MHT CET 2026 Physics Question Paper with Answer and Solution

(C) When two waves of the same frequency and velocity travel in opposite directions, they form a standing wave.
For a wave, the relationship between velocity $(v)$, frequency $(n)$, and wavelength $(\lambda)$ is given by $v = n \lambda$.
Given $v = 18\text{ m/s}$, we have $18 = n \lambda$, which implies $\lambda = 18/n$.
In a standing wave, the distance between two consecutive antinodes is equal to half the wavelength $(\lambda/2)$.
Therefore, the distance between two consecutive antinodes $= \lambda/2 = (18/n) / 2 = 9/n\text{ m}$.

(D) The standard equation of a stationary wave is given by $y = A \cos(kx) \sin(\omega t)$.
Comparing the given equation $y = 12 \cos \left( \frac{\pi x}{10} \right) \sin(36\pi t)$ with the standard form, we get the wave number $k = \frac{\pi}{10} \text{ rad/cm}$.
The wave number $k$ is related to the wavelength $\lambda$ by the formula $k = \frac{2\pi}{\lambda}$.
Substituting the value of $k$, we have $\frac{\pi}{10} = \frac{2\pi}{\lambda}$, which gives $\lambda = 20 \text{ cm}$.
The distance between a node and the successive antinode in a stationary wave is equal to $\frac{\lambda}{4}$.
Therefore, the required distance is $\frac{20 \text{ cm}}{4} = 5 \text{ cm}$.

(B) The resultant amplitude $R$ of two interfering waves with individual amplitudes $A_1$ and $A_2$ and phase difference $\phi$ is given by $R = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos \phi}$.
Given $A_1 = A_2 = A$.
The path difference $\Delta x = \frac{\lambda}{4}$.
The phase difference $\phi$ is related to path difference by $\phi = \frac{2\pi}{\lambda} \Delta x$.
Substituting the values: $\phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2}$.
Now, calculate the resultant amplitude $R$:
$R = \sqrt{A^2 + A^2 + 2A^2 \cos(\frac{\pi}{2})}$
Since $\cos(\frac{\pi}{2}) = 0$, we get:
$R = \sqrt{A^2 + A^2 + 0} = \sqrt{2A^2} = \sqrt{2}A$.

(C) The standard equation of a progressive wave is $Y = A \sin \left( \omega t - kx + \phi \right)$.
Comparing the given equation $Y = 3 \sin \left[ \pi \left( \frac{t}{3} - \frac{x}{5} \right) + \frac{\pi}{4} \right]$ with the standard form:
$Y = 3 \sin \left( \frac{\pi t}{3} - \frac{\pi x}{5} + \frac{\pi}{4} \right)$.
Here, Amplitude $A = 3 \text{ m} = 300 \text{ cm}$.
Angular frequency $\omega = \frac{\pi}{3} \text{ rad/s}$.
Wave number $k = \frac{\pi}{5} \text{ m}^{-1}$.
$1$. Frequency $f = \frac{\omega}{2\pi} = \frac{\pi/3}{2\pi} = \frac{1}{6} \text{ Hz} \approx 0.167 \text{ Hz}$.
$2$. Wavelength $\lambda = \frac{2\pi}{k} = \frac{2\pi}{\pi/5} = 10 \text{ m}$.
$3$. Velocity $v = \frac{\omega}{k} = \frac{\pi/3}{\pi/5} = \frac{5}{3} \approx 1.67 \text{ m/s}$.
Comparing these results with the options, option $C$ is correct.

(C) The energy $E$ of a sound wave is proportional to the square of its amplitude $A$ and the square of its frequency $n$. Mathematically, $E \propto A^2 n^2$.
Since the energies of the waves produced by $P$ and $Q$ are equal, we have $E_P = E_Q$.
Therefore, $A_P^2 n_P^2 = A_Q^2 n_Q^2$.
Given $A_P = A$, $n_P = n$, and $n_Q = n/4$, we substitute these values into the equation:
$A^2 n^2 = A_Q^2 (n/4)^2$.
$A^2 n^2 = A_Q^2 (n^2 / 16)$.
$A^2 = A_Q^2 / 16$.
$A_Q^2 = 16 A^2$.
Taking the square root of both sides, we get $A_Q = 4A$.

(C) The velocity of the wave is given as $v = 10 \text{ m/s}$.
The distance between two consecutive crests is the wavelength, $\lambda = 40 \text{ m}$.
The time interval between two consecutive waves striking the boat is equal to the time period $(T)$ of the wave.
The relationship between velocity, wavelength, and time period is given by $v = \frac{\lambda}{T}$.
Rearranging for $T$, we get $T = \frac{\lambda}{v}$.
Substituting the given values: $T = \frac{40 \text{ m}}{10 \text{ m/s}} = 4 \text{ s}$.
Therefore, the consecutive waves strike the boat after $4 \text{ s}$.

(D) The given equations are $y_1 = a \sin(\omega t + kx + 0.57)$ and $y_2 = a \cos(\omega t + kx)$.
To find the phase difference, we must express both waves in terms of the same trigonometric function (sine).
We know that $\cos(\theta) = \sin(\theta + \pi/2)$.
Therefore, $y_2 = a \sin(\omega t + kx + \pi/2)$.
Given $\pi = 3.14$, then $\pi/2 = 3.14 / 2 = 1.57 \text{ radians}$.
So, $y_2 = a \sin(\omega t + kx + 1.57)$.
The phase of $y_1$ is $\phi_1 = \omega t + kx + 0.57$ and the phase of $y_2$ is $\phi_2 = \omega t + kx + 1.57$.
The phase difference $\Delta\phi = |\phi_2 - \phi_1| = |1.57 - 0.57| = 1.0 \text{ radian}$.

(B) The given equation of the wave is $y = A \sin(120\pi t + 3x)$.
Comparing this with the standard wave equation $y = A \sin(\omega t + kx)$, we get the wave number $k = 3 \text{ rad/m}$.
The relationship between phase difference $(\Delta \phi)$ and path difference $(\Delta x)$ is given by $\Delta \phi = k \cdot \Delta x$.
Given $\Delta \phi = \frac{\pi}{3} \text{ radian}$ and $k = 3 \text{ rad/m}$.
Substituting the values: $\frac{\pi}{3} = 3 \cdot \Delta x$.
Therefore, $\Delta x = \frac{\pi}{3 \times 3} = \frac{\pi}{9} \text{ m}$.

