5000 \ Å$:
$n(6000) = (n+1)(5000)$.
Dividing both sides by $1000$:
$6n = 5(n+1)$.
$6n = 5n + 5$.
$n = 5$.