TS EAMCET 2025 Physics Question Paper with Answer and Solution

(A) The beat frequency is given by the difference between the two frequencies: $f_{beat} = |f_2 - f_1| = |323 \ Hz - 320 \ Hz| = 3 \ Hz$.
This means there are $3$ beats per second.
The time interval between two successive beats (one maximum and one adjacent minimum) is half of the time period of one beat cycle.
The time period of the beat is $T = \frac{1}{f_{beat}} = \frac{1}{3} \ s$.
The time interval between a maximum sound and its adjacent minimum sound is $\Delta t = \frac{T}{2} = \frac{1}{2} \times \frac{1}{3} \ s = \frac{1}{6} \ s$.

(D) Let $v$ be the speed of sound,$v_o$ be the speed of the observer,and $f$ be the frequency of the source.
When the observer moves towards the stationary source,the observed frequency is $n_1 = f \left( \frac{v + v_o}{v} \right)$.
When the observer moves away from the stationary source,the observed frequency is $n_2 = f \left( \frac{v - v_o}{v} \right)$.
Given the ratio $n_1 / n_2 = 71 / 65$,we have:
$\frac{v + v_o}{v - v_o} = \frac{71}{65}$.
Cross-multiplying gives: $65(v + v_o) = 71(v - v_o)$.
$65v + 65v_o = 71v - 71v_o$.
$136v_o = 6v$.
$v_o = \frac{6}{136} v = \frac{3}{68} v$.
Given $v = 340 \ m/s$,we get $v_o = \frac{3}{68} \times 340 = 3 \times 5 = 15 \ m/s$.
To convert $m/s$ to $km/h$,multiply by $18/5$: $v_o = 15 \times \frac{18}{5} = 3 \times 18 = 54 \ km/h$.

(A) Let $v$ be the speed of sound $(336 \ m/s)$ and $u$ be the speed of the car.
The frequency of the sound emitted by the horn is $n$.
The frequency of the echo heard by the driver is $n' = n \left( \frac{v + u}{v - u} \right)$.
The difference in frequencies is $n' - n = 0.1n$.
Substituting $n'$,we get $n \left( \frac{v + u}{v - u} - 1 \right) = 0.1n$.
Simplifying the expression: $\frac{v + u - (v - u)}{v - u} = 0.1$.
This leads to $\frac{2u}{v - u} = 0.1$.
$2u = 0.1(v - u) \implies 2u = 0.1v - 0.1u$.
$2.1u = 0.1v \implies u = \frac{0.1}{2.1} v = \frac{1}{21} v$.
Given $v = 336 \ m/s$,$u = \frac{336}{21} = 16 \ m/s$.

(A) The resultant intensity $I_R$ of two waves with individual intensities $I_1$ and $I_2$ and a phase difference $\phi$ is given by the formula:
$I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\phi)$
Given that $I_1 = I_2 = I$ and $\phi = \frac{\pi}{2}$,we substitute these values into the formula:
$I_R = I + I + 2\sqrt{I \cdot I} \cos(\frac{\pi}{2})$
Since $\cos(\frac{\pi}{2}) = 0$,the expression simplifies to:
$I_R = 2I + 2I(0) = 2I$
Therefore,the resultant intensity is $2I$.

(B) Let $L_o$ and $L_c$ be the lengths of the open and closed pipes respectively. Given $L_o / L_c = 2 / 3$.
For an open pipe, the frequency of the $n^{th}$ harmonic is given by $f_{o,n} = n(v / 2L_o)$. The third harmonic $(n=3)$ is $f_{o,3} = 3(v / 2L_o)$.
For a closed pipe, the frequency of the $m^{th}$ harmonic is given by $f_{c,m} = m(v / 4L_c)$, where $m$ must be an odd integer. The fifth harmonic $(m=5)$ is $f_{c,5} = 5(v / 4L_c)$.
The ratio of the frequencies is $R = f_{o,3} / f_{c,5} = [3(v / 2L_o)] / [5(v / 4L_c)]$.
$R = (3v / 2L_o) \times (4L_c / 5v) = (3 \times 4 \times L_c) / (2 \times 5 \times L_o) = (12 / 10) \times (L_c / L_o)$.
Since $L_o / L_c = 2 / 3$, then $L_c / L_o = 3 / 2$.
Substituting this value: $R = (1.2) \times (1.5) = 1.8 = 18 / 10 = 9 / 5$.
Thus, the ratio is $9 : 5$.