(C) The angular frequency $\omega$ is given by $\omega = 2\pi f$, where $f = 50\text{ Hz}$.
$\omega = 2 \times \pi \times 50 = 100\pi\text{ rad/s}$.
The change in phase $\Delta\phi$ for a time interval $\Delta t$ is given by $\Delta\phi = \omega \Delta t$.
Given $\Delta t = 0.01\text{ s}$.
$\Delta\phi = 100\pi \times 0.01 = \pi\text{ rad}$.
Thus, the change in phase is $\pi\text{ rad}$.

(A) Given:
Number of waves $(n)$ = $54$
Time $(t)$ = $1 \text{ minute} = 60 \text{ seconds}$
Wavelength $(\lambda)$ = $5 \text{ m}$
Step $1$: Calculate the frequency $(f)$ of the waves.
Frequency is the number of waves per second.
$f = \frac{n}{t} = \frac{54}{60} \text{ Hz} = 0.9 \text{ Hz}$
Step $2$: Calculate the velocity $(v)$ of the waves.
The relationship between velocity, frequency, and wavelength is given by:
$v = f \times \lambda$
$v = 0.9 \text{ Hz} \times 5 \text{ m} = 4.5 \text{ m/s}$
Therefore, the velocity of the waves is $4.5 \text{ m/s}$.

(D) The frequency of a wave remains constant when it travels from one medium to another.
Let the frequency be $f$.
The relationship between velocity $(v)$, frequency $(f)$, and wavelength $(\lambda)$ is given by $v = f \lambda$.
For medium $P$: $\lambda_P = 150 \text{ mm} = 0.15 \text{ m}$ and $v_P = 80 \text{ cm s}^{-1} = 0.8 \text{ m s}^{-1}$.
Frequency $f = \frac{v_P}{\lambda_P} = \frac{0.8}{0.15} = \frac{80}{15} = \frac{16}{3} \text{ Hz}$.
For medium $Q$: $\lambda_Q = 0.30 \text{ m}$.
The velocity in medium $Q$ is $v_Q = f \lambda_Q$.
$v_Q = \left( \frac{16}{3} \right) \times 0.30 = \frac{16}{3} \times \frac{30}{100} = 16 \times 0.1 = 1.6 \text{ m s}^{-1}$.

(C) The given wave equation is $y = 6 \cos(100t - 4x)$.
Comparing this with the standard wave equation $y = A \cos(\omega t - kx)$, we get:
Amplitude $A = 6 \mu\text{m} = 6 \times 10^{-6} \text{ m}$.
Angular frequency $\omega = 100 \text{ rad/s}$.
Wave number $k = 4 \text{ m}^{-1}$.
The maximum particle velocity $v_p$ is given by $v_p = A\omega$.
$v_p = (6 \times 10^{-6} \text{ m}) \times (100 \text{ rad/s}) = 6 \times 10^{-4} \text{ m/s}$.
The wave velocity $v$ is given by $v = \frac{\omega}{k}$.
$v = \frac{100}{4} = 25 \text{ m/s}$.
The ratio of maximum particle velocity to wave velocity is $\frac{v_p}{v} = \frac{6 \times 10^{-4}}{25}$.
$\frac{v_p}{v} = 0.24 \times 10^{-4} = 2.4 \times 10^{-5}$.

(C) First, express both waves in the same trigonometric function (sine).
$y_1 = a_1 \sin \left( \omega t - \frac{2\pi x}{\lambda} \right)$
$y_2 = a_2 \cos \left( \omega t - \frac{2\pi x}{\lambda} + \frac{\pi}{6} \right) = a_2 \sin \left( \omega t - \frac{2\pi x}{\lambda} + \frac{\pi}{6} + \frac{\pi}{2} \right)$
$y_2 = a_2 \sin \left( \omega t - \frac{2\pi x}{\lambda} + \frac{2\pi}{3} \right)$
The phase difference $\phi$ between the two waves is the difference in their arguments:
$\phi = \left( \omega t - \frac{2\pi x}{\lambda} + \frac{2\pi}{3} \right) - \left( \omega t - \frac{2\pi x}{\lambda} \right) = \frac{2\pi}{3}$
The relationship between path difference $\Delta x$ and phase difference $\phi$ is given by $\phi = \frac{2\pi}{\lambda} \Delta x$.
Substituting the values: $\frac{2\pi}{3} = \frac{2\pi}{\lambda} \Delta x$
$\Delta x = \frac{\lambda}{3}$

(C) In an adiabatic process, there is no exchange of heat with the surroundings, i.e.,$dQ = 0$.
According to the first law of thermodynamics, $dQ = dU + dW$.
Since $dQ = 0$, we have $dU = -dW$.
When a system is compressed, work is done on the system, so $dW < 0$.
Therefore, $dU = -dW > 0$, which means the internal energy of the system increases.
Since the internal energy of an ideal gas is directly proportional to its temperature $(U \propto T)$, an increase in internal energy leads to an increase in the temperature of the system.

(B) An isochoric process is a thermodynamic process in which the volume of the system remains constant $(V = \text{constant})$.
In a pressure-volume $(P-V)$ diagram, a constant volume process is represented by a vertical line, because for any change in pressure, the volume value on the $x$-axis remains the same.
Looking at the given options:
Option $A$ represents a process where $P \propto V$ (e.g.,an ideal gas at constant temperature if $P$ vs $1/V$ were plotted, or a specific linear relationship).
Option $B$ represents a vertical line where $V$ is constant, which corresponds to an isochoric process.
Option $C$ represents an isothermal process for an ideal gas $(PV = \text{constant})$.
Option $D$ represents an isobaric process where pressure remains constant $(P = \text{constant})$.
Therefore, the correct graph for an isochoric process is shown in option $B$.

(B) Free expansion is an irreversible process in which a gas expands into a vacuum.
$1$. Since the gas expands into a vacuum, the external pressure $P_{ext} = 0$, so the work done $W = \int P_{ext} dV = 0$.
$2$. The process is adiabatic because it occurs in an isolated system where no heat is exchanged $(Q = 0)$.
$3$. According to the first law of thermodynamics, $\Delta U = Q - W$. Since $Q = 0$ and $W = 0$, the internal energy change $\Delta U = 0$.
$4$. Because the process is rapid and irreversible, the system does not pass through a sequence of equilibrium states. Therefore, it cannot be represented as a continuous path on a $p-V$ diagram.
Thus, the statement that a free expansion can be plotted on a $p-V$ diagram is incorrect.

