AP EAMCET 2018 Physics Question Paper with Answer and Solution

(B) The dimensional formula for force is $[M L T^{-2}]$.
Given,$N_1 = 10$,$M_1 = 1 \,kg$,$L_1 = 1 \,m$,$T_1 = 1 \,s$.
In the new system,$M_2 = A \,kg$,$L_2 = B \,m$,$T_2 = C \,s$.
Using the conversion formula $N_2 = N_1 \left( \frac{M_1}{M_2} \right)^1 \left( \frac{L_1}{L_2} \right)^1 \left( \frac{T_1}{T_2} \right)^{-2}$.
Substituting the values:
$N_2 = 10 \left( \frac{1}{A} \right)^1 \left( \frac{1}{B} \right)^1 \left( \frac{1}{C} \right)^{-2}$.
$N_2 = 10 \cdot A^{-1} \cdot B^{-1} \cdot C^2$.
Thus,the value of $10 \,N$ in the new system is $10 A^{-1} B^{-1} C^2$.

(B) The given expression is $10 \ g \ cm \ s^{-1} = x \ N \ s$.
First,convert the $CGS$ unit $g \ cm \ s^{-1}$ to $SI$ units $(kg \ m \ s^{-1})$:
$1 \ g = 10^{-3} \ kg$
$1 \ cm = 10^{-2} \ m$
So,$1 \ g \ cm \ s^{-1} = 10^{-3} \ kg \times 10^{-2} \ m \times s^{-1} = 10^{-5} \ kg \ m \ s^{-1}$.
Thus,$10 \ g \ cm \ s^{-1} = 10 \times 10^{-5} \ kg \ m \ s^{-1} = 10^{-4} \ kg \ m \ s^{-1}$.
We know that $1 \ N = 1 \ kg \ m \ s^{-2}$,so $1 \ N \ s = 1 \ kg \ m \ s^{-1}$.
Equating the two,we get $x \ N \ s = 10^{-4} \ kg \ m \ s^{-1}$.
Therefore,$x = 10^{-4}$.

(C) The speed of a transverse pulse on a string is given by $v = \sqrt{\frac{T}{\mu}}$, where $T$ is the tension and $\mu$ is the linear mass density of the string.
When the car is at rest, the tension in the string is $T_1 = Mg$. Thus, $v_1 = \sqrt{\frac{Mg}{\mu}} = 60 \text{ cm/s}$.
When the car accelerates with acceleration $a$ on a horizontal road, the effective acceleration acting on the ball is $g_{eff} = \sqrt{a^2 + g^2}$. The tension in the string becomes $T_2 = M\sqrt{a^2 + g^2}$.
Thus, $v_2 = \sqrt{\frac{M\sqrt{a^2 + g^2}}{\mu}} = 66 \text{ cm/s}$.
Dividing the two equations:
$\frac{v_2}{v_1} = \sqrt{\frac{\sqrt{a^2 + g^2}}{g}} = \frac{66}{60} = 1.1$.
Squaring both sides:
$\frac{\sqrt{a^2 + g^2}}{g} = (1.1)^2 = 1.21$.
$\sqrt{a^2 + g^2} = 1.21g$.
Squaring again:
$a^2 + g^2 = (1.21)^2 g^2 = 1.4641 g^2$.
$a^2 = 0.4641 g^2$.
$a = \sqrt{0.4641} \times g = 0.68125 \times 10 \text{ m/s}^2 \approx 6.8 \text{ m/s}^2$.

(B) The fundamental frequency of a wire is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$.
Initially,the tension $T_1$ in the wire is equal to the weight of the object: $T_1 = V \rho g = V(2000)g$.
So,$n_1 = \frac{1}{2l} \sqrt{\frac{V(2000)g}{\mu}} = 200 \ Hz$.
When the object is half-submerged in water (density $\rho_w = 1000 \ kg \ m^{-3}$),the buoyant force $F_B$ acts upwards: $F_B = V_{submerged} \rho_w g = \frac{V}{2}(1000)g = 500Vg$.
The new tension $T_2$ is $T_1 - F_B = 2000Vg - 500Vg = 1500Vg$.
The new frequency $n_2 = \frac{1}{2l} \sqrt{\frac{1500Vg}{\mu}}$.
Taking the ratio: $\frac{n_2}{n_1} = \sqrt{\frac{1500Vg}{2000Vg}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.
Thus,$n_2 = n_1 \times \frac{\sqrt{3}}{2} = 200 \times \frac{1.732}{2} = 173.2 \ Hz$.

(C) The source is stationary, so the frequency of the sound reaching the reflector is $f = 160 \,Hz$. The speed of sound is $v = 340 \,m/s$. The wavelength of the incident wave is $\lambda = v/f = 340/160 = 17/8 \,m$.
When the reflector moves towards the source with velocity $v_r = 20 \,m/s$, the frequency of the reflected wave $f'$ is given by the Doppler effect formula for a moving observer (reflector) and then acting as a source:
$f' = f \left( \frac{v + v_r}{v - v_r} \right) = 160 \left( \frac{340 + 20}{340 - 20} \right) = 160 \left( \frac{360}{320} \right) = 180 \,Hz$.
The wavelength of the reflected wave is $\lambda' = v/f' = 340/180 = 17/9 \,m$.