(D) For a tube closed at one end,the fundamental frequency is given by $f_c = \frac{v}{4L}$.
Given that the beat frequency is $f_{c1} - f_{c2} = 2$,where $f_{c1} = \frac{v}{4L_1}$ and $f_{c2} = \frac{v}{4L_2}$.
Thus,$\frac{v}{4L_1} - \frac{v}{4L_2} = 2$.
For a tube open at both ends,the fundamental frequency is given by $f_o = \frac{v}{2L}$.
The new beat frequency is $f_{o1} - f_{o2} = \frac{v}{2L_1} - \frac{v}{2L_2}$.
We can rewrite this as $2 \times (\frac{v}{4L_1} - \frac{v}{4L_2})$.
Substituting the known value,the new beat frequency is $2 \times 2 = 4$ beats per second.

(C) For a tube closed at one end,the frequencies of harmonics are given by $f_n = n \frac{v}{4L}$,where $n = 1, 3, 5, ...$ are odd integers. The fifth harmonic corresponds to $n = 5$.
The wavelength $\lambda$ is given by $\lambda = \frac{4L}{n} = \frac{4 \times 50 \ cm}{5} = 40 \ cm$.
The standing wave equation for a closed pipe (taking the open end at $x = 0$) is $y = A \sin(kx) \cos(\omega t)$,where $k = \frac{2\pi}{\lambda} = \frac{2\pi}{40} = \frac{\pi}{20} \ rad/cm$.
The particle at the open end $(x = 0)$ is at a node of the displacement wave (or antinode of pressure),but in terms of displacement amplitude,the open end is an antinode. However,the standard equation $y = A \sin(kx) \cos(\omega t)$ assumes $x=0$ is a node. For an open end at $x=0$,the displacement is $y = A \cos(kx) \cos(\omega t)$.
At $x = 0$,$y_1 = A \cos(0) \cos(\omega t) = A \cos(\omega t)$.
At $x = 42 \ cm$,$y_2 = A \cos(k \times 42) \cos(\omega t) = A \cos(\frac{\pi}{20} \times 42) \cos(\omega t) = A \cos(2.1\pi) \cos(\omega t) = A \cos(2\pi + 0.1\pi) \cos(\omega t) = A \cos(0.1\pi) \cos(\omega t)$.
Since both particles oscillate in phase (the cosine terms have the same sign),the phase difference is $0^{\circ}$.

(B) The general equation of a progressive wave is $s(x, t) = A \sin(kx - \omega t)$.
$1$. Amplitude $(A)$: The particle moves $10 \text{ cm}$ between two extreme points,which is the total path length $(2A)$. Thus,$2A = 10 \text{ cm} = 0.10 \text{ m}$,so $A = 0.05 \text{ m}$.
$2$. Angular frequency $(\omega)$: $\omega = 2\pi f = 2 \times \pi \times 210 = 420\pi \approx 1319.47 \approx 1320 \text{ rad/s}$.
$3$. Wave number $(k)$: $v = \frac{\omega}{k} \implies k = \frac{\omega}{v} = \frac{420\pi}{330} = \frac{42\pi}{33} = \frac{14\pi}{11} \approx 4 \text{ rad/m}$.
$4$. Substituting these values into the wave equation: $s(x, t) = 0.05 \sin(4x - 1320t) \text{ m}$.

(D) The fundamental frequency $f$ of a stretched string is given by the formula: $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $L$ is the length,$T$ is the tension,and $\mu$ is the linear mass density.
Given $f_1 = 300 \ Hz$,$L_1 = L$,and $T_1 = T_1$. Thus,$300 = \frac{1}{2L} \sqrt{\frac{T_1}{\mu}}$.
Given $f_2 = 100 \ Hz$,$L_2 = 2L$,and $T_2 = T_2$. Thus,$100 = \frac{1}{2(2L)} \sqrt{\frac{T_2}{\mu}} = \frac{1}{4L} \sqrt{\frac{T_2}{\mu}}$.
Dividing the two equations: $\frac{300}{100} = \frac{\frac{1}{2L} \sqrt{\frac{T_1}{\mu}}}{\frac{1}{4L} \sqrt{\frac{T_2}{\mu}}}$.
$3 = \frac{4L}{2L} \sqrt{\frac{T_1}{T_2}} = 2 \sqrt{\frac{T_1}{T_2}}$.
$1.5 = \sqrt{\frac{T_1}{T_2}}$.
Squaring both sides: $2.25 = \frac{T_1}{T_2}$,which is $\frac{9}{4} = \frac{T_1}{T_2}$.
Therefore,$\frac{T_2}{T_1} = \frac{4}{9}$.

(D) The fundamental frequency of a vibrating 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 linear mass density.
Initially,$f_1 = \frac{1}{2L_1} \sqrt{\frac{T_1}{\mu}}$.
Given that the length is decreased by $25 \%$,the new length $L_2 = L_1 - 0.25L_1 = 0.75L_1 = \frac{3}{4}L_1$.
The new frequency $f_2$ increases by $100 \%$,so $f_2 = f_1 + 1.00f_1 = 2f_1$.
Using the frequency formula for the new state: $f_2 = \frac{1}{2L_2} \sqrt{\frac{T_2}{\mu}}$.
Substituting $L_2$ and $f_2$: $2f_1 = \frac{1}{2(\frac{3}{4}L_1)} \sqrt{\frac{T_2}{\mu}} = \frac{4}{6L_1} \sqrt{\frac{T_2}{\mu}} = \frac{2}{3L_1} \sqrt{\frac{T_2}{\mu}}$.
Dividing the expression for $f_2$ by $f_1$: $\frac{2f_1}{f_1} = \frac{\frac{2}{3L_1} \sqrt{\frac{T_2}{\mu}}}{\frac{1}{2L_1} \sqrt{\frac{T_1}{\mu}}}$.
$2 = \frac{2}{3} \cdot 2 \cdot \sqrt{\frac{T_2}{T_1}} = \frac{4}{3} \sqrt{\frac{T_2}{T_1}}$.
$\sqrt{\frac{T_2}{T_1}} = 2 \cdot \frac{3}{4} = \frac{3}{2}$.
Squaring both sides,$\frac{T_2}{T_1} = (\frac{3}{2})^2 = \frac{9}{4}$.

(A) For a rod clamped at its midpoint,the midpoint acts as a node and the two ends act as antinodes for the fundamental longitudinal vibration.
Let $L$ be the length of the rod. The distance between two consecutive antinodes is $\lambda/2$. Here,the rod is clamped at the center,so the length $L$ corresponds to the distance between the two ends (antinodes),which is $\lambda/2$.
Thus,$L = \lambda/2$,which implies $\lambda = 2L$.
The fundamental frequency $f$ is given by $f = v/\lambda$,where $v$ is the speed of sound.
Substituting $\lambda = 2L$,we get $f = v/(2L)$.
Given: $L = 125 \ cm = 1.25 \ m$ and $v = 5000 \ ms^{-1}$.
$f = 5000 / (2 \times 1.25) = 5000 / 2.5 = 2000 \ Hz$.
$2000 \ Hz = 2 \ kHz$.

(B) Given: Mass $m = 500 \ g = 0.5 \ kg$, Height $h = 3.2 \ m$, Initial velocity $u = 0 \ ms^{-1}$, Final velocity $v = 6 \ ms^{-1}$, Acceleration due to gravity $g = 10 \ ms^{-2}$.
The initial potential energy of the body at height $h$ is $PE_i = mgh = 0.5 \times 10 \times 3.2 = 16 \ J$.
The initial kinetic energy is $KE_i = \frac{1}{2}mu^2 = 0 \ J$.
Total initial mechanical energy $E_i = PE_i + KE_i = 16 \ J$.
The final potential energy at the ground is $PE_f = 0 \ J$.
The final kinetic energy is $KE_f = \frac{1}{2}mv^2 = \frac{1}{2} \times 0.5 \times (6)^2 = 0.25 \times 36 = 9 \ J$.
Total final mechanical energy $E_f = PE_f + KE_f = 9 \ J$.
The energy lost due to air resistance is the difference between the initial and final mechanical energy:
$Loss = E_i - E_f = 16 \ J - 9 \ J = 7 \ J$.

(D) The kinetic energy $(K)$ of a body is given by the formula $K = \frac{1}{2}mv^2$, where $m$ is the mass and $v$ is the magnitude of the velocity.
First, calculate the magnitude of the velocity vector $\vec{v} = (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) \text{ m/s}$.
$v = |\vec{v}| = \sqrt{2^2 + 3^2 + (-4)^2} = \sqrt{4 + 9 + 16} = \sqrt{29} \text{ m/s}$.
Thus, $v^2 = 29 \text{ m}^2/\text{s}^2$.
Given $K = 87 \text{ J}$, substitute the values into the kinetic energy formula:
$87 = \frac{1}{2} \times m \times 29$.
$87 = 14.5 \times m$.
$m = \frac{87}{14.5} = 6 \text{ kg}$.
Therefore, the mass of the body is $6 \text{ kg}$.