(C) For an isothermal process, the temperature remains constant. According to Boyle's Law, $PV = \text{constant}$.
Initial state: $(P, V)$.
Final state: $(P_i, 2V)$.
So, $PV = P_i(2V) \implies P_i = P/2$.
For an adiabatic process, $PV^\gamma = \text{constant}$.
Initial state: $(P, V)$.
Final state: $(P_a, 2V)$.
So, $PV^\gamma = P_a(2V)^\gamma \implies P_a = P(V/2V)^\gamma = P(1/2)^\gamma = P/2^\gamma$.
Now, the ratio $P_i/P_a$ is:
$P_i/P_a = (P/2) / (P/2^\gamma) = 2^\gamma / 2 = 2^{\gamma-1}$.

(C) The work done in a cyclic process is equal to the area enclosed by the $P-V$ loop.
For a clockwise cycle, the work done is positive, and for a counter-clockwise cycle, the work done is negative.
In the given diagram, the cycle $ABCDA$ is traversed in a counter-clockwise direction.
Area of the rectangle $ABCD = \text{length} \times \text{width} = (3V_0 - V_0) \times (2P_0 - P_0) = 2V_0 \times P_0 = 2P_0V_0$.
Since the cycle is counter-clockwise, the work done is $W = -(\text{Area}) = -2P_0V_0$.

(D) The work done by a gas during a thermodynamic process is given by the formula $W = \int P \, dV$, where $P$ is the pressure and $dV$ is the change in volume.
For no work to be done by or on the gas, the condition $W = 0$ must be satisfied.
This implies that $\int P \, dV = 0$, which occurs when the change in volume $dV = 0$, meaning the volume remains constant.
$A$ process in which the volume of the system remains constant is known as an isochoric process.
Therefore, in an isochoric process, no work is done by or on the gas.

(A) For an adiabatic process, the relation between temperature and volume is given by $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.
Given, initial temperature $T_1 = 273 \ K$ (normal temperature).
Final volume $V_2 = V_1 / 4$.
Given $\gamma = 1.5$, so $\gamma - 1 = 0.5$.
Substituting the values: $273 \times V_1^{0.5} = T_2 \times (V_1 / 4)^{0.5}$.
$T_2 = 273 \times (V_1 / (V_1 / 4))^{0.5} = 273 \times (4)^{0.5} = 273 \times 2 = 546 \ K$.
The increase in temperature is $\Delta T = T_2 - T_1 = 546 - 273 = 273 \ K$.

(C) For an adiabatic process, the relationship between temperature $(T)$ and volume $(V)$ is given by $TV^{\gamma-1} = \text{constant}$.
Given the equation $TV^x = \text{constant}$, we can compare the exponents of $V$.
Therefore, $x = \gamma - 1$.
Given that for a rigid diatomic gas, $\gamma = 7/5 = 1.4$.
Substituting the value of $\gamma$ into the equation:
$x = 1.4 - 1 = 0.4$.
Thus, the value of $x$ is $0.4$.

(C) For an adiabatic process, the relationship between temperature $(T)$ and volume $(V)$ is given by $T V^{\gamma - 1} = \text{constant}$.
Thus, $T_i V_i^{\gamma - 1} = T_f V_f^{\gamma - 1}$.
Given $\gamma = 7/5$, so $\gamma - 1 = 7/5 - 1 = 2/5$.
The final volume $V_f = V_i/32$.
Substituting these values: $T_i V_i^{2/5} = (x T_i) (V_i/32)^{2/5}$.
Dividing both sides by $T_i V_i^{2/5}$: $1 = x (1/32)^{2/5}$.
$1 = x (1/(2^5))^{2/5}$.
$1 = x (1/2^2) = x (1/4)$.
Therefore, $x = 4$.

(C) For isothermal expansion, the process follows $PV = \text{constant}$.
Given initial state is $(P, V)$ and final state is $(P_I, 3V)$.
So, $P \cdot V = P_I \cdot (3V) \implies P_I = P/3$.
For adiabatic expansion, the process follows $PV^\gamma = \text{constant}$.
Given initial state is $(P, V)$ and final state is $(P_A, 3V)$.
So, $P \cdot V^\gamma = P_A \cdot (3V)^\gamma \implies P_A = P \cdot (V/3V)^\gamma = P \cdot (1/3)^\gamma = P \cdot 3^{-\gamma}$.
Now, the ratio $P_A/P_I = (P \cdot 3^{-\gamma}) / (P/3) = 3^{-\gamma} / 3^{-1} = 3^{1-\gamma}$.

(D) The linear mass density of the string is $\mu = \frac{m}{L} = \frac{10^{-3} \text{ kg}}{0.5 \text{ m}} = 2 \times 10^{-3} \text{ kg/m}$.
The velocity of the wave is $v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{0.8}{2 \times 10^{-3}}} = \sqrt{400} = 20 \text{ m/s}$.
The time taken for a pulse to travel the length of the string is $t = \frac{L}{v} = \frac{0.5 \text{ m}}{20 \text{ m/s}} = 0.025 \text{ s}$.
For constructive interference to occur between successive pulses, the first pulse must reflect from the fixed end and return to the starting point before the next pulse is generated.
Therefore, the minimum time interval is $\Delta t = 2t = 2 \times 0.025 \text{ s} = 0.05 \text{ s}$.

(C) The standard equation of a traveling wave is given by $y(x,t) = A \sin(kx - \omega t)$.
Comparing the given equation $y(x,t) = 0.05 \sin(8x - 4t)$ with the standard equation, we get:
Wave number $k = 8 \text{ rad/m}$
Angular frequency $\omega = 4 \text{ rad/s}$
The velocity of the wave $v$ is given by the formula $v = \frac{\omega}{k}$.
Substituting the values, we get $v = \frac{4}{8} = 0.5 \text{ m/s}$.

(D) The relationship between phase difference $\Delta \phi$ and path difference $\Delta x$ is given by $\Delta \phi = (2\pi / \lambda) \Delta x$.
Given that the phase difference $\Delta \phi = 60^{\circ} = \pi/3$ radians and the path difference $\Delta x = x$ meters.
Substituting these values into the formula: $\pi/3 = (2\pi / \lambda) x$.
Solving for wavelength $\lambda$: $\lambda = (2\pi / \pi) \times 3x = 6x$.
The velocity of the wave $v$ is given by $v = n\lambda$.
Substituting the value of $\lambda$: $v = n(6x) = 6nx$ meters per second.