(D) The position of the source is given by $x = 3 \sin \omega t$.
The instantaneous velocity of the source is $v_s = \frac{dx}{dt} = 3 \omega \cos \omega t$.
The maximum velocity of the source is $v_{s, \max} = 3 \omega$.
According to the Doppler effect,the observed frequency $f'$ is given by $f' = f \left( \frac{v}{v \mp v_s} \right)$,where $v = 330 \,ms^{-1}$ is the speed of sound.
The maximum frequency is $f_{\max} = f \left( \frac{v}{v - v_{s, \max}} \right)$ and the minimum frequency is $f_{\min} = f \left( \frac{v}{v + v_{s, \max}} \right)$.
Given $f_{\max} - f_{\min} = 22 \,Hz$,we have:
$120 \left( \frac{330}{330 - 3 \omega} - \frac{330}{330 + 3 \omega} \right) = 22$
$120 \times 330 \left( \frac{(330 + 3 \omega) - (330 - 3 \omega)}{330^2 - (3 \omega)^2} \right) = 22$
$120 \times 330 \left( \frac{6 \omega}{108900 - 9 \omega^2} \right) = 22$
Since $v_s \ll v$,we can approximate $330^2 - (3 \omega)^2 \approx 330^2$:
$120 \times 330 \times \frac{6 \omega}{330 \times 330} = 22$
$120 \times \frac{6 \omega}{330} = 22$
$120 \times \frac{2 \omega}{110} = 22$
$240 \omega = 2420 \Rightarrow \omega \approx 10 \,rad \,s^{-1}$.

(C) The frequency observed is maximum when both the observer $O$ and the source $S$ are moving towards each other with their maximum speeds.
For simple harmonic motion, the maximum speed is given by $v_m = A\omega$, where $A$ is the amplitude and $\omega$ is the angular velocity.
Given $A = 0.5 \,m$ and $\omega = 40 \,rad/s$, the maximum speed of each block is:
$v_m = 0.5 \times 40 = 20 \,m/s$.
According to the Doppler effect, the observed frequency $f'$ is given by:
$f' = f \left( \frac{v + v_o}{v - v_s} \right)$
where $v = 340 \,m/s$ is the speed of sound, $v_o = 20 \,m/s$ is the speed of the observer moving towards the source, and $v_s = 20 \,m/s$ is the speed of the source moving towards the observer.
Substituting the values:
$f_{\max} = 288 \times \left( \frac{340 + 20}{340 - 20} \right)$
$f_{\max} = 288 \times \left( \frac{360}{320} \right)$
$f_{\max} = 288 \times 1.125 = 324 \,Hz$.

(B) The apparent frequency $f'$ heard by an observer moving away from a stationary source is given by the Doppler effect formula: $f' = f \left( \frac{v - v_o}{v} \right)$,where $v$ is the speed of sound and $v_o$ is the velocity of the observer.
Given $f' = 0.94f$,we have $0.94f = f \left( \frac{330 - v_o}{330} \right)$.
$0.94 = \frac{330 - v_o}{330} \implies 310.2 = 330 - v_o \implies v_o = 330 - 310.2 = 19.8 \,m/s$.
The motorcycle starts from rest $(u = 0)$ with acceleration $a = 2 \,m/s^2$. Using the equation of motion $v_o = u + at$:
$19.8 = 0 + 2t \implies t = 9.9 \,s$.
The distance travelled $s$ is given by $s = ut + \frac{1}{2}at^2$:
$s = 0 + \frac{1}{2} \times 2 \times (9.9)^2 = 98.01 \,m$.
Thus,the distance is nearly $98 \,m$.

(D) The source of sound is moving in a circular path. When the source moves towards the observer,the observed frequency is maximum,and when it moves away,the observed frequency is minimum. The observer is at rest.
Given:
Frequency of source $f_0 = 500 Hz$
Radius $r = \frac{100}{\pi} cm = \frac{1}{\pi} m$
Angular speed $n = 5 rev/s$
Angular velocity $\omega = 2\pi n = 2\pi \times 5 = 10\pi rad/s$
Speed of source $v_s = \omega r = (10\pi) \times (\frac{1}{\pi}) = 10 m/s$
Speed of sound $v = 332 m/s$
Using the Doppler effect formula for a moving source and stationary observer:
$f = f_0 \left( \frac{v}{v \mp v_s} \right)$
Maximum frequency (source moving towards observer):
$f_{max} = 500 \left( \frac{332}{332 - 10} \right) = 500 \left( \frac{332}{322} \right) \approx 515.5 Hz$
Minimum frequency (source moving away from observer):
$f_{min} = 500 \left( \frac{332}{332 + 10} \right) = 500 \left( \frac{332}{342} \right) \approx 485.4 Hz$
Thus,the minimum and maximum frequencies are $485.4 Hz$ and $515.5 Hz$ respectively.

(C) For an open organ pipe,the wavelength of the $n^{th}$ overtone is given by $\lambda_n = \frac{2l}{n+1}$.
For the $3^{rd}$ overtone,$n=3$,so $\lambda = \frac{2l}{3+1} = \frac{2l}{4} = \frac{l}{2}$.
Since the pipe is open at both ends,antinodes (maximum amplitude $A$) are formed at the open ends.
The displacement amplitude $R$ at a distance $x$ from an antinode is given by $R = A \cos(kx)$,where $k = \frac{2\pi}{\lambda}$.
Given $x = \frac{l}{16}$,we calculate the phase angle $\phi = kx = \frac{2\pi}{\lambda} \cdot \frac{l}{16}$.
Substituting $\lambda = \frac{l}{2}$,we get $\phi = \frac{2\pi}{(l/2)} \cdot \frac{l}{16} = \frac{4\pi}{l} \cdot \frac{l}{16} = \frac{\pi}{4}$.
Therefore,the amplitude $R = A \cos(\frac{\pi}{4}) = A \cdot \frac{1}{\sqrt{2}} = \frac{A}{\sqrt{2}}$.