(A) The power supplied to the body is given by $P = \frac{dK}{dt} = 3t^2 + 3$,where $K$ is the kinetic energy.
Integrating with respect to time $t$ from $0$ to $3 \ s$:
$\int_{0}^{K} dK = \int_{0}^{3} (3t^2 + 3) dt$
$K = [t^3 + 3t]_{0}^{3} = (3^3 + 3(3)) - 0 = 27 + 9 = 36 \ J$.
Since the initial velocity is zero,the initial kinetic energy is zero.
Using the formula for kinetic energy $K = \frac{1}{2}mv^2$:
$36 = \frac{1}{2} \times 0.5 \times v^2$
$36 = 0.25 \times v^2$
$v^2 = \frac{36}{0.25} = 144$
$v = \sqrt{144} = 12 \ ms^{-1}$.

(A) The work done $W$ by a variable force $F(y)$ in displacing a particle from $y_1$ to $y_2$ is given by the integral: $W = \int_{y_1}^{y_2} F(y) dy$.
Given $F(y) = -\frac{K}{y^2}$,$y_1 = a$,and $y_2 = 2a$.
Substituting these values: $W = \int_{a}^{2a} (-\frac{K}{y^2}) dy$.
$W = -K \int_{a}^{2a} y^{-2} dy$.
Integrating $y^{-2}$ gives $-y^{-1} = -\frac{1}{y}$.
$W = -K [-\frac{1}{y}]_{a}^{2a}$.
$W = K [\frac{1}{y}]_{a}^{2a}$.
$W = K (\frac{1}{2a} - \frac{1}{a})$.
$W = K (\frac{1-2}{2a}) = K (-\frac{1}{2a}) = -\frac{K}{2a}$.

(C) Given: Force $F = 10 \ N$,mass $m = 1.5 \ kg$,coefficient of kinetic friction $\mu_k = 0.2$,$g = 10 \ m/s^2$,initial velocity $u = 0$,time $t = 6 \ s$.
First,calculate the kinetic friction force: $f_k = \mu_k \cdot m \cdot g = 0.2 \times 1.5 \times 10 = 3 \ N$.
The net force acting on the block is $F_{net} = F - f_k = 10 - 3 = 7 \ N$.
The acceleration of the block is $a = F_{net} / m = 7 / 1.5 = 14 / 3 \ m/s^2$.
The displacement $s$ in $t = 6 \ s$ is given by $s = ut + (1/2)at^2 = 0 + (1/2) \times (14/3) \times (6)^2 = (1/2) \times (14/3) \times 36 = 7 \times 12 = 84 \ m$.
The work done by the applied force is $W = F \times s = 10 \times 84 = 840 \ J$.

(B) Let the prism angle be $A = 90^{\circ}$ and refractive index be $\mu = \sqrt{2}$.
For total internal reflection $(TIR)$ at the second face, the angle of incidence at the second face $r_2$ must be greater than or equal to the critical angle $C$.
The critical angle $C$ is given by $\sin C = \frac{1}{\mu} = \frac{1}{\sqrt{2}}$, which gives $C = 45^{\circ}$.
Thus, for $TIR$, $r_2 > 45^{\circ}$.
From the geometry of the prism, $r_1 + r_2 = A = 90^{\circ}$.
Substituting $r_2 > 45^{\circ}$, we get $r_1 < 90^{\circ} - 45^{\circ} = 45^{\circ}$.
Using Snell's law at the first face: $\sin i = \mu \sin r_1$.
Since $r_1 < 45^{\circ}$, $\sin r_1 < \sin 45^{\circ} = \frac{1}{\sqrt{2}}$.
Therefore, $\sin i < \sqrt{2} \times \frac{1}{\sqrt{2}} = 1$, which implies $i < 90^{\circ}$.
However, for the light to enter the prism and satisfy the condition, we look for the boundary condition where $r_2 = 45^{\circ}$, which implies $r_1 = 45^{\circ}$.
Then $\sin i = \sqrt{2} \sin 45^{\circ} = \sqrt{2} \times \frac{1}{\sqrt{2}} = 1$, so $i = 90^{\circ}$.
Since the question asks for the condition for $TIR$, any angle of incidence $i < 90^{\circ}$ will result in $r_1 > 45^{\circ}$ which leads to $r_2 < 45^{\circ}$ (no $TIR$). Thus, the boundary is at $i = 90^{\circ}$.

(B) When light passes through a thick lens,it behaves similarly to a prism.
Different wavelengths (colors) of light travel at different speeds within the lens material,leading to different refractive indices for each color.
This phenomenon,where white light splits into its constituent colors due to the variation of refractive index with wavelength,is known as dispersion.
Consequently,the lens focuses different colors at slightly different points,resulting in colored fringes or blurred images,a defect known as chromatic aberration.
Therefore,the formation of colored images due to thick lenses is explained by the property of dispersion.

(A) For a small-angled prism,the angle of deviation $\delta$ is given by the formula: $\delta = (\mu - 1)A$,where $\mu$ is the refractive index and $A$ is the angle of the prism.
The angle of dispersion (angular dispersion) $\theta$ is the difference between the angles of deviation for violet and red light: $\theta = \delta_v - \delta_r$.
Substituting the formula for deviation: $\theta = (\mu_v - 1)A - (\mu_r - 1)A = (\mu_v - \mu_r)A$.
Given values: $\mu_v = 1.52$,$\mu_r = 1.48$,and $A = 6^{\circ}$.
Calculation: $\theta = (1.52 - 1.48) \times 6^{\circ} = 0.04 \times 6^{\circ} = 0.24^{\circ}$.
Therefore,the angle of dispersion is $0.24^{\circ}$.

(C) For an equilateral prism,the angle of the prism $A = 60^{\circ}$.
Given that the angle of minimum deviation $\delta_m = A = 60^{\circ}$.
The formula for the refractive index $n$ of a prism is given by $n = \frac{\sin((A + \delta_m) / 2)}{\sin(A / 2)}$.
Substituting the values,we get $n = \frac{\sin((60^{\circ} + 60^{\circ}) / 2)}{\sin(60^{\circ} / 2)}$.
$n = \frac{\sin(120^{\circ} / 2)}{\sin(30^{\circ})} = \frac{\sin(60^{\circ})}{\sin(30^{\circ})}$.
Since $\sin(60^{\circ}) = \frac{\sqrt{3}}{2}$ and $\sin(30^{\circ}) = \frac{1}{2}$,we have $n = \frac{\sqrt{3} / 2}{1 / 2} = \sqrt{3}$.
The value of $\sqrt{3}$ is approximately $1.732$.

(C) For a small-angle prism,the relationship between the angle of incidence $i$,the angle of refraction $r_1$ at the first face,the angle of refraction $r_2$ at the second face,and the angle of emergence $e$ is given by:
$A = r_1 + r_2$
Given that the ray emerges normally from the second face,the angle of emergence $e = 0^{\circ}$,which implies that $r_2 = 0^{\circ}$.
Since the refracting angle $A = 6^{\circ}$,we have $r_1 = A - r_2 = 6^{\circ} - 0^{\circ} = 6^{\circ}$.
According to Snell's Law at the first face,$n_1 \sin(i) = n_2 \sin(r_1)$,where $n_1 = 1$ (air) and $n_2 = n$ (prism material).
$1 \cdot \sin(9.3^{\circ}) = n \cdot \sin(6^{\circ})$.
Using the small-angle approximation $\sin(\theta) \approx \theta$ (in radians) or simply the ratio of angles for small values:
$n = \frac{\sin(9.3^{\circ})}{\sin(6^{\circ})} \approx \frac{9.3}{6} = 1.55$.
Thus,the refractive index of the material of the prism is $1.55$.

(A) The initial kinetic energy of the electron is $K_i = \frac{1}{2}mv^2$.
As the electron moves through the barrier potential $V$,it loses energy equal to $eV$.
The final kinetic energy $K_f$ is given by $K_f = K_i - eV = \frac{1}{2}mv_f^2$.
Substituting the values: $K_i = \frac{1}{2} \times (9 \times 10^{-31}) \times (5 \times 10^5)^2 = 0.5 \times 9 \times 10^{-31} \times 25 \times 10^{10} = 112.5 \times 10^{-21} \ J$.
The energy lost is $eV = 1.6 \times 10^{-19} \times 0.45 = 0.72 \times 10^{-19} = 72 \times 10^{-21} \ J$.
So,$K_f = 112.5 \times 10^{-21} - 72 \times 10^{-21} = 40.5 \times 10^{-21} \ J$.
Now,$\frac{1}{2}mv_f^2 = 40.5 \times 10^{-21} \implies v_f^2 = \frac{2 \times 40.5 \times 10^{-21}}{9 \times 10^{-31}} = 9 \times 10^{10} \ m^2/s^2$.
Therefore,$v_f = 3 \times 10^5 \ m/s$.

(A) Given: Power gain $(A_p)$ = $1800$, Voltage gain $(A_v)$ = $60$, Change in emitter current $(\Delta I_e)$ = $0.62 \text{ mA}$.
We know that Power gain $(A_p)$ = Current gain $(\beta)$ $\times$ Voltage gain $(A_v)$.
Therefore, $\beta = A_p / A_v = 1800 / 60 = 30$.
We also know that the current gain in common emitter configuration is $\beta = \Delta I_c / \Delta I_b$, and $\Delta I_e = \Delta I_b + \Delta I_c$.
Substituting $\Delta I_b = \Delta I_c / \beta$, we get $\Delta I_e = \Delta I_c / \beta + \Delta I_c = \Delta I_c (1 + 1/\beta) = \Delta I_c ((\beta + 1) / \beta)$.
Rearranging for $\Delta I_c$: $\Delta I_c = \Delta I_e \times (\beta / (\beta + 1))$.
Substituting the values: $\Delta I_c = 0.62 \times (30 / 31) = 0.62 \times (0.9677) \approx 0.60 \text{ mA}$.

(D) The emitter current $I_E$ is given by the rate of flow of charge: $I_E = \frac{q}{t} = \frac{n \cdot e}{t}$.
Given $n = 10^{10}$, $e = 1.6 \times 10^{-19} \text{ C}$, and $t = 0.4 \times 10^{-6} \text{ s}$.
$I_E = \frac{10^{10} \times 1.6 \times 10^{-19}}{0.4 \times 10^{-6}} = \frac{1.6 \times 10^{-9}}{0.4 \times 10^{-6}} = 4 \times 10^{-3} \text{ A} = 4 \text{ mA}$.
Since $5 \%$ of the electrons are lost in the base, the collector current $I_C$ is $95 \%$ of the emitter current $I_E$.
$I_C = 0.95 \times I_E = 0.95 \times 4 \text{ mA} = 3.8 \text{ mA}$.

(A) transistor acts as a switch when it is operated between two states: the $OFF$ state and the $ON$ state.
In the $OFF$ state,the transistor is in the cutoff region,where the collector current is zero.
In the $ON$ state,the transistor is in the saturation region,where the collector current is maximum and the output voltage is very low.
The active region is used for amplification,not for switching.
Therefore,the transistor acts as a switch in the cutoff or saturation region.

(B) In a common emitter transistor configuration,the transfer characteristic curve shows three distinct regions:
$1$. Cutoff Region: When the input voltage $(V_{i})$ is low,the transistor is in the cutoff region,and the output voltage $(V_{o})$ is high (near $V_{CC}$). This corresponds to Region $I$.
$2$. Active Region: As $(V_{i})$ increases,the transistor enters the active region where the output voltage $(V_{o})$ decreases linearly with $(V_{i})$. This corresponds to Region $II$.
$3$. Saturation Region: When $(V_{i})$ is high,the transistor enters the saturation region,and the output voltage $(V_{o})$ becomes very low (near $0$). This corresponds to Region $III$.
Therefore,the active,saturation,and cutoff regions are represented by $II, III$ and $I$ respectively.

(D) In a common emitter configuration,the current amplification factor is given by $\beta = 80$.
The relationship between emitter current $(I_E)$,collector current $(I_C)$,and base current $(I_B)$ is $I_E = I_C + I_B$.
We also know that $\beta = \frac{I_C}{I_B}$,which implies $I_C = \beta I_B$.
Substituting $I_C$ in the first equation: $I_E = \beta I_B + I_B = I_B(\beta + 1)$.
Given $I_E = 2.43 \text{ mA} = 2430 \mu \text{A}$ and $\beta = 80$.
$I_B = \frac{I_E}{\beta + 1} = \frac{2430 \mu \text{A}}{80 + 1} = \frac{2430 \mu \text{A}}{81}$.
$I_B = 30 \mu \text{A}$.

(B) Negative feedback in an amplifier is a technique where a portion of the output signal is fed back to the input in phase opposition to the original input signal.
This process reduces the overall gain of the amplifier but provides several benefits.
Specifically,negative feedback significantly reduces the noise and distortion introduced by the amplifier circuitry itself.
It also increases the bandwidth and stabilizes the gain against variations in temperature or component aging.
Therefore,the correct effect of negative feedback is the reduction of both noise and distortion.

(D) Let us analyze the circuit step by step:
$1$. The top $NAND$ gate has inputs $1$ and $0$. The output is $(1 \cdot 0)' = 0' = 1$.
$2$. The middle $AND$ gate has inputs $1$ and $1$. The output is $1 \cdot 1 = 1$.
$3$. The bottom $OR$ gate has inputs $0$ and $1$. The output is $0 + 1 = 1$.
$4$. The top $NOR$ gate receives inputs from the $NAND$ gate $(1)$ and the $AND$ gate $(1)$. Its output $y_1 = (1 + 1)' = 1' = 0$.
$5$. The bottom $NOR$ gate receives inputs from the $AND$ gate $(1)$ and the $OR$ gate $(1)$. Its output $y_2 = (1 + 1)' = 1' = 0$.
$6$. The final $AND$ gate receives inputs $y_1 = 0$ and $y_2 = 0$. Its output $y_3 = 0 \cdot 0 = 0$.
Therefore,the values are $y_1 = 0, y_2 = 0, y_3 = 0$.

(D) universal gate is a logic gate that can be used to implement any other logic gate or Boolean function without the need for any other type of gate.
$NAND$ and $NOR$ gates are known as universal gates.
In the given options,$NAND$ is a universal gate.
Therefore,the correct option is $D$.

(A) The circuit consists of a $NAND$ gate,a $NOR$ gate,and an $AND$ gate.
$1$. The $NAND$ gate has inputs $A=1$ and $B=0$. The output $y_1$ of a $NAND$ gate is given by $y_1 = \overline{A \cdot B}$. Thus,$y_1 = \overline{1 \cdot 0} = \overline{0} = 1$.
$2$. The $NOR$ gate has inputs $B=0$ and $C=1$. The output $y_2$ of a $NOR$ gate is given by $y_2 = \overline{B + C}$. Thus,$y_2 = \overline{0 + 1} = \overline{1} = 0$.
$3$. The $AND$ gate has inputs $y_1=1$ and $y_2=0$. The output $y_3$ of an $AND$ gate is given by $y_3 = y_1 \cdot y_2$. Thus,$y_3 = 1 \cdot 0 = 0$.
Therefore,the values are $y_1=1, y_2=0, y_3=0$.

(D) At absolute zero temperature $(T = 0 \ K)$,all the valence electrons are tightly bound to their respective atoms in the crystal lattice.
There is no thermal energy available to excite electrons from the valence band to the conduction band.
Consequently,the conduction band remains completely empty,and the valence band is completely filled.
Due to the absence of free charge carriers,the semiconductor acts as a perfect insulator.

(B) The energy of the emitted photon is given by the relation $E = \frac{hc}{\lambda}$.
Given the energy gap $E = 1.9 \ eV$.
We know that $hc \approx 1240 \ eV \cdot nm$.
Therefore,the wavelength $\lambda = \frac{hc}{E} = \frac{1240 \ eV \cdot nm}{1.9 \ eV} \approx 652.6 \ nm$.
The wavelength range for red light is approximately $620 \ nm$ to $750 \ nm$.
Since $652.6 \ nm$ falls within this range,the color of the light emitted by the $LED$ is red.

(A) The resonant frequency of an $LC$ circuit is given by the formula $\omega = \frac{1}{\sqrt{LC}}$.
Here,$\omega$ represents the angular frequency,which has the dimensions of $[T^{-1}]$.
Therefore,$(LC)^{-\frac{1}{2}} = \frac{1}{\sqrt{LC}} = \omega$.
The dimensional formula for angular frequency is $[M^0 L^0 T^{-1}]$.
Thus,the correct option is $A$.

(A) The Fresnel distance $(z_F)$ is the distance up to which ray optics is a good approximation. It is given by the formula:
$z_F = \frac{a^2}{\lambda}$
where $a$ is the aperture size and $\lambda$ is the wavelength of light.
Given:
Aperture $a = 5 \times 10^{-3} \ m$
Fresnel distance $z_F = 50 \ m$
Substituting the values into the formula:
$50 = \frac{(5 \times 10^{-3})^2}{\lambda}$
$50 = \frac{25 \times 10^{-6}}{\lambda}$
$\lambda = \frac{25 \times 10^{-6}}{50}$
$\lambda = 0.5 \times 10^{-6} \ m$
$\lambda = 5 \times 10^{-7} \ m$
Converting to $\text{Å}$s $(\text{Å})$:
$1 \ \text{Å} = 10^{-10} \ m$
$\lambda = 5 \times 10^{-7} \times 10^{10} \ \text{Å} = 5000 \ \text{Å}$
Therefore, the correct option is $A$.

(B) The Fresnel distance $(z_F)$ is the distance at which ray optics is a good approximation for an aperture of size $a$ and wavelength $\lambda$. It is given by the formula: $z_F = \frac{a^2}{\lambda}$.
Given:
Aperture $a = 0.3 \ cm = 0.3 \times 10^{-2} \ m = 3 \times 10^{-3} \ m$.
Wavelength $\lambda = 6000 \ Å = 6000 \times 10^{-10} \ m = 6 \times 10^{-7} \ m$.
Substituting the values into the formula:
$z_F = \frac{(3 \times 10^{-3})^2}{6 \times 10^{-7}}$
$z_F = \frac{9 \times 10^{-6}}{6 \times 10^{-7}}$
$z_F = 1.5 \times 10^1 = 15 \ m$.
Therefore,the correct option is $B$.

(B) Let the intensity of each wave be $I_0$. The maximum intensity $I_{max}$ for two waves of intensity $I_1$ and $I_2$ is given by $I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2$.
Given $I_1 = I_2 = I_0$,the maximum intensity is $I = (\sqrt{I_0} + \sqrt{I_0})^2 = (2\sqrt{I_0})^2 = 4I_0$. Thus,$I_0 = I/4$.
Now,the intensity of one wave is quadrupled,so $I_1' = 4I_0 = 4(I/4) = I$ and $I_2' = I_0 = I/4$.
The new maximum intensity $I_{max}'$ is $(\sqrt{I_1'} + \sqrt{I_2'})^2 = (\sqrt{I} + \sqrt{I/4})^2 = (\sqrt{I} + \frac{\sqrt{I}}{2})^2 = (\frac{3\sqrt{I}}{2})^2 = \frac{9I}{4}$.

(A) According to Brewster's Law,when light is incident at the polarizing angle $(i_p)$,the reflected light is totally plane-polarized.
At this angle,the reflected ray and the refracted ray are perpendicular to each other.
Brewster's Law states that $\tan(i_p) = \mu$.
Given $\mu = \sqrt{3}$,we have $\tan(i_p) = \sqrt{3}$,which implies $i_p = 60^{\circ}$.
According to Snell's Law,$\mu_1 \sin(i_p) = \mu_2 \sin(r)$,where $r$ is the angle of refraction.
Here,$\mu_1 = 1$ (air) and $\mu_2 = \sqrt{3}$.
$1 \cdot \sin(60^{\circ}) = \sqrt{3} \cdot \sin(r)$.
$\frac{\sqrt{3}}{2} = \sqrt{3} \cdot \sin(r)$.
$\sin(r) = \frac{1}{2}$.
Therefore,$r = 30^{\circ}$.

(D) The intensity at any point in an interference pattern is given by $I_p = I_{max} \cos^2(\frac{\phi}{2})$,where $\phi$ is the phase difference.
Given that $I_{max} = I$ (intensity of the central bright fringe).
The relationship between phase difference $\phi$ and path difference $\Delta x$ is $\phi = \frac{2\pi}{\lambda} \Delta x$.
Given $\Delta x = \frac{\lambda}{3}$,we have $\phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{3} = \frac{2\pi}{3}$.
Substituting this into the intensity formula:
$I_p = I \cos^2(\frac{2\pi/3}{2}) = I \cos^2(\frac{\pi}{3})$.
Since $\cos(\frac{\pi}{3}) = \frac{1}{2}$,we get $I_p = I (\frac{1}{2})^2 = \frac{I}{4}$.

(C) The position of the $n^{\text{th}}$ bright fringe is given by $y_n = n \beta$, where $\beta = \frac{\lambda D}{d}$.
The position of the $n^{\text{th}}$ dark fringe is given by $y'_n = (n - 0.5) \beta$.
Given the distance between the $5^{\text{th}}$ bright fringe $(y_5 = 5 \beta)$ and the $7^{\text{th}}$ dark fringe $(y'_7 = (7 - 0.5) \beta = 6.5 \beta)$ is $3 \text{ mm}$:
$|6.5 \beta - 5 \beta| = 3 \text{ mm} \implies 1.5 \beta = 3 \text{ mm} \implies \beta = 2 \text{ mm}$.
Now, we need to find the distance between the $5^{\text{th}}$ dark fringe $(y'_5 = (5 - 0.5) \beta = 4.5 \beta)$ and the $7^{\text{th}}$ bright fringe $(y_7 = 7 \beta)$:
Distance $= |7 \beta - 4.5 \beta| = 2.5 \beta$.
Substituting $\beta = 2 \text{ mm}$:
Distance $= 2.5 \times 2 \text{ mm} = 5 \text{ mm}$.