(C) The frequency $f$ of a wave remains constant when it travels from one medium to another.
Using the relation $\lambda = v/f$, where $v$ is the speed and $f$ is the frequency:
For warm air: $\lambda_1 = v_1 / f = 350 / 700 = 0.5 \text{ m}$.
For brass: $\lambda_2 = v_2 / f = 3500 / 700 = 5 \text{ m}$.
The ratio of the wavelengths is $\lambda_2 / \lambda_1 = 5 / 0.5 = 10$.
Therefore, the wavelength increases by a factor of $10$.

(D) The standard equation of a traveling wave is given by $Y = A \sin(\omega t - kx + \phi)$.
Comparing the given equation $Y = 6 \sin (12\pi t - 0.02\pi x + \pi/3)$ with the standard equation, we get:
Angular frequency $\omega = 12\pi \text{ rad/s}$
Wave number $k = 0.02\pi \text{ rad/m}$
The velocity of the wave $v$ is given by the formula $v = \omega / k$.
Substituting the values, $v = (12\pi) / (0.02\pi) = 12 / 0.02 = 600 \text{ m/s}$.

(C) The speed of a wave in a stretched wire is given by $v = \sqrt{\frac{T}{\mu}}$, where $T$ is the tension and $\mu$ is the linear mass density.
From this relation, we see that $v \propto \sqrt{T}$.
Let the initial tension be $T_1 = 2 \text{ kg wt}$ and the initial speed be $v_1 = v$.
Let the final tension be $T_2$ and the final speed be $v_2 = 3v$.
Using the proportionality $v \propto \sqrt{T}$, we have $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$.
Substituting the values: $\frac{3v}{v} = \sqrt{\frac{T_2}{2}} \implies 3 = \sqrt{\frac{T_2}{2}}$.
Squaring both sides: $9 = \frac{T_2}{2} \implies T_2 = 18 \text{ kg wt}$.
The increase in tension is $\Delta T = T_2 - T_1 = 18 - 2 = 16 \text{ kg wt}$.

(D) Let the velocity of sound be $v$. The frequencies of the two sources are given by $f_1 = v / \lambda_1$ and $f_2 = v / \lambda_2$.
Given $\lambda_1 = 50 \text{ cm} = 0.5 \text{ m}$ and $\lambda_2 = 50.5 \text{ cm} = 0.505 \text{ m}$.
The difference in frequencies is $f_1 - f_2 = 6 \text{ Hz}$.
Substituting the values: $v / 0.5 - v / 0.505 = 6$.
$v(2 - 1.9802) \approx 6$.
$v(0.0198) = 6$.
$v = 6 / 0.0198 \approx 303 \text{ m/s}$.
However, using the approximation $v(1/0.5 - 1/0.505) = 6 \implies v(2 - 1.98) = 6 \implies v(0.02) = 6 \implies v = 300 \text{ m/s}$.

(A) The given wave equation is $Y = 0.01 \sin [2\pi (t/0.02 - x/0.50)] \text{ m}$.
Comparing this with the standard wave equation $Y = A \sin (\omega t - kx)$, we get:
$\omega = 2\pi / 0.02 = 100\pi \text{ rad/s}$
$k = 2\pi / 0.50 = 4\pi \text{ m}^{-1}$
The wave speed $v$ is given by $v = \omega / k = 100\pi / 4\pi = 25 \text{ m/s}$.
The speed of a wave on a string is also given by $v = \sqrt{T/\mu}$, where $T$ is the tension and $\mu$ is the linear mass density.
Given $\mu = 0.02 \text{ kg m}^{-1}$, we have:
$25 = \sqrt{T / 0.02}$
Squaring both sides: $625 = T / 0.02$
$T = 625 \times 0.02 = 12.5 \text{ N}$.

(A) Given: Voltage $V = (6.0 \pm 0.3) \text{ V}$ and Current $I = (4.0 \pm 0.1) \text{ A}$.
Power $P = V \times I = 6.0 \times 4.0 = 24.0 \text{ W}$.
The relative error in power is given by $\frac{\Delta P}{P} = \frac{\Delta V}{V} + \frac{\Delta I}{I}$.
Substituting the values: $\frac{\Delta P}{24.0} = \frac{0.3}{6.0} + \frac{0.1}{4.0}$.
$\frac{\Delta P}{24.0} = 0.05 + 0.025 = 0.075$.
$\Delta P = 24.0 \times 0.075 = 1.8 \text{ W}$.
Thus, the power consumed is $(24.0 \pm 1.8) \text{ W}$.

(C) The $SI$ base unit of luminous intensity is the $Candela$ $(cd)$.
$Ampere$ is the unit of electric current.
$Coulomb$ is the unit of electric charge.
$Ohm$ is the unit of electrical resistance.
Therefore, the correct option is $C$.

(A) The de Broglie wavelength $(\lambda)$ is given by the formula: $\lambda = \frac{h}{mv}$.
Given:
Planck's constant $(h)$ = $6.626 \times 10^{-34} \text{ J s}$.
Mass of electron $(m)$ = $9.1 \times 10^{-31} \text{ kg}$.
Velocity of electron $(v)$ = $2.2 \times 10^6 \text{ m s}^{-1}$.
Substituting these values into the formula:
$\lambda = \frac{6.626 \times 10^{-34}}{(9.1 \times 10^{-31}) \times (2.2 \times 10^6)}$.
$\lambda = \frac{6.626 \times 10^{-34}}{20.02 \times 10^{-25}}$.
$\lambda = 0.33096 \times 10^{-9} \text{ m}$.
$\lambda = 3.31 \times 10^{-10} \text{ m}$.
Thus, the correct option is $A$.

(A) According to Heisenberg's uncertainty principle, the product of uncertainty in position $(\Delta x)$ and uncertainty in momentum $(\Delta p)$ is given by: $\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$.
Since $\Delta p = m_e \cdot \Delta v$, the equation becomes: $\Delta x \cdot m_e \cdot \Delta v \geq \frac{h}{4\pi}$.
Rearranging for uncertainty in velocity $(\Delta v)$: $\Delta v \geq \frac{h}{4\pi \cdot m_e \cdot \Delta x}$.
Given: $\Delta x = 50 \text{ pm} = 50 \times 10^{-12} \text{ m}$, $m_e = 9.1 \times 10^{-31} \text{ kg}$, $h = 6.63 \times 10^{-34} \text{ Js}$, $\pi = 3.142$.
Substituting the values: $\Delta v = \frac{6.63 \times 10^{-34}}{4 \times 3.142 \times 9.1 \times 10^{-31} \times 50 \times 10^{-12}}$.
$\Delta v = \frac{6.63 \times 10^{-34}}{1715.332 \times 10^{-43}} = \frac{6.63}{1715.332} \times 10^9 \approx 0.003865 \times 10^9 \text{ ms}^{-1} = 3.865 \times 10^6 \text{ ms}^{-1}$.
Re-evaluating the calculation: $\Delta v = \frac{6.63 \times 10^{-34}}{4 \times 3.142 \times 9.1 \times 10^{-31} \times 50 \times 10^{-12}} = \frac{6.63 \times 10^{-34}}{5717.64 \times 10^{-43}} = 0.001159 \times 10^9 \approx 1.16 \times 10^6 \text{ ms}^{-1}$.
Given the options provided, the closest value is $0.98 \times 10^6 \text{ ms}^{-1}$.

(D) The energy $(E)$ of a photon is given by the equation $E = \frac{hc}{\lambda}$, where $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength.
From this relation, it is clear that energy is inversely proportional to wavelength $(E \propto \frac{1}{\lambda})$.
Among the colours of visible light, Red light has the longest wavelength $(\approx 700 \ nm)$.
Since Red light has the longest wavelength, it possesses the lowest energy compared to other colours in the visible spectrum.

(B) The energy $E$ of a photon is given by the equation $E = \frac{hc}{\lambda}$, where $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength.
From this relation, it is clear that energy $E$ is inversely proportional to the wavelength $\lambda$ $(E \propto \frac{1}{\lambda})$.
Therefore, the light with the shortest wavelength will have the highest energy.
Comparing the given wavelengths: $410 \text{ nm}$ (violet), $470 \text{ nm}$ (blue), $580 \text{ nm}$ (yellow), and $750 \text{ nm}$ (red).
The shortest wavelength is $410 \text{ nm}$, which corresponds to violet light.
Thus, violet light has the highest energy.

(D) The Rydberg formula for the wave number of a spectral line is given by: $\bar{\nu} = R_H (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.
For the Paschen series, the transition occurs to the energy level $n_1 = 3$.
The lowest energy transition corresponds to the transition from the immediate next energy level, which is $n_2 = 4$.
Substituting these values into the formula:
$\bar{\nu} = R_H (\frac{1}{3^2} - \frac{1}{4^2})$
$\bar{\nu} = R_H (\frac{1}{9} - \frac{1}{16})$
$\bar{\nu} = R_H (\frac{16 - 9}{144})$
$\bar{\nu} = R_H (\frac{7}{144}) \text{ cm}^{-1}$.

(A) The Rydberg formula for the wavelength of emitted radiation is given by: $\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the Lyman series, the transition occurs to the ground state, so $n_1 = 1$.
For the shortest wavelength, the transition must occur from the highest possible energy level, i.e.,$n_2 = \infty$.
Substituting these values into the formula: $\frac{1}{\lambda} = R_H \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R_H (1 - 0) = R_H$.
Therefore, $\lambda = \frac{1}{R_H} = \frac{1}{109677 \text{ cm}^{-1}}$.
Calculating the value: $\lambda \approx 9.117 \times 10^{-6} \text{ cm} = 911.7 \times 10^{-8} \text{ cm}$.

(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by the formula: $E_n = -2.18 \times 10^{-18} \times \frac{Z^2}{n^2} \text{ J}$.
For the $H$-atom, the atomic number $Z = 1$.
For the first orbit $(n = 1)$, $E_1 = -2.18 \times 10^{-18} \times \frac{1^2}{1^2} = -2.18 \times 10^{-18} \text{ J}$.
For the second orbit $(n = 2)$, $E_2 = -2.18 \times 10^{-18} \times \frac{1^2}{2^2} = -2.18 \times 10^{-18} \times \frac{1}{4} \text{ J}$.
$E_2 = -0.545 \times 10^{-18} \text{ J}$.

(A) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{mv}$.
Given:
$h = 6.626 \times 10^{-34} \text{ J s}$
$m = 9.10938 \times 10^{-31} \text{ kg}$
$v = 2.19 \times 10^6 \text{ m s}^{-1}$
Substituting the values:
$\lambda = \frac{6.626 \times 10^{-34}}{(9.10938 \times 10^{-31}) \times (2.19 \times 10^6)}$
$\lambda = \frac{6.626 \times 10^{-34}}{19.9495 \times 10^{-25}}$
$\lambda \approx 0.33214 \times 10^{-9} \text{ m}$
$\lambda \approx 332.14 \times 10^{-12} \text{ m}$
Since $1 \text{ pm} = 10^{-12} \text{ m}$, the wavelength is approximately $332 \text{ pm}$.

(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula: $E_n = -\frac{2.18 \times 10^{-18} \text{ J}}{n^2}$.
For the stationary state corresponding to $n = 2$, we substitute the value of $n$ into the formula:
$E_2 = -\frac{2.18 \times 10^{-18} \text{ J}}{(2)^2}$
$E_2 = -\frac{2.18 \times 10^{-18} \text{ J}}{4}$
$E_2 = -0.545 \times 10^{-18} \text{ J}$
$E_2 = -5.45 \times 10^{-19} \text{ J}$.
Therefore, the energy of the electron in the $n = 2$ state is $-5.45 \times 10^{-19} \text{ J}$.

(C) In Bohr's model of the hydrogen atom, the energy of the $n^{th}$ orbit is given by the formula: $E_n = \frac{E_1}{n^2}$, where $E_1$ is the energy of the ground state $(n=1)$.
Given that the ground state energy $E_1 = -E \text{ J}$.
For the third stationary orbit, $n = 3$.
Substituting the values into the formula: $E_3 = \frac{E_1}{3^2} = \frac{-E}{9} \text{ J}$.
Therefore, the energy of the third stationary orbit is $-E/9 \text{ J}$.

(D) Viscosity is the property of fluids (liquids and gases) that offers resistance to flow.
$(A)$ Gradation of lubricant oils depends on their viscosity.
$(B)$ Blood flow velocity and viscosity are used to indicate cardiovascular health.
$(C)$ The cleansing action of soap involves reducing surface tension and altering the viscosity of the medium to remove dirt.
$(D)$ The thickening of glass panes at the bottom in old buildings is a common misconception; glass is an amorphous solid, not a supercooled liquid, and this phenomenon is not related to viscosity.

(D) To obtain an $n$-type semiconductor, a pentavalent impurity (Group $15$ element) is added to a tetravalent semiconductor like $Ge$ (Germanium).
$P$ (Phosphorus), $As$ (Arsenic), and $Sb$ (Antimony) are all Group $15$ elements and can be used as dopants to create $n$-type semiconductors.
$B$ (Boron) is a trivalent element (Group $13$). Adding a trivalent impurity to $Ge$ results in a $p$-type semiconductor, not an $n$-type semiconductor.
Therefore, $B$ is the dopant that is $NOT$ used to obtain an $n$-type semiconductor.

(B, D) Magnetic substances are classified based on their response to an external magnetic field:
$1$. Paramagnetic substances are weakly attracted by an external magnetic field.
$2$. Diamagnetic substances are weakly repelled by an external magnetic field.
$3$. Ferromagnetic substances are strongly attracted by an external magnetic field.
Comparing these facts with the given options:
- Option $(A)$ is true.
- Option $(B)$ is false because diamagnetic substances are repelled, not attracted.
- Option $(C)$ is true.
- Option $(D)$ is false because diamagnetic substances are not strongly attracted; they are weakly repelled.
Therefore, both $(B)$ and $(D)$ are false statements.

(D) To form an $n$-type semiconductor, a pentavalent impurity (an element from Group $15$ of the periodic table) must be added to an intrinsic semiconductor like germanium $(Ge)$.
$B$ (Boron), $In$ (Indium), and $Ga$ (Gallium) are trivalent elements (Group $13$), which are used to create $p$-type semiconductors.
$P$ (Phosphorus) is a pentavalent element (Group $15$), which provides an extra electron when doped into germanium, thus creating an $n$-type semiconductor.
Therefore, the correct dopant is $P$.

(C) In an optical fiber communication system, an Erbium-doped fiber amplifier $(EDFA)$ is widely used to amplify optical signals. Erbium $(Er)$ ions are doped into the silica fiber core. When these ions are pumped with light at specific wavelengths (typically $980 \ nm$ or $1480 \ nm$), they reach an excited state and provide optical gain through stimulated emission at the $1550 \ nm$ wavelength, which is the low-loss window for optical fibers. Therefore, the correct element is Erbium $(Er)$.

(C) The rate of disintegration $R$ at time $t$ is given by the law of radioactive decay: $R = R_0 e^{-\lambda t}$.
Here, $R_0 = 16,000 \text{ disintegrations/minute}$, $R = 2000 \text{ disintegrations/minute}$, and $t = 4 \text{ minutes}$.
Substituting these values into the equation: $2000 = 16,000 e^{-\lambda \times 4}$.
Dividing both sides by $16,000$: $\frac{2000}{16,000} = e^{-4\lambda}$, which simplifies to $\frac{1}{8} = e^{-4\lambda}$.
Taking the natural logarithm $(log_e)$ on both sides: $\log_e(1/8) = -4\lambda$.
Since $\log_e(1/8) = \log_e(2^{-3}) = -3 \log_e 2$, we have $-3 \log_e 2 = -4\lambda$.
Solving for $\lambda$: $\lambda = \frac{3}{4} \log_e 2 = 0.75 \log_e 2 \text{ min}^{-1}$.

(A) The density of a nucleus is given by the formula $\rho = \frac{M}{V}$, where $M$ is the mass of the nucleus and $V$ is its volume.
The mass of a nucleus with mass number $A$ is approximately $M = A \times m_p$, where $m_p$ is the mass of a proton.
The volume of a nucleus is given by $V = \frac{4}{3} \pi R^3$, where $R = R_0 A^{1/3}$ is the radius of the nucleus.
Substituting the expression for $R$, we get $V = \frac{4}{3} \pi (R_0 A^{1/3})^3 = \frac{4}{3} \pi R_0^3 A$.
Now, the density $\rho = \frac{A \times m_p}{\frac{4}{3} \pi R_0^3 A} = \frac{3 m_p}{4 \pi R_0^3}$.
Since the density $\rho$ is independent of the mass number $A$, the density of all nuclei is constant.
Therefore, the ratio of the density of oxygen nucleus to helium nucleus is $1 : 1$.

(C) The given circuit consists of a $NOR$ gate followed by a $NOT$ gate.
$1$. The first gate is a $NOR$ gate with inputs $A$ and $B$. The output of the $NOR$ gate is $C = \overline{A + B}$.
$2$. The second gate is a $NOT$ gate (inverter) which takes the input $C$ and produces the output $Y = \overline{C}$.
$3$. Substituting the value of $C$, we get $Y = \overline{(\overline{A + B})}$.
$4$. According to the law of double negation, $\overline{\overline{X}} = X$. Therefore, $Y = A + B$.
$5$. $A$ circuit that performs the operation $Y = A + B$ is an $OR$ gate.
Thus, the resultant gate is an $OR$ gate and its Boolean expression is $A + B$.

(C) The logic gate described is the $X-OR$ (Exclusive-$OR$) gate.
For an $X-OR$ gate with two inputs $A$ and $B$, the output $Y$ is given by the Boolean expression $Y = A \oplus B = A\overline{B} + \overline{A}B$.
The truth table for an $X-OR$ gate is:
- If $A=0, B=0$, then $Y=0$.
- If $A=1, B=0$, then $Y=1$.
- If $A=0, B=1$, then $Y=1$.
- If $A=1, B=1$, then $Y=0$.
This confirms that the output is high $(1)$ when either of the inputs is high, but not when both inputs are high.

(C) The given circuit consists of an $OR$ gate with two inputs. One input is $A$ and the other input is the output of gate $G$, which takes $A$ and $B$ as inputs. Let the output of gate $G$ be $X$. Then the final output $Y = A + X$.
From the truth table:
For $A=0, B=0$, $Y=0$. Since $Y = A + X$, $0 = 0 + X \implies X=0$.
For $A=0, B=1$, $Y=0$. Since $Y = A + X$, $0 = 0 + X \implies X=0$.
For $A=1, B=0$, $Y=1$. Since $Y = A + X$, $1 = 1 + X \implies X$ can be $0$ or $1$.
For $A=1, B=1$, $Y=1$. Since $Y = A + X$, $1 = 1 + X \implies X$ can be $0$ or $1$.
Looking at the values of $X$ derived from the inputs $A$ and $B$: when $A=0, B=0 \implies X=0$; when $A=0, B=1 \implies X=0$; when $A=1, B=0 \implies X=0$ (consistent with $Y=1$); when $A=1, B=1 \implies X=1$ (consistent with $Y=1$).
This truth table for $X$ (inputs $A, B$ and output $X$) corresponds to the $AND$ gate, where $X = A \cdot B$.

(D) Let the output of the $NAND$ gate be $Y_1 = \overline{A \cdot B}$.
Let the output of the $NOT$ gate be $A' = \overline{A}$.
Let the output of the $NOR$ gate be $Y_2 = \overline{A' + B} = \overline{\overline{A} + B} = A \cdot \overline{B}$.
The final output $Y$ is the output of a $NOR$ gate with inputs $Y_1$ and $Y_2$, so $Y = \overline{Y_1 + Y_2} = \overline{\overline{A \cdot B} + A \cdot \overline{B}}$.
Using De Morgan's law, $Y = (A \cdot B) \cdot (\overline{A \cdot \overline{B}}) = (A \cdot B) \cdot (\overline{A} + B) = (A \cdot B \cdot \overline{A}) + (A \cdot B \cdot B) = 0 + (A \cdot B) = A \cdot B$.
For $Y = 1$, we must have $A = 1$ and $B = 1$.

(B) Let us analyze the output for each logic gate:
$(A)$ $NAND$ gate with inputs $1$ and $1$: The output of an $AND$ gate is $1 \cdot 1 = 1$. The $NAND$ gate inverts this, so the output is $0$.
$(B)$ $NAND$ gate with inputs $1$ and $0$: The output of an $AND$ gate is $1 \cdot 0 = 0$. The $NAND$ gate inverts this, so the output is $1$.
$(C)$ $NOR$ gate with inputs $0$ and $1$: The output of an $OR$ gate is $0 + 1 = 1$. The $NOR$ gate inverts this, so the output is $0$.
$(D)$ $XOR$ gate with inputs $0$ and $0$: The output of an $XOR$ gate is $0 \oplus 0 = 0$.
Therefore, the logic gate that gives an output of '$1$' is the $NAND$ gate with inputs $1$ and $0$.

(A) $1$. The circuit consists of a $NAND$ gate, a $NOR$ gate, and an $OR$ gate.
$2$. The inputs to the $NAND$ gate are $A$ and $B$. Therefore, its output is $\overline{A \cdot B}$.
$3$. The inputs to the $NOR$ gate are $B$ and $C$. Therefore, its output is $\overline{B + C}$.
$4$. These two outputs are fed into an $OR$ gate. The output of an $OR$ gate is the sum of its inputs.
$5$. Thus, the final output $Y = \overline{A \cdot B} + \overline{B + C}$.

(C) Let the inputs be $A$ and $B$, and the output be $Y$.
Given conditions:
$1$. When $A = 0$ and $B = 0$, $Y = 1$.
$2$. When $A = 1$ and $B = 0$, $Y = 0$.
Checking the truth tables:
- For a $NOR$ gate: $0 NOR 0 = 1$ and $1 NOR 0 = 0$. This matches the given conditions.
- For a $NAND$ gate: $0 NAND 0 = 1$ and $1 NAND 0 = 1$. This does not match.
- For an $OR$ gate: $0 OR 0 = 0$ and $1 OR 0 = 1$. This does not match.
- For an $XOR$ gate: $0 XOR 0 = 0$ and $1 XOR 0 = 1$. This does not match.
Therefore, the logic gate is a $NOR$ gate.

(D) $1$. The circuit consists of an $AND$ gate, a $NOT$ gate, and an $OR$ gate.
$2$. The inputs $A$ and $B$ are fed into an $AND$ gate. The output of the $AND$ gate is $A \cdot B$.
$3$. The input $C$ is fed into a $NOT$ gate. The output of the $NOT$ gate is $\bar{C}$.
$4$. The outputs of the $AND$ gate $(A \cdot B)$ and the $NOT$ gate $(\bar{C})$ are fed into an $OR$ gate.
$5$. The final output $Y$ of the $OR$ gate is the sum of its inputs: $Y = (A \cdot B) + \bar{C}$.

(D) The given circuit consists of a $NOT$ gate, an $AND$ gate, and an $OR$ gate.
From the circuit diagram, the output of the $NOT$ gate is $\bar{A}$.
The output of the $AND$ gate is $B \cdot C$.
These two outputs are fed into an $OR$ gate, so the final output $Y$ is given by the Boolean expression: $Y = \bar{A} + (B \cdot C)$.
We want to find the inputs for which $Y = 1$.
Let's test the options:
$(A)$ $A = 1, B = 0, C = 0$: $Y = \bar{1} + (0 \cdot 0) = 0 + 0 = 0$.
$(B)$ $A = 1, B = 0, C = 1$: $Y = \bar{1} + (0 \cdot 1) = 0 + 0 = 0$.
$(C)$ $A = 1, B = 1, C = 0$: $Y = \bar{1} + (1 \cdot 0) = 0 + 0 = 0$.
$(D)$ $A = 1, B = 1, C = 1$: $Y = \bar{1} + (1 \cdot 1) = 0 + 1 = 1$.
Thus, for the inputs $A = 1, B = 1, C = 1$, the output $Y$ is $1$.

(D) The Boolean expression $A + B = Y$ represents the $OR$ gate operation.
In Boolean algebra, the $+$ sign denotes the logical $OR$ operation.
The truth table for an $OR$ gate is as follows:
- If $A = 0$ and $B = 0$, then $Y = 0$.
- If $A = 1$ and $B = 0$, then $Y = 1$.
- If $A = 0$ and $B = 1$, then $Y = 1$.
- If $A = 1$ and $B = 1$, then $Y = 1$.
This means that the output $Y$ is $1$ (exists) if at least one of the inputs $A$ or $B$ is $1$ (exists). Therefore, $Y$ exists when $A$ exists, or $B$ exists, or both $A$ and $B$ exist.

(D) The power gain $(A_p)$ of a transistor amplifier is given by the product of current gain $(\beta)$ and voltage gain $(A_v)$.
$A_p = \beta \times A_v$
Given, $A_p = 5 \times 10^6$ and $\beta = 50$.
$5 \times 10^6 = 50 \times A_v$
$A_v = \frac{5 \times 10^6}{50} = 10^5$.
Voltage gain is also defined as $A_v = \beta \times \frac{R_{out}}{R_{in}}$, where $R_{out}$ is the output resistance and $R_{in}$ is the input resistance.
$10^5 = 50 \times \frac{500 \times 10^3 \Omega}{R_{in}}$
$R_{in} = \frac{50 \times 500 \times 10^3}{10^5} = \frac{25000 \times 10^3}{10^5} = 250 \Omega$.
Thus, the input resistance is $250 \Omega$.

(B) Given:
Common emitter current gain, $\beta = 20$
Emitter current, $I_E = 7 \text{ mA}$
We know the relationship between emitter current and collector current in common emitter mode is $I_E = I_C + I_B$.
Also, $\beta = I_C / I_B$, which implies $I_B = I_C / \beta$.
Substituting $I_B$ in the first equation:
$I_E = I_C + (I_C / \beta) = I_C(1 + 1/\beta) = I_C((\beta + 1) / \beta)$.
Rearranging for $I_C$:
$I_C = I_E \times (\beta / (\beta + 1))$.
Substituting the values:
$I_C = 7 \times (20 / (20 + 1)) = 7 \times (20 / 21)$.
$I_C = 7 \times (20 / 21) = 20 / 3 \text{ mA}$.

(A) In a common-emitter transistor amplifier configuration, the input current is the base current $(I_B)$ and the output current is the collector current $(I_C)$.
The current gain, denoted by $\beta$, is defined as the ratio of the change in collector current to the change in base current, keeping the collector-emitter voltage constant.
Mathematically, $\beta = \Delta I_C / \Delta I_B$.

(A) Given:
$d.c. \text{ current gain } (\beta) = 20$
$Emitter current (I_E) = 7 \text{ mA}$
We know the relationship between current gain $(\beta)$ and collector current $(I_C)$ is given by $I_C = \beta I_B$, where $I_B$ is the base current.
Also, $I_E = I_B + I_C$.
Substituting $I_B = I_C / \beta$ into the equation:
$I_E = I_C / \beta + I_C$
$I_E = I_C (1/\beta + 1)$
$I_E = I_C (1 + \beta) / \beta$
Rearranging to solve for $I_C$:
$I_C = I_E \times \beta / (1 + \beta)$
$I_C = 7 \times 20 / (1 + 20)$
$I_C = 140 / 21$
Dividing both numerator and denominator by $7$:
$I_C = 20 / 3 \text{ mA}$.

(A) Given: $\alpha = 0.9$ and change in base current $\Delta I_B = 0.4 \text{ mA}$.
First, we calculate the current gain $\beta$ for the $CE$ configuration using the relation: $\beta = \frac{\alpha}{1 - \alpha}$.
Substituting the value of $\alpha$: $\beta = \frac{0.9}{1 - 0.9} = \frac{0.9}{0.1} = 9$.
The relationship between collector current change $\Delta I_C$ and base current change $\Delta I_B$ is given by: $\Delta I_C = \beta \times \Delta I_B$.
Substituting the values: $\Delta I_C = 9 \times 0.4 \text{ mA} = 3.6 \text{ mA}$.
Therefore, the change in collector current is $3.6 \text{ mA}$.

(D) The current gain $\beta$ is given by the ratio of the change in collector current $\Delta I_C$ to the change in base current $\Delta I_B$.
$\beta = \frac{\Delta I_C}{\Delta I_B} = \frac{5 \text{ mA}}{0.2 \text{ mA}} = 25$.
The voltage gain $A_V$ is given by the formula $A_V = \beta \times \frac{R_L}{R_{in}}$, where $R_L$ is the load resistance and $R_{in}$ is the input resistance.
Given $A_V = 75$, $\beta = 25$, and $R_{in} = 2 \text{ k}\Omega$.
Substituting the values: $75 = 25 \times \frac{R_L}{2 \text{ k}\Omega}$.
$3 = \frac{R_L}{2 \text{ k}\Omega}$.
$R_L = 3 \times 2 \text{ k}\Omega = 6 \text{ k}\Omega$.

(B) The current gain $\beta$ is defined as the ratio of the change in collector current $(\Delta I_C)$ to the change in base current $(\Delta I_B)$.
$\beta = \frac{\Delta I_C}{\Delta I_B} = \frac{5 \text{ mA}}{0.2 \text{ mA}} = 25$.
The voltage gain $A_V$ is given by the formula $A_V = \beta \times \frac{R_L}{R_{in}}$, where $R_L$ is the load resistance and $R_{in}$ is the input resistance.
Given $A_V = 75$, $\beta = 25$, and $R_{in} = 2 \text{ k}\Omega$.
Substituting the values: $75 = 25 \times \frac{R_L}{2 \text{ k}\Omega}$.
$3 = \frac{R_L}{2 \text{ k}\Omega}$.
$R_L = 3 \times 2 \text{ k}\Omega = 6 \text{ k}\Omega$.
Therefore, the load resistance is $6 \text{ k}\Omega$.

(A) In a common emitter transistor amplifier:
$1$. The voltage gain $A_V$ is defined as the ratio of the output voltage to the input voltage.
$A_V = \frac{V_{out}}{V_{in}} = \frac{I_c R_L}{I_b r_i} = \beta \frac{R_L}{r_i}$, where $\beta = \frac{I_c}{I_b}$ is the a.c. current gain.
$2$. The power gain $A_P$ is defined as the product of the current gain and the voltage gain.
$A_P = \beta \times A_V = \beta \times (\beta \frac{R_L}{r_i}) = \beta^2 \frac{R_L}{r_i}$.
Therefore, the voltage gain is $\beta (R_L/r_i)$ and the power gain is $\beta^2 (R_L/r_i)$.

(A) In a transistor, the emitter current $(I_E)$ is the sum of the collector current $(I_C)$ and the base current $(I_B)$, i.e.,$I_E = I_C + I_B$.
Given that $80\%$ of the electrons emitted reach the collector, we have $I_C = 0.80 \times I_E$.
Given $I_C = 28 \text{ mA}$, we can find $I_E$ as:
$I_E = \frac{I_C}{0.80} = \frac{28}{0.80} = 35 \text{ mA}$.
Now, using the relation $I_B = I_E - I_C$, we get:
$I_B = 35 \text{ mA} - 28 \text{ mA} = 7 \text{ mA}$.
Therefore, the base current is $7 \text{ mA}$.

(B) In a bipolar junction transistor $(BJT)$, the three regions are doped differently to optimize their functions:
$1$. The $Emitter$ is heavily doped to provide a large number of charge carriers.
$2$. The $Base$ is very thin and lightly doped to allow most of the charge carriers from the emitter to pass through to the collector.
$3$. The $Collector$ is moderately doped compared to the emitter and base.
Therefore, the part that is heavily doped is the $Emitter$, and the part that is lightly doped is the $Base$.

(C) The voltage