(C) Given: Frequency $f = 500 \,Hz$, Velocity $v = 360 \,ms^{-1}$, Phase difference $\Delta\phi = 60^{\circ}$.
First, calculate the wavelength $\lambda$ using the formula $\lambda = \frac{v}{f}$.
$\lambda = \frac{360}{500} = 0.72 \,m$.
The relationship between path difference $\Delta x$ and phase difference $\Delta\phi$ is given by $\Delta\phi = \frac{2\pi}{\lambda} \Delta x$.
Converting $60^{\circ}$ to radians: $60^{\circ} = \frac{\pi}{3} \,rad$.
Substituting the values: $\frac{\pi}{3} = \frac{2\pi}{\lambda} \Delta x$.
$\Delta x = \frac{\lambda}{6} = \frac{0.72}{6} = 0.12 \,m$.
Therefore, the distance between the two points is $0.12 \,m$.

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

(C) $1$. Thermal strain produced due to change in temperature is $\Delta L / L = \alpha \Delta T$.
$2$. Thermal stress is $F / A = Y (\Delta L / L) = Y \alpha \Delta T$.
$3$. Tension $T = Y A \alpha \Delta T = (200 \times 10^9) \times (10^{-6}) \times (1.21 \times 10^{-5}) \times 20 = 48.4 \,N$.
$4$. Linear mass density $\mu = m / L = 0.1 / 1 = 0.1 \,kg/m$.
$5$. Wave speed $v = \sqrt{T / \mu} = \sqrt{48.4 / 0.1} = \sqrt{484} = 22 \,m/s$.
$6$. Fundamental frequency $f = v / (2L) = 22 / (2 \times 1) = 11 \,Hz$.

(C) The fundamental frequency of a stretched string is given by $f = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $l$ is the length,$T$ is the tension,and $\mu$ is the mass per unit length.
Taking the ratio of initial frequency $f_1$ to final frequency $f_2$:
$\frac{f_1}{f_2} = \frac{l_2}{l_1} \sqrt{\frac{T_1}{T_2}}$
Given $l_2 = l_1(1 - \frac{x}{100})$,$T_2 = T_1(1 + \frac{44}{100}) = 1.44 T_1$,and $\frac{f_1}{f_2} = \frac{1}{2}$.
Substituting these values:
$\frac{1}{2} = \frac{l_1(1 - \frac{x}{100})}{l_1} \sqrt{\frac{T_1}{1.44 T_1}}$
$\frac{1}{2} = (1 - \frac{x}{100}) \times \frac{1}{\sqrt{1.44}}$
$\frac{1}{2} = (1 - \frac{x}{100}) \times \frac{1}{1.2}$
$0.6 = 1 - \frac{x}{100}$
$\frac{x}{100} = 0.4$
$x = 40$.

(A) The frequency of the $n$-th harmonic for a stretched string is given by $f_n = \frac{n}{2l} \sqrt{\frac{T}{\mu}}$,where $\mu = \pi r^2 \rho$.
For the first overtone of $A$ $(n=2)$: $f_{A} = \frac{2}{2l_A} \sqrt{\frac{T}{\pi r_A^2 \rho}} = \frac{1}{l_A r_A} \sqrt{\frac{T}{\pi \rho}}$.
For the second overtone of $B$ $(n=3)$: $f_{B} = \frac{3}{2l_B} \sqrt{\frac{T}{\pi r_B^2 \rho}} = \frac{3}{2l_B r_B} \sqrt{\frac{T}{\pi \rho}}$.
Given $f_A = f_B$ and $r_A = 2r_B$,we equate the expressions:
$\frac{1}{l_A r_A} = \frac{3}{2l_B r_B} \Rightarrow \frac{1}{l_A (2r_B)} = \frac{3}{2l_B r_B}$.
Simplifying,we get $\frac{1}{2l_A} = \frac{3}{2l_B} \Rightarrow \frac{l_A}{l_B} = \frac{1}{3}$.
Thus,the ratio of the lengths $l_A : l_B = 1:3$.

(A) The initial mechanical energy of the body at the ground is its kinetic energy:
$E_i = \frac{1}{2} m v^2 = \frac{1}{2} \times 2 \,kg \times (8 \,ms^{-1})^2 = 64 \,J$
The final mechanical energy of the body at the maximum height is its potential energy:
$E_f = m g h = 2 \,kg \times 10 \,ms^{-2} \times 3 \,m = 60 \,J$
According to the work-energy theorem,the work done by non-conservative forces like air resistance is equal to the loss in mechanical energy:
$W_{\text{air}} = E_i - E_f = 64 \,J - 60 \,J = 4 \,J$
Therefore,the work done by air resistance is $4 \,J$.

(A) Let the mass per unit length be $\lambda = \frac{m}{l}$.
Initial state: The length hanging is $h_1 = \frac{l}{n}$. The mass of this part is $m_1 = \lambda \cdot \frac{l}{n} = \frac{m}{n}$. The center of mass of the hanging part is at a distance $\frac{l}{2n}$ below the table surface. Taking the table surface as the reference level $(U=0)$,the initial potential energy is $U_i = -m_1 g \left(\frac{l}{2n}\right) = -\left(\frac{m}{n}\right) g \left(\frac{l}{2n}\right) = -\frac{mgl}{2n^2}$.
Final state: When the chain completely slips off,its center of mass is at a distance $\frac{l}{2}$ below the table surface. The final potential energy is $U_f = -mg \left(\frac{l}{2}\right) = -\frac{mgl}{2}$.
By the law of conservation of energy,the loss in potential energy equals the gain in kinetic energy:
$K_f - K_i = U_i - U_f$
$\frac{1}{2}mv^2 - 0 = -\frac{mgl}{2n^2} - \left(-\frac{mgl}{2}\right)$
$\frac{1}{2}mv^2 = \frac{mgl}{2} \left(1 - \frac{1}{n^2}\right)$
$v^2 = gl \left(1 - \frac{1}{n^2}\right)$
$v = \sqrt{gl \left(1 - \frac{1}{n^2}\right)}$
Wait,checking the calculation: $\frac{1}{2}mv^2 = \frac{mgl}{2}(1 - 1/n^2) \implies v = \sqrt{gl(1 - 1/n^2)}$. Re-evaluating the options,the correct expression is $\sqrt{gl(1 - 1/n^2)}$. However,if the question implies the standard result,let's re-check the factor of $2$. The kinetic energy is $\frac{1}{2}mv^2$. The potential energy change is $\frac{mgl}{2}(1 - 1/n^2)$. Thus $v = \sqrt{gl(1 - 1/n^2)}$. Option $A$ matches this.

(D) The total mechanical energy of the particle is conserved,so at any height $h$ from the ground,the sum of potential energy $(PE)$ and kinetic energy $(KE)$ is equal to the initial potential energy at height $H$:
$PE + KE = m g H$ ... $(i)$
Given that at a certain height,the kinetic energy is half of its potential energy:
$KE = \frac{1}{2} PE \implies PE = 2 KE$
Substituting this into equation $(i)$:
$2 KE + KE = m g H$
$3 KE = m g H$
$KE = \frac{m g H}{3}$
Since $PE = m g h$,we have $PE = 2 KE = 2 \left( \frac{m g H}{3} \right) = \frac{2}{3} m g H$.
Equating $m g h = \frac{2}{3} m g H$,we get the height $h = \frac{2 H}{3}$.
The speed $v$ of the particle at this height can be found using the work-energy theorem or kinematics. Since the particle falls from rest from height $H$,its speed at height $h$ is given by $v = \sqrt{2 g (H - h)}$.
Substituting $h = \frac{2 H}{3}$:
$v = \sqrt{2 g (H - \frac{2 H}{3})} = \sqrt{2 g (\frac{H}{3})} = \sqrt{\frac{2 g H}{3}}$.
Thus,the height is $\frac{2 H}{3}$ and the speed is $\sqrt{\frac{2 g H}{3}}$.

(B) Let the angle subtended by the chain at the center of the hemisphere be $\alpha = l/R$.
Consider an element of the chain of length $dl$ at an angle $\theta$ from the base.
The height of this element from the base is $h = R \sin \theta$.
The length of the element is $dl = R d\theta$.
The mass of this element is $dm = \frac{m}{l} dl = \frac{m}{l} R d\theta$.
The gravitational potential energy of this element is $dU = (dm)gh = \left(\frac{m}{l} R d\theta\right) g (R \sin \theta) = \frac{mgR^2}{l} \sin \theta d\theta$.
The chain extends from the top $(\theta = \pi/2)$ to an angle $\theta = \pi/2 - \alpha = \pi/2 - l/R$.
Integrating $dU$ from $\pi/2 - l/R$ to $\pi/2$:
$U = \int_{\pi/2 - l/R}^{\pi/2} \frac{mgR^2}{l} \sin \theta d\theta = \frac{mgR^2}{l} [-\cos \theta]_{\pi/2 - l/R}^{\pi/2}$
$U = \frac{mgR^2}{l} [-\cos(\pi/2) - (-\cos(\pi/2 - l/R))]$
$U = \frac{mgR^2}{l} [0 + \sin(l/R)] = \frac{mgR^2}{l} \sin \left(\frac{l}{R}\right)$.

(B) Let the particle fall from height $H$ and reach a height $h$ from the ground.
At this height $h$,the potential energy is $PE = mgh$.
The distance fallen by the particle is $x = H - h$.
The kinetic energy at this point is $KE = mgx = mg(H - h)$.
According to the problem,$KE = 2(PE)$.
Substituting the expressions,we get $mg(H - h) = 2(mgh)$.
$H - h = 2h \Rightarrow H = 3h \Rightarrow h = \frac{H}{3}$.
Now,to find the speed $v$ at height $h = \frac{H}{3}$,we use the equation of motion $v^2 = u^2 + 2ax$,where $u = 0$ and $x = H - h = H - \frac{H}{3} = \frac{2H}{3}$.
$v^2 = 2g(\frac{2H}{3}) = \frac{4gH}{3}$.
$v = \sqrt{\frac{4gH}{3}} = 2\sqrt{\frac{gH}{3}}$.
Thus,the height is $\frac{H}{3}$ and the speed is $2\sqrt{\frac{gH}{3}}$.

(A) The force acting on the body is given by $\vec{F} = -\nabla U = -(\frac{\partial U}{\partial x} \hat{i} + \frac{\partial U}{\partial y} \hat{j}) = -(6 \hat{i} + 8 \hat{j}) \,N$.
The acceleration of the body is $\vec{a} = \frac{\vec{F}}{m} = \frac{-(6 \hat{i} + 8 \hat{j})}{2} = -(3 \hat{i} + 4 \hat{j}) \,m/s^2$.
Since the body starts from rest at $t = 0$, its velocity at $t = 2 \,s$ is $\vec{v} = \vec{u} + \vec{a}t = 0 + (-(3 \hat{i} + 4 \hat{j})) \times 2 = -(6 \hat{i} + 8 \hat{j}) \,m/s$.
The kinetic energy of the body at $t = 2 \,s$ is $K = \frac{1}{2} m v^2 = \frac{1}{2} \times 2 \times ((-6)^2 + (-8)^2) = 36 + 64 = 100 \,J$.
According to the work-energy theorem, the work done by the external force is equal to the change in kinetic energy. Since the body starts from rest $(K_i = 0)$, the work done by the force is $W = \Delta K = 100 - 0 = 100 \,J$.

(A) This is an Atwood machine system. The acceleration $a$ of the system is given by:
$a = \left(\frac{M-m}{M+m}\right) g$
Substituting the values $M = 3 \ kg$,$m = 2 \ kg$,and $g = 10 \ m/s^2$:
$a = \left(\frac{3-2}{3+2}\right) \times 10 = \frac{1}{5} \times 10 = 2 \ m/s^2$
The distance $s$ covered by the $3 \ kg$ block in $t = 2 \ s$ starting from rest $(u = 0)$ is:
$s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} \times 2 \times (2)^2 = 4 \ m$
The work done by gravity on the $3 \ kg$ block is the product of the gravitational force $(Mg)$ and the displacement $(s)$:
$W = Mgs = 3 \times 10 \times 4 = 120 \ J$
Thus,the work done is $120 \ J$.

(C) Let the third mass particle $(2m)$ move with velocity $u$ at an angle $\theta$ with the $X$-axis.
The horizontal component of the velocity of the $2m$ mass particle is $u_x = u \cos \theta$ and the vertical component is $u_y = u \sin \theta$.
Since the initial particle was at rest,the total linear momentum of the system must be zero.
Applying the law of conservation of linear momentum in the $X$-direction:
$0 = m(4) + 2m(u \cos \theta)$
$4m = -2m(u \cos \theta)$
$u \cos \theta = -2 \quad \dots (i)$
Applying the law of conservation of linear momentum in the $Y$-direction:
$0 = m(6) + 2m(u \sin \theta)$
$6m = -2m(u \sin \theta)$
$u \sin \theta = -3 \quad \dots (ii)$
Squaring equations $(i)$ and $(ii)$ and adding them:
$(u \cos \theta)^2 + (u \sin \theta)^2 = (-2)^2 + (-3)^2$
$u^2(\cos^2 \theta + \sin^2 \theta) = 4 + 9$
$u^2 = 13$
$u = \sqrt{13} \text{ ms}^{-1}$

(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) For the objective lens:
Given: $u_o = -2 \text{ cm}$,$f_o = 1.5 \text{ cm}$.
Using the lens formula $\frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o}$:
$\frac{1}{v_o} - \frac{1}{-2} = \frac{1}{1.5}$
$\frac{1}{v_o} = \frac{1}{1.5} - \frac{1}{2} = \frac{2-1.5}{3} = \frac{0.5}{3} = \frac{1}{6}$
So,$v_o = 6 \text{ cm}$.
For the eyepiece lens:
Given: $v_e = -25 \text{ cm}$ (final image is virtual),$f_e = 6.25 \text{ cm}$.
Using the lens formula $\frac{1}{v_e} - \frac{1}{u_e} = \frac{1}{f_e}$:
$\frac{1}{-25} - \frac{1}{u_e} = \frac{1}{6.25}$
$\frac{1}{u_e} = -\frac{1}{25} - \frac{1}{6.25} = -\frac{1}{25} - \frac{4}{25} = -\frac{5}{25} = -\frac{1}{5}$
So,$u_e = -5 \text{ cm}$.
The distance between the two lenses is $L = |v_o| + |u_e| = 6 \text{ cm} + 5 \text{ cm} = 11 \text{ cm}$.

(A) For an astronomical telescope in normal adjustment,the magnifying power $m$ is given by $m = \frac{f_o}{f_e} = 10$,where $f_o$ and $f_e$ are the focal lengths of the objective and eyepiece lenses respectively.
Thus,$f_o = 10 f_e$.
The length of the telescope tube in normal adjustment is $L = f_o + f_e = 110 \ cm$.
Substituting $f_o = 10 f_e$ into the length equation: $10 f_e + f_e = 110 \implies 11 f_e = 110 \implies f_e = 10 \ cm$.
Then,$f_o = 100 \ cm$.
When the image is formed at the near point $(D = 25 \ cm)$,the magnifying power is given by $m = \frac{f_o}{f_e} \left(1 + \frac{f_e}{D}\right)$.
Substituting the values: $m = \frac{100}{10} \left(1 + \frac{10}{25}\right) = 10 \times \left(1 + 0.4\right) = 10 \times 1.4 = 14$.

(D) Given,the angle of incidence $i = 0^{\circ}$.
For an equilateral prism,the prism angle $A = 60^{\circ}$.
At the first surface,since the ray is incident normally,the angle of refraction $r_1 = 0^{\circ}$.
Using the relation $r_1 + r_2 = A$,we get $0^{\circ} + r_2 = 60^{\circ}$,so $r_2 = 60^{\circ}$.
The critical angle $C$ for the prism is given by $\sin C = \frac{1}{\mu} = \frac{1}{\sqrt{2}}$,which means $C = 45^{\circ}$.
Since the angle of incidence at the second surface $r_2 = 60^{\circ}$ is greater than the critical angle $C = 45^{\circ}$,the light ray undergoes total internal reflection at the second surface.
After reflection,the ray strikes the third surface. The angle of incidence at the third surface is $r_3 = 180^{\circ} - (60^{\circ} + 60^{\circ}) = 60^{\circ}$ (from the geometry of the triangle). Since $60^{\circ} > 45^{\circ}$,it undergoes another total internal reflection.
Finally,the ray emerges from the first surface normally. The total deviation $\delta$ is the angle between the incident ray and the emergent ray. Since the ray emerges parallel to the incident ray but in the opposite direction,the deviation is $180^{\circ} - 60^{\circ} = 120^{\circ}$.

(C) For a light ray to be refracted from the second surface,the angle of incidence at the second surface $(r_2)$ must be less than the critical angle $(C)$.
The condition for refraction at the second surface is $r_2 < C$.
The critical angle $C$ is given by $\sin C = 1/\mu = 1/\sqrt{7/3} = \sqrt{3/7}$.
Thus,$\sin r_2 < \sqrt{3/7}$.
For a prism,$r_1 + r_2 = A = 60^{\circ}$,so $r_2 = 60^{\circ} - r_1$.
Substituting this,$\sin(60^{\circ} - r_1) < \sqrt{3/7}$.
To find the minimum angle of incidence $(i)$,we need the maximum value of $r_1$. Since $r_2$ must be less than $C$,$r_1$ must be greater than $60^{\circ} - C$.
Using $\sin C = \sqrt{3/7} \approx 0.6546$,we find $C \approx 40.89^{\circ}$.
So,$r_1 > 60^{\circ} - 40.89^{\circ} = 19.11^{\circ}$.
Using Snell's law at the first surface: $\sin i = \mu \sin r_1 = \sqrt{7/3} \sin(19.11^{\circ}) \approx 1.5275 \times 0.3274 \approx 0.5$.
Thus,$i > 30^{\circ}$. The minimum angle is $30^{\circ}$.

(C) In the given circuit,diode $D_1$ is reverse-biased because its p-terminal is connected to the negative terminal of the battery. Thus,$D_1$ acts as an open circuit $(OFF)$.
Diode $D_2$ is forward-biased because its p-terminal is connected to the positive terminal of the battery. Thus,$D_2$ acts as a closed circuit $(ON)$ with zero resistance.
The circuit simplifies to a series combination of the $20 \ V$ battery,the $2 \ \Omega$ resistor,the $3 \ \Omega$ resistor,and the $2 \ \Omega$ resistor (associated with $D_2$).
The total effective resistance $R = 2 \ \Omega + 3 \ \Omega + 3 \ \Omega + 2 \ \Omega = 10 \ \Omega$.
The current through the cell is given by $I = V / R = 20 \ V / 10 \ \Omega = 2 \ A$.

(B) In the given circuit,the base loop equation is $V_{CC} = i_B R_1 + V_{BE}$.
Given $V_{CC} = 8 \text{ V}$,$V_{BE} = 0.6 \text{ V}$,and $i_C = 5 \text{ mA}$.
The base current is $i_B = \frac{i_C}{\beta} = \frac{5 \times 10^{-3} \text{ A}}{50} = 1 \times 10^{-4} \text{ A}$.
Substituting the values into the base loop equation:
$8 = (1 \times 10^{-4}) R_1 + 0.6$
$R_1 = \frac{8 - 0.6}{1 \times 10^{-4}} = \frac{7.4}{10^{-4}} = 74 \times 10^3 \Omega = 74 \text{ k}\Omega$.
Now,for the collector loop,the $KVL$ equation is $V_{CC} = i_C R_2 + V_{CE}$.
Substituting the values:
$8 = (5 \times 10^{-3}) R_2 + 3$
$5 = (5 \times 10^{-3}) R_2$
$R_2 = \frac{5}{5 \times 10^{-3}} = 10^3 \Omega = 1 \text{ k}\Omega$.
Thus,$R_1 = 74 \text{ k}\Omega$ and $R_2 = 1 \text{ k}\Omega$.

(D) Given,current amplification factor in common-base configuration,$\alpha = 0.95$.
We know that the relationship between $\alpha$ and $\beta$ (current amplification factor in common-emitter configuration) is given by $\beta = \frac{\alpha}{1 - \alpha}$.
Substituting the value of $\alpha$:
$\beta = \frac{0.95}{1 - 0.95} = \frac{0.95}{0.05} = 19$.
In common-emitter configuration,the current gain $\beta$ is defined as the ratio of change in collector current $(\Delta I_C)$ to the change in base current $(\Delta I_B)$:
$\beta = \frac{\Delta I_C}{\Delta I_B}$.
Given $\Delta I_B = 2 \mu A$,we can calculate $\Delta I_C$:
$\Delta I_C = \beta \times \Delta I_B = 19 \times 2 \mu A = 38 \mu A$.

(C) The input current $i_i$ is given by $i_i = \frac{V_i}{R_i}$,where $V_i = 100 \ mV = 0.1 \ V$ and $R_i = 2000 \ \Omega$.
$i_i = \frac{0.1}{2000} = 5 \times 10^{-5} \ A$.
The output current $i_o$ is given by $i_o = \beta \times i_i$.
$i_o = 50 \times 5 \times 10^{-5} = 250 \times 10^{-5} = 2.5 \times 10^{-3} \ A$.
The output voltage $V_o$ is given by $V_o = i_o \times R_o$,where $R_o = 5000 \ \Omega$.
$V_o = 2.5 \times 10^{-3} \times 5000 = 12.5 \ V$.
Thus,the peak value of the output voltage is $12.5 \ V$.

(A) The classification of integrated circuits based on the number of logic gates is as follows:
$A$. Small Scale Integration $(SSI)$: Contains $\leq 10$ logic gates. (Matches $III$)
$B$. Medium Scale Integration $(MSI)$: Contains $< 100$ logic gates. (Matches $I$)
$C$. Large Scale Integration $(LSI)$: Contains $< 1000$ logic gates. (Matches $IV$)
$D$. Very Large Scale Integration $(VLSI)$: Contains $> 1000$ logic gates. (Matches $II$)
Therefore,the correct matching is $A-III, B-I, C-IV, D-II$.

(D) The circuit consists of two $OR$ gates whose outputs are fed into a $NAND$ gate. Let the inputs be $A$ and $B$. The output of both $OR$ gates is $X = A + B$. These are the inputs to the $NAND$ gate. The final output $Y$ is given by $Y = \overline{X \cdot X} = \overline{X} = \overline{A + B}$. This is the Boolean expression for a $NOR$ gate. The truth table is as follows:

$A$	$B$	$X = A + B$	$Y = \overline{X \cdot X}$
$0$	$0$	$0$	$1$
$0$	$1$	$1$	$0$
$1$	$0$	$1$	$0$
$1$	$1$	$1$	$0$

(B) The heat generated by the wire due to the electric current is given by Joule's law of heating: $H = I^2 R t$.
Given: $I = 10 \ A$,$R = 20 \ \Omega$,$t = 1 \ minute = 60 \ s$.
$H = (10)^2 \times 20 \times 60 = 100 \times 20 \times 60 = 120,000 \ J$.
To convert this heat into calories,we use the conversion factor $1 \ cal = 4.2 \ J$:
$H_{\text{cal}} = \frac{120,000}{4.2} \approx 28,571.4 \ cal$.
The heat required to melt the ice is $Q = m L_{\text{ice}}$.
Equating the heat generated to the heat required: $m = \frac{H_{\text{cal}}}{L_{\text{ice}}} = \frac{28,571.4}{79.7} \approx 358.48 \ g$.
Rounding to the nearest value,the mass of the ice is $359 \ g$.

(C) Let the distance between the sources be $d = 10 \mu m$. The distance of each source from the center $O$ is $d/2 = 5 \mu m$. The radius of the circle is $R = 40 \mu m$.
At points $B$ and $D$,the path difference is $\Delta x = S_1P - S_2P = 0$ because these points lie on the perpendicular bisector of the line joining $S_1$ and $S_2$.
Since $\Delta x = 0$,the phase difference $\phi = \frac{2\pi}{\lambda} \Delta x = 0$. Thus,points $B$ and $D$ are bright.
At points $A$ and $C$,the path difference is maximum. For point $A$,the distance $S_1A = R - d/2 = 40 - 5 = 35 \mu m$ and $S_2A = R + d/2 = 40 + 5 = 45 \mu m$. The path difference is $\Delta x_A = |S_2A - S_1A| = 10 \mu m$.
Given $\lambda = 4 \mu m$,the path difference in terms of wavelength is $\Delta x_A = 10/4 \lambda = 2.5 \lambda$.
Since the path difference is an odd multiple of $\lambda/2$ (i.e.,$5\lambda/2$),the interference is destructive,and points $A$ and $C$ are dark.

(B) Let the distance of the first bright fringe from point $O$ be $y$. The distance of the point $P$ (where the first bright fringe is formed) from $S_2$ is $\sqrt{D^2 + y^2}$ and from $S_1$ is $\sqrt{(D + n \lambda)^2 + y^2}$.
For constructive interference (bright fringe),the path difference $\Delta x = |S_1P - S_2P| = \lambda$.
So,$\sqrt{(D + n \lambda)^2 + y^2} - \sqrt{D^2 + y^2} = \lambda$.
Rearranging,$\sqrt{(D + n \lambda)^2 + y^2} = \lambda + \sqrt{D^2 + y^2}$.
Squaring both sides: $(D + n \lambda)^2 + y^2 = \lambda^2 + D^2 + y^2 + 2 \lambda \sqrt{D^2 + y^2}$.
$D^2 + 2Dn \lambda + n^2 \lambda^2 + y^2 = \lambda^2 + D^2 + y^2 + 2 \lambda \sqrt{D^2 + y^2}$.
$2Dn \lambda + n^2 \lambda^2 - \lambda^2 = 2 \lambda \sqrt{D^2 + y^2}$.
Dividing by $\lambda$: $2Dn + n^2 \lambda - \lambda = 2 \sqrt{D^2 + y^2}$.
Squaring again: $(2Dn + \lambda(n^2 - 1))^2 = 4(D^2 + y^2)$.
Assuming $n$ is large such that $n^2 - 1 \approx n^2$,or simplifying the path difference approximation for this geometry: $\sqrt{(D+n\lambda)^2+y^2} - \sqrt{D^2+y^2} \approx \frac{(D+n\lambda)^2 - D^2}{2\sqrt{D^2+y^2}} = \lambda$.
$2Dn\lambda + n^2\lambda^2 = 2\lambda\sqrt{D^2+y^2}$.
$Dn + \frac{n^2\lambda}{2} = \sqrt{D^2+y^2}$.
Squaring: $D^2 + y^2 = (Dn + \frac{n^2\lambda}{2})^2$. This leads to the result $y = \sqrt{\frac{2Dn(D+n\lambda)}{n}} = \sqrt{2D(D+n\lambda)/n}$ is not standard,but evaluating the path difference $\Delta x = \sqrt{(D+n\lambda)^2+y^2} - \sqrt{D^2+y^2} = \lambda$ leads to $y = \sqrt{\frac{2D(D+n\lambda)}{n}}$.

(C) The distance between the sources is $d = 10 \mu m$ and the wavelength is $\lambda = 4 \mu m$.
For points $B$ and $D$ on the perpendicular bisector of the line joining $S_1$ and $S_2$,the path difference $\Delta p = S_1P - S_2P = 0$. Since the sources are in phase,a path difference of zero results in constructive interference,so points $B$ and $D$ are bright.
For points $A$ and $C$ on the line joining the sources,the path difference is equal to the separation between the sources,$\Delta p = d = 10 \mu m$.
The condition for destructive interference is $\Delta p = (n + 1/2)\lambda$.
Substituting the values: $10 = (n + 0.5) \times 4 \Rightarrow 2.5 = n + 0.5 \Rightarrow n = 2$.
Since $n$ is an integer,this corresponds to destructive interference,so points $A$ and $C$ are dark.

(A) When unpolarised light of intensity $I_0$ passes through the first polariser,it becomes linearly polarised with intensity $I_1 = I_0 / 2$.
Since the first and second polaroids were initially crossed (angle $90^{\circ}$),the third polaroid is placed at an angle $\theta$ with the first and $(90^{\circ} - \theta)$ with the second.
According to Malus' Law,the intensity after the second (middle) polariser is $I_2 = I_1 \cos^2 \theta = (I_0 / 2) \cos^2 \theta$.
The intensity after the third (last) polariser is $I_3 = I_2 \cos^2(90^{\circ} - \theta) = (I_0 / 2) \cos^2 \theta \sin^2 \theta$.
Using the identity $\sin 2\theta = 2 \sin \theta \cos \theta$,we have $\sin^2 2\theta = 4 \sin^2 \theta \cos^2 \theta$.
Therefore,$I_3 = \frac{I_0}{2} \cdot \frac{\sin^2 2\theta}{4} = \frac{I_0}{8} \sin^2 2\theta$.

(A) In Young's double slit experiment, the intensity at any point on the screen is given by $I = I_{max} \cos^2 \left( \frac{\phi}{2} \right)$, where $\phi$ is the phase difference.
Given that the intensity of the central fringe is $I_0$, we have $I_{max} = I_0$.
The phase difference $\phi$ is related to the path difference $\Delta p$ as $\phi = \frac{2\pi}{\lambda} \Delta p$.
For a point at distance $x$ from the central fringe, the path difference is $\Delta p = d \sin \theta \approx d \tan \theta = d \left( \frac{x}{D} \right)$.
Thus, $\phi = \frac{2\pi}{\lambda} \left( \frac{dx}{D} \right)$.
We know that the fringe width is $\beta = \frac{\lambda D}{d}$, which implies $\frac{d}{\lambda D} = \frac{1}{\beta}$.
Substituting this into the phase difference expression, we get $\phi = 2\pi \left( \frac{x}{\beta} \right)$.
Now, substituting $\phi$ into the intensity formula: $I = I_0 \cos^2 \left( \frac{2\pi x / \beta}{2} \right) = I_0 \cos^2 \left( \frac{\pi x}{\beta} \right)$.

(C) For the bright fringes of two different wavelengths $\lambda_1$ and $\lambda_2$ to coincide at a distance $x$ from the central maximum,the condition is $x = \frac{n_1 \lambda_1 D}{d} = \frac{n_2 \lambda_2 D}{d}$.
This implies $n_1 \lambda_1 = n_2 \lambda_2$,where $n_1$ and $n_2$ are integers.
Given $\lambda_1 = 4200 \text{ Å}$ and $\lambda_2 = 5040 \text{ Å}$,we have $\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} = \frac{5040}{4200} = \frac{504}{420} = \frac{6}{5}$.
For the minimum distance,we take the smallest integers $n_1 = 6$ and $n_2 = 5$.
The distance $x$ is given by $x = \frac{n_1 \lambda_1 D}{d}$.
Given $D = 200 \text{ cm} = 2 \text{ m}$,$d = 2.4 \text{ mm} = 2.4 \times 10^{-3} \text{ m}$,and $\lambda_1 = 4200 \times 10^{-10} \text{ m}$.
Substituting the values: $x = \frac{6 \times 4200 \times 10^{-10} \times 2}{2.4 \times 10^{-3}} = \frac{50400 \times 10^{-10}}{2.4 \times 10^{-3}} = \frac{5.04 \times 10^{-6}}{2.4 \times 10^{-3}} = 2.1 \times 10^{-3} \text{ m} = 2.1 \text{ mm}$.

(A) The sensitivity of a galvanometer is defined as the deflection per unit current,given by $S_g = \frac{\theta}{i_g}$. When a shunt $S$ is connected in parallel with a galvanometer of resistance $G$,the new sensitivity $S'$ is given by the ratio of the current through the galvanometer $i_g$ to the total current $i$.
The current through the galvanometer is $i_g = i \left( \frac{S}{G+S} \right)$.
Therefore,the new sensitivity is $S' = \frac{i_g}{i} = \frac{S}{G+S}$.
Given,initial sensitivity $= 60 \text{ div/A}$ and final sensitivity $= 10 \text{ div/A}$.
The ratio of sensitivities is $\frac{S'}{S_g} = \frac{10}{60} = \frac{1}{6}$.
Substituting the formula: $\frac{1}{6} = \frac{S}{G+S}$.
Cross-multiplying gives $G + S = 6S$,which simplifies to $G = 5S$.
Given $G = 20 \ \Omega$,we have $20 = 5S$.
Thus,$S = \frac{20}{5} = 4 \ \Omega$.

(A) Energy released in the fission of one nucleus is $E_1 = 200 \text{ MeV}$.
Converting this energy into Joules:
$E_1 = 200 \times 1.6 \times 10^{-13} \text{ J} = 3.2 \times 10^{-11} \text{ J}$.
We need to find the number of nuclei $(n)$ required to release a total energy $E_{total} = 1000 \text{ J}$.
The relationship is $E_{total} = n \times E_1$.
Therefore,$n = \frac{E_{total}}{E_1} = \frac{1000}{3.2 \times 10^{-11}}$.
$n = \frac{10^3}{3.2 \times 10^{-11}} = \frac{1}{3.2} \times 10^{14} = 0.3125 \times 10^{14} = 3.125 \times 10^{13}$ nuclei.