AP EAMCET 2019 Physics Question Paper with Answer and Solution

(B) Given,frequency of source,$f_s = 5 \text{ kHz}$.
For the passenger in train $A$,the observed frequency is $f_A = 5.5 \text{ kHz}$.
Using the Doppler effect formula for an observer moving towards a stationary source: $f_A = f_s \left( \frac{v + v_A}{v} \right)$,where $v$ is the speed of sound.
$5.5 = 5 \left( 1 + \frac{v_A}{v} \right) \implies 1.1 = 1 + \frac{v_A}{v} \implies \frac{v_A}{v} = 0.1$ --- $(i)$
For the passenger in train $B$,the observed frequency is $f_B = 6 \text{ kHz}$.
Similarly,$f_B = f_s \left( \frac{v + v_B}{v} \right)$.
$6 = 5 \left( 1 + \frac{v_B}{v} \right) \implies 1.2 = 1 + \frac{v_B}{v} \implies \frac{v_B}{v} = 0.2$ --- $(ii)$
Dividing equation $(ii)$ by equation $(i)$:
$\frac{v_B / v}{v_A / v} = \frac{0.2}{0.1} = 2$.
Thus,the ratio of the speed of train $B$ to that of train $A$ is $2$.

(C) Given, amplitude of spring, $A = 50 \,cm = 0.5 \,m$.
Mass of sound source, $m = 100 \,g = 0.1 \,kg$.
Speed of sound, $v = 340 \,ms^{-1}$.
The maximum frequency heard by the observer is $n_{\max} = n_{\min} + 0.125 n_{\min} = 1.125 n_{\min}$.
Using the Doppler effect formula for a moving source:
$n_{\max} = \frac{n_0 v}{v - v_{s\max}} \quad (i)$
$n_{\min} = \frac{n_0 v}{v + v_{s\max}} \quad (ii)$
Dividing $(i)$ by $(ii)$:
$\frac{n_{\max}}{n_{\min}} = \frac{v + v_{s\max}}{v - v_{s\max}} = 1.125$
$v + v_{s\max} = 1.125(v - v_{s\max})$
$v + v_{s\max} = 1.125v - 1.125v_{s\max}$
$2.125 v_{s\max} = 0.125v$
$v_{s\max} = \frac{0.125 \times 340}{2.125} = \frac{42.5}{2.125} = 20 \,ms^{-1}$.
For a spring-mass system, $v_{s\max} = A\omega = A\sqrt{\frac{k}{m}}$.
$20 = 0.5 \sqrt{\frac{k}{0.1}}$
$40 = \sqrt{\frac{k}{0.1}}$
$1600 = \frac{k}{0.1}$
$k = 160 \,N m^{-1}$.
Thus, the correct option is $(c)$.

(B) Let the position of the train be $T$,the crossing be $C$,and the observer be $O$. The distance $TC = 288 \ m$ and $CO = 384 \ m$. The distance $TO = \sqrt{288^2 + 384^2} = \sqrt{82944 + 147456} = \sqrt{230400} = 480 \ m$.
The component of the train's velocity along the line joining the train and the observer is the effective velocity of the source moving towards the observer.
Let $v_s$ be the velocity of the train $(120 \ km/h = 120 \times \frac{5}{18} \ m/s = \frac{100}{3} \ m/s)$.
The angle $\theta$ is between the track and the line $TO$. From the geometry,$\cos \theta = \frac{TC}{TO} = \frac{288}{480} = \frac{3}{5} = 0.6$.
The velocity of the source towards the observer is $u = v_s \cos \theta = \frac{100}{3} \times 0.6 = 20 \ m/s$.
Using the Doppler effect formula for a source moving towards a stationary observer:
$f' = f \left( \frac{v}{v - u} \right)$
Where $f = 576 \ Hz$,$v = 340 \ m/s$,and $u = 20 \ m/s$.
$f' = 576 \left( \frac{340}{340 - 20} \right) = 576 \left( \frac{340}{320} \right) = 576 \times 1.0625 = 612 \ Hz$.

(B) Given: Velocity of the police car, $v_{s_1} = 22 \,ms^{-1}$.
Speed of sound in air, $v = 330 \,ms^{-1}$.
Frequency of the police car horn, $f_1 = 176 \,Hz$.
Frequency of the stationary siren, $f_2 = 165 \,Hz$.
Let the velocity of the motorcyclist be $v_0$.
The apparent frequency $f_1^{\prime}$ heard by the motorcyclist from the police car (source moving towards observer, observer moving away from source) is:
$f_1^{\prime} = \left( \frac{v - v_0}{v - v_{s_1}} \right) f_1 = \left( \frac{330 - v_0}{330 - 22} \right) \times 176 = \left( \frac{330 - v_0}{308} \right) \times 176$
The apparent frequency $f_2^{\prime}$ heard by the motorcyclist from the stationary siren (source stationary, observer moving towards source) is:
$f_2^{\prime} = \left( \frac{v + v_0}{v} \right) f_2 = \left( \frac{330 + v_0}{330} \right) \times 165$
Since the number of beats heard is zero, $f_1^{\prime} = f_2^{\prime}$:
$\left( \frac{330 - v_0}{308} \right) \times 176 = \left( \frac{330 + v_0}{330} \right) \times 165$
$\left( 330 - v_0 \right) \times \frac{176}{308} = \left( 330 + v_0 \right) \times \frac{165}{330}$
$\left( 330 - v_0 \right) \times 0.5714 = \left( 330 + v_0 \right) \times 0.5$
$188.56 - 0.5714 v_0 = 165 + 0.5 v_0$
$23.56 = 1.0714 v_0$
$v_0 = 22 \,ms^{-1}$
Thus, the speed of the motorcycle is $22 \,ms^{-1}$. The correct option is $(b)$.

(B) Let the length of the tube be $L$.
The fundamental frequency for an open organ pipe is given by:
$f = \frac{V}{2L}$
When the tube is dipped vertically in water such that $60 \%$ of its length is submerged,the remaining length of the air column above the water surface acts as a closed organ pipe (closed at one end by the water surface).
The length of this air column is:
$l' = (100 \% - 60 \%) L = 40 \% L = 0.4L = \frac{2L}{5}$
The fundamental frequency of a closed organ pipe is given by:
$f' = \frac{V}{4l'}$
Substituting the value of $l'$:
$f' = \frac{V}{4(\frac{2L}{5})} = \frac{5V}{8L}$
We can rewrite this in terms of $f$:
$f' = \frac{5}{4} \left( \frac{V}{2L} \right) = \frac{5}{4} f$

(C) The frequency of an air column in a resonance tube is proportional to the speed of sound,which is proportional to the square root of the absolute temperature $(v \propto \sqrt{T})$.
Let $n$ be the frequency of the tuning fork.
At $T_1 = 273 + 51 = 324 \ K$,the frequency of the air column is $n_1 = n + 3$.
At $T_2 = 273 + 16 = 289 \ K$,the frequency of the air column is $n_2 = n - 3$.
Using the relation $\frac{n_1}{n_2} = \sqrt{\frac{T_1}{T_2}}$,we get:
$\frac{n+3}{n-3} = \sqrt{\frac{324}{289}} = \frac{18}{17}$.
Cross-multiplying gives: $17(n + 3) = 18(n - 3)$.
$17n + 51 = 18n - 54$.
$n = 51 + 54 = 105 \ Hz$.

(A) The total mechanical energy $E$ of a system is the sum of its potential energy $U$ and kinetic energy $KE$. That is,$E = U + KE$.
Given,$E = 2 \ J$ and $m = 1 \ kg$.
To find the maximum speed,we need the maximum kinetic energy,which occurs when the potential energy is at its minimum.
$U(x) = \frac{x^2}{2} - x$.
To find the minimum potential energy,we set $\frac{dU}{dx} = 0$:
$\frac{d}{dx}(\frac{x^2}{2} - x) = x - 1 = 0 \implies x = 1 \ m$.
$U_{\min} = U(1) = \frac{1^2}{2} - 1 = -0.5 \ J$.
Since $E = U + KE$,we have $KE_{\max} = E - U_{\min} = 2 - (-0.5) = 2.5 \ J$.
Using $KE = \frac{1}{2}mv^2$:
$2.5 = \frac{1}{2}(1)v^2 \implies v^2 = 5 \implies v = \sqrt{5} \ ms^{-1}$.

(B) Let the total height be $H = 30 \ m$ and the velocity of the body be $v$ at a point where potential energy $(PE)$ is twice the kinetic energy $(KE)$.
According to the law of conservation of energy,the total energy at any point is equal to the initial potential energy:
$m g H = PE + KE$
Given that $PE = 2 \times KE$,we substitute this into the equation:
$m g H = 2 \times KE + KE = 3 \times KE$
Since $KE = \frac{1}{2} m v^2$,we have:
$m g H = 3 \times (\frac{1}{2} m v^2)$
$g H = \frac{3}{2} v^2$
$v^2 = \frac{2}{3} g H$
Substituting the values $g = 10 \ m \ s^{-2}$ and $H = 30 \ m$:
$v^2 = \frac{2}{3} \times 10 \times 30$
$v^2 = 2 \times 10 \times 10 = 200$
$v = \sqrt{200} = 10 \sqrt{2} \ m \ s^{-1}$

(B) Let the mass of the man be $m$. Then the mass of the boy is $m_b = \frac{m}{2}$.
Let the initial speed of the man be $v_m$ and the speed of the boy be $v_b$.
According to the problem,the kinetic energy of the man is half the kinetic energy of the boy:
$\frac{1}{2} m v_m^2 = \frac{1}{2} (\frac{1}{2} m_b v_b^2) = \frac{1}{2} (\frac{1}{2} \cdot \frac{m}{2} v_b^2) = \frac{1}{8} m v_b^2$.
Simplifying,$v_m^2 = \frac{1}{4} v_b^2$,which gives $v_m = \frac{v_b}{2}$.
When the man speeds up by $1 \,ms^{-1}$,his new speed is $v_m' = v_m + 1 = \frac{v_b}{2} + 1$.
Now,his kinetic energy equals the boy's kinetic energy:
$\frac{1}{2} m (\frac{v_b}{2} + 1)^2 = \frac{1}{2} (\frac{m}{2}) v_b^2$.
Dividing both sides by $\frac{m}{2}$,we get $(\frac{v_b}{2} + 1)^2 = \frac{v_b^2}{2}$.
Taking the square root of both sides: $\frac{v_b}{2} + 1 = \frac{v_b}{\sqrt{2}}$.
$1 = v_b (\frac{1}{\sqrt{2}} - \frac{1}{2}) = v_b (\frac{2 - \sqrt{2}}{2\sqrt{2}})$.
$v_b = \frac{2\sqrt{2}}{2 - \sqrt{2}} = \frac{2\sqrt{2}(2 + \sqrt{2})}{4 - 2} = \frac{4\sqrt{2} + 4}{2} = 2\sqrt{2} + 2 = 2(\sqrt{2} + 1) \,ms^{-1}$.

(C) Given:
Depth of the well,$d = 30 \,m$
Volume of water per minute,$V = 1800 \,L = 1800 \times 10^{-3} \,m^3 = 1.8 \,m^3$
Mass of water per minute,$m = \rho \times V = 1000 \,kg/m^3 \times 1.8 \,m^3 = 1800 \,kg$
Cross-sectional area of pipe,$A = 30 \,cm^2 = 30 \times 10^{-4} \,m^2$
Time,$t = 60 \,s$
Velocity of water jet,$v = \frac{V}{A \times t} = \frac{1.8}{30 \times 10^{-4} \times 60} = \frac{1.8}{0.18} = 10 \,m/s$
Total work done by the engine is the sum of potential energy and kinetic energy:
$W = mgd + \frac{1}{2}mv^2$
$W = (1800 \times 10 \times 30) + (\frac{1}{2} \times 1800 \times 10^2)$
$W = 540000 + 90000 = 630000 \,J$
Power of the engine,$P = \frac{W}{t} = \frac{630000}{60} = 10500 \,W$
$P = 10.5 \,kW$

(B) The total depth of the well is $6 \ m + 14 \ m = 20 \ m$. Let a small volume element $dx$ be at a distance $x$ from the top surface. The mass of this element is $dm = \rho dV = \rho \pi r^2 dx$.
The potential energy required to lift this element to the top is $dU = dm \cdot g \cdot x = \rho \pi r^2 g x dx$.
Integrating from $x = 6 \ m$ to $x = 20 \ m$ to find the total work done $W$:
$W = \int_{6}^{20} \rho \pi r^2 g x dx = \rho \pi r^2 g \left[ \frac{x^2}{2} \right]_{6}^{20} = \rho \pi r^2 g \left( \frac{400 - 36}{2} \right) = \rho \pi r^2 g (182)$.
Given $\rho = 10^3 \ kg/m^3$,$r = 2.5 \ m$,$g = 10 \ m/s^2$:
$W = 10^3 \times 3.14 \times (2.5)^2 \times 10 \times 182 = 35.71 \times 10^6 \ J$.
Power $P = 10 \ HP = 10 \times 746 \ W = 7460 \ W$.
Time $t = \frac{W}{P} = \frac{35.71 \times 10^6}{7460} \approx 4787 \ s$.
Converting to minutes: $t = \frac{4787}{60} \approx 79.78 \ min \approx 80 \ min$.
Thus,the correct option is $B$.

(D) Given: Power $P = 7 \,W$, mass $m = 15 \,kg$, initial velocity $v_i = 3 \,ms^{-1}$, and final velocity $v_f = 5 \,ms^{-1}$.
According to the work-energy theorem, the work done $W$ is equal to the change in kinetic energy:
$W = \Delta K = \frac{1}{2} m (v_f^2 - v_i^2)$
$W = \frac{1}{2} \times 15 \times (5^2 - 3^2) = \frac{1}{2} \times 15 \times (25 - 9) = \frac{1}{2} \times 15 \times 16 = 120 \,J$.
Since power is constant, $P = \frac{W}{t}$, so the time taken is $t = \frac{W}{P} = \frac{120}{7} \,s$.
Using the equation of motion $v_f = v_i + at$, we find the acceleration $a$:
$a = \frac{v_f - v_i}{t} = \frac{5 - 3}{120/7} = \frac{2 \times 7}{120} = \frac{14}{120} = \frac{7}{60} \,ms^{-2}$.
Now, using the kinematic equation $v_f^2 - v_i^2 = 2as$, the distance $s$ is:
$s = \frac{v_f^2 - v_i^2}{2a} = \frac{25 - 9}{2 \times (7/60)} = \frac{16 \times 60}{14} = \frac{8 \times 60}{7} = \frac{480}{7} \approx 68.57 \,m$.
Rounding to the nearest integer provided in the options, the distance is $70 \,m$.

(C) Let $m = 0.1 \,kg$, $\mu = 0.2$, $g = 10 \,ms^{-2}$, $\theta = 30^{\circ}$, and $s_2 = 1 \,m$ (horizontal distance).
On the horizontal plane, the retardation $a_2 = \mu g = 0.2 \times 10 = 2 \,ms^{-2}$.
Using $v^2 = u^2 - 2a_2s_2$, where $v=0$ (comes to rest), we get $0 = u^2 - 2(2)(1)$, so $u^2 = 4$, or $u = 2 \,ms^{-1}$. This is the velocity at the bottom of the incline.
On the inclined plane, the net acceleration $a_1 = g(\sin \theta - \mu \cos \theta) = 10(\sin 30^{\circ} - 0.2 \cos 30^{\circ}) = 10(0.5 - 0.2 \times 0.866) = 10(0.5 - 0.1732) = 3.268 \,ms^{-2}$.
Using $v^2 = u_0^2 + 2a_1s_1$, where $u_0=0$ and $v=2 \,ms^{-1}$, we get $4 = 2(3.268)s_1$, so $s_1 = 4 / 6.536 \approx 0.612 \,m$.
The work done by friction on the incline is $W_1 = -f_1 s_1 = -(\mu mg \cos 30^{\circ}) s_1 = -(0.2 \times 0.1 \times 10 \times 0.866) \times 0.612 \approx -0.106 \,J$.
The work done by friction on the horizontal plane is $W_2 = -f_2 s_2 = -(\mu mg) s_2 = -(0.2 \times 0.1 \times 10) \times 1 = -0.2 \,J$.
The total work done by friction is $W = W_1 + W_2 = -0.106 - 0.2 = -0.306 \,J$. The magnitude is $0.306 \,J$.

(D) The given force is $F = K \left[ \frac{x}{(x^2+y^2)^{3/2}} \hat{i} + \frac{y}{(x^2+y^2)^{3/2}} \hat{j} \right]$.
In polar coordinates,$x = r \cos \theta$ and $y = r \sin \theta$,where $r = \sqrt{x^2+y^2}$.
Substituting these into the force expression:
$F = K \left[ \frac{r \cos \theta}{r^3} \hat{i} + \frac{r \sin \theta}{r^3} \hat{j} \right] = \frac{K}{r^2} (\cos \theta \hat{i} + \sin \theta \hat{j})$.
This force is a central force directed radially outward,i.e.,$F = \frac{K}{r^2} \hat{r}$.
The work done by a central force is given by $W = \int F \cdot dr = \int \frac{K}{r^2} dr$.
Since the particle moves along a circular path of radius $a$,the distance $r$ from the origin is constant $(r = a)$.
For a circular path,the displacement vector $d\vec{l}$ is always perpendicular to the radial force vector $\vec{F}$.
Therefore,the dot product $\vec{F} \cdot d\vec{l} = 0$ at every point on the path.
Thus,the total work done $W = \int \vec{F} \cdot d\vec{l} = 0$.

(C) The work done $W$ by a variable force $\vec{F}$ is given by the integral $W = \int_{x_i}^{x_f} \vec{F} \cdot d\vec{r}$.
Given $\vec{F} = -5x^4 \hat{i}$ and $d\vec{r} = dx \hat{i}$,the work done is:
$W = \int_{2}^{-2} (-5x^4) dx$
$W = -5 \int_{2}^{-2} x^4 dx$
$W = -5 \left[ \frac{x^5}{5} \right]_{2}^{-2}$
$W = -[x^5]_{2}^{-2}$
$W = -[(-2)^5 - (2)^5]$
$W = -[-32 - 32]$
$W = -[-64] = 64 \text{ J}$.

(A) Given:
Mass of block $1$, $m_1 = 0.36 \,kg$
Mass of block $2$, $m_2 = 0.72 \,kg$
Acceleration due to gravity, $g = 10 \,m/s^2$
Time, $t = 1 \,s$
The acceleration $a$ of the system is given by:
$a = \frac{(m_2 - m_1)g}{m_1 + m_2} = \frac{(0.72 - 0.36) \times 10}{0.72 + 0.36} = \frac{0.36 \times 10}{1.08} = \frac{3.6}{1.08} = \frac{10}{3} \,m/s^2$
The tension $T$ in the string is:
$T = m_1(g + a) = 0.36 \times (10 + \frac{10}{3}) = 0.36 \times \frac{40}{3} = 0.12 \times 40 = 4.8 \,N$
The displacement $s$ of the block $m_1$ in $t = 1 \,s$ is:
$s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} \times \frac{10}{3} \times (1)^2 = \frac{5}{3} \,m$
The work done by the string on block $m_1$ is:
$W = T \times s \times \cos(0^\circ) = 4.8 \times \frac{5}{3} = 1.6 \times 5 = 8 \,J$

(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 $\mu = \frac{1}{3} \tan 60^{\circ} = \frac{1}{3} \times \sqrt{3} = \frac{1}{\sqrt{3}}$.

(B) For the output loop of the common emitter circuit,applying Kirchhoff's voltage law $(KVL)$:
$15 \text{ V} - I_C \times (2 \text{ k}\Omega) - V_{CE} = 0$
Given $V_{CE} = 7 \text{ V}$,we have:
$15 - I_C \times 2000 - 7 = 0$
$8 = I_C \times 2000$
$I_C = \frac{8}{2000} \text{ A} = 4 \times 10^{-3} \text{ A} = 4 \text{ mA}$
Now,using the current gain relation $\beta = \frac{I_C}{I_B}$:
$100 = \frac{4 \text{ mA}}{I_B}$
$I_B = \frac{4 \text{ mA}}{100} = 0.04 \text{ mA}$

(A) Given, voltage gains of three amplifiers are $A_1 = 10, A_2 = 20$, and $A_3 = 30$.
The peak value of the input signal is $V_i = 1 mV = 10^{-3} \,V$.
When amplifiers are connected in series, the total voltage gain $A$ is the product of individual gains:
$A = A_1 \times A_2 \times A_3 = 10 \times 20 \times 30 = 6000$.
The relationship between output voltage $V_0$ and input voltage $V_i$ is given by $A = \frac{V_0}{V_i}$.
Therefore, the peak output voltage is $V_0 = A \times V_i = 6000 \times 10^{-3} \,V = 6 \,V$.

(C) Let the output of the $OR$ gate be $P = A + B$. Given $A = 1$ and $B = 1$,we have $P = 1 + 1 = 1$.
Let the output of the $NAND$ gate be $Q = \overline{A \cdot B}$. Given $A = 1$ and $B = 1$,we have $Q = \overline{1 \cdot 1} = \overline{1} = 0$.
Now,$Y_1$ is the output of an $AND$ gate with inputs $P$ and $Q$. Thus,$Y_1 = P \cdot Q = 1 \cdot 0 = 0$.
$Y_2$ is the output of an $OR$ gate with inputs $P$ and $Q$. Thus,$Y_2 = P + Q = 1 + 0 = 1$.
Therefore,the values are $Y_1 = 0$ and $Y_2 = 1$. The correct option is $(C)$.

(B) The circuit consists of two branches feeding into a $NOR$ gate. Each branch has a $NAND$ gate followed by a $NOT$ gate (which together form an $AND$ gate).
Let the output of the top branch be $Y_1$. The top branch has inputs $A$ and $B$ to a $NAND$ gate,followed by a $NOT$ gate. This is equivalent to an $AND$ gate. So,$Y_1 = A \cdot B = 1 \cdot 0 = 0$.
Similarly,the bottom branch has inputs $A$ and $B$ to a $NAND$ gate,followed by a $NOT$ gate. This is also equivalent to an $AND$ gate. So,the output of the bottom branch is $A \cdot B = 1 \cdot 0 = 0$.
Now,$Y_1$ is the output of the top branch,so $Y_1 = 0$.
The final gate is a $NOR$ gate with both inputs as $0$.
The output $Y_2$ of a $NOR$ gate is given by $\overline{0 + 0} = \overline{0} = 1$.
Thus,$Y_1 = 0$ and $Y_2 = 1$.

(D) Given energy required to produce a hole,$E = 50 \text{ meV} = 50 \times 10^{-3} \times 1.6 \times 10^{-19} \text{ J} = 8.0 \times 10^{-21} \text{ J}$.
Planck's constant,$h = 6.6 \times 10^{-34} \text{ Js}$.
Speed of light,$c = 3 \times 10^8 \text{ m/s}$.
The energy of a photon is given by $E = \frac{hc}{\lambda}$.
To find the maximum wavelength $\lambda$,we use $\lambda = \frac{hc}{E}$.
Substituting the values:
$\lambda = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{8.0 \times 10^{-21}}$
$\lambda = \frac{19.8 \times 10^{-26}}{8.0 \times 10^{-21}}$
$\lambda = 2.475 \times 10^{-5} \text{ m} = 24.75 \times 10^{-6} \text{ m} = 24.75 \mu \text{m}$.
Rounding to the nearest option,the maximum wavelength is $24.8 \mu \text{m}$.

(C) The visible region of the electromagnetic spectrum corresponds to wavelengths approximately between $400 \ nm$ and $700 \ nm$. The energy $E$ of a photon is given by $E = \frac{hc}{\lambda}$.
For $\lambda = 700 \ nm = 700 \times 10^{-9} \ m$:
$E_{\min} = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{700 \times 10^{-9} \times 1.6 \times 10^{-19}} \ eV \approx 1.77 \ eV$.
For $\lambda = 400 \ nm = 400 \times 10^{-9} \ m$:
$E_{\max} = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9} \times 1.6 \times 10^{-19}} \ eV \approx 3.1 \ eV$.
Thus,the energy band gap required for an $LED$ to emit visible light is in the range of $1.7 \ eV$ to $3.1 \ eV$. The correct option is $(c)$.

(D) For a single slit diffraction, the condition for the $n^{th}$ order dark fringe is given by $d \sin \theta = n \lambda$.
Given: $n = 5$, $\theta = 12^{\circ}$, $d = 9 \mu m = 9 \times 10^{-6} \text{ m}$.
Using the small angle approximation $\sin \theta \approx \theta$ (in radians), we have:
$\theta = 12^{\circ} = 12 \times \frac{\pi}{180} \text{ rad} = \frac{\pi}{15} \text{ rad}$.
Substituting the values into the formula:
$d \theta = n \lambda$
$(9 \times 10^{-6}) \times (\frac{\pi}{15}) = 5 \times \lambda$
$\lambda = \frac{9 \times 10^{-6} \times \pi}{15 \times 5} = \frac{9 \times 3.14159 \times 10^{-6}}{75} \approx 3.77 \times 10^{-7} \text{ m}$.
Converting to $\mathring{A}$s: $\lambda \approx 3770 Å$.
Given the options, the closest value is $3768 Å$.

(D) Given, refractive index $\mu = 1.414 = \sqrt{2}$.
For complete polarisation, the angle of incidence is the Brewster's angle $i_p$, where $\tan i_p = \mu = \sqrt{2}$.
Thus, $\sin i_p = \sqrt{\frac{2}{3}}$ and $\cos i_p = \frac{1}{\sqrt{3}}$.
From Snell's law, $\sin i_p = \mu \sin r$, so $\sin r = \frac{\sin i_p}{\mu} = \frac{\sqrt{2/3}}{\sqrt{2}} = \frac{1}{\sqrt{3}}$.
Therefore, $r = \sin^{-1} \left(\frac{1}{\sqrt{3}}\right)$. This matches $(iii)$.
The angle of reflection $\theta$ is equal to the angle of incidence $i_p$. Since $\cos i_p = \frac{1}{\sqrt{3}}$, we have $i_p = \cos^{-1} \left(\frac{1}{\sqrt{3}}\right)$. This matches $(iv)$.
The angle of deviation $\delta$ of the refracted ray is $i_p - r = \sin^{-1} \left(\sqrt{\frac{2}{3}}\right) - \sin^{-1} \left(\frac{1}{\sqrt{3}}\right)$. This matches $(ii)$.
The angle $\phi$ between the incident ray and the reflected (polarised) ray is $i_p + \theta = 2i_p = 2 \sin^{-1} \left(\sqrt{\frac{2}{3}}\right)$. This matches $(i)$.
The correct matching is $A-(iv), B-(iii), C-(i), D-(ii)$.

(C) For interference maxima on the screen,the path difference condition is given by:
$d \sin \theta = n \lambda$
where $d$ is the slit separation,$\theta$ is the angle,$n$ is the order of maxima,and $\lambda$ is the wavelength.
Given that $d = 2 \lambda$,the equation becomes:
$2 \lambda \sin \theta = n \lambda$
$2 \sin \theta = n$
Since the maximum value of $\sin \theta$ is $1$,the maximum value of $n$ is:
$n_{max} = 2 \times 1 = 2$
The possible values for $n$ are integers such that $-2 \le n \le 2$.
These values are $\{-2, -1, 0, 1, 2\}$.
Counting these,we get a total of $5$ interference maxima.

(B) The fringe width $\beta$ in a medium with refractive index $\mu$ is given by the formula: $\beta = \frac{\lambda' D}{d}$,where $\lambda' = \frac{\lambda}{\mu}$.
Thus,$\beta = \frac{\lambda D}{\mu d}$.
Given values:
$\lambda = 6300 \ \mathring{A} = 6300 \times 10^{-10} \ m$
$D = 1.33 \ m$
$d = 1 \ mm = 10^{-3} \ m$
$\mu = 1.33$
Substituting these values into the formula:
$\beta = \frac{6300 \times 10^{-10} \times 1.33}{1.33 \times 10^{-3}}$
$\beta = 6300 \times 10^{-7} \ m$
$\beta = 6.3 \times 10^{-4} \ m = 0.63 \ mm$.

(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 $0.8 \ \text{mm}$.

(A) The work done $W$ by a force $\vec{F}$ over a displacement $d\vec{r}$ is given by the dot product: $W = \int \vec{F} \cdot d\vec{r}$.
When a point charge moves in a circular path around another charge,the electrostatic force is always directed radially (towards or away from the center),while the displacement vector is always tangential to the circular path.
Since the angle between the radial force and the tangential displacement is $90^{\circ}$,the dot product $\vec{F} \cdot d\vec{r} = F dr \cos 90^{\circ} = 0$.
Thus,the work done is zero. Both the assertion and the reason are true,and the reason correctly explains why the work done is zero.

(C) For a balanced Wheatstone bridge,the ratio of resistances in the arms is equal: $\frac{A}{B} = \frac{C}{D}$.
Given,in the first case: $\frac{A}{B} = \frac{100}{D} \quad ... (1)$
When $A$ and $B$ are interchanged,the new ratio becomes $\frac{B}{A} = \frac{121}{D} \quad ... (2)$
Multiplying equation $(1)$ and $(2)$:
$\left(\frac{A}{B}\right) \times \left(\frac{B}{A}\right) = \left(\frac{100}{D}\right) \times \left(\frac{121}{D}\right)$
$1 = \frac{12100}{D^2}$
$D^2 = 12100$
$D = \sqrt{12100} = 110 \ \Omega$.

(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:
$\frac{1.2}{\beta_2} = 1.5$.
$\beta_2 = \frac{1.2}{1.5} = 0.8 \text{ mm}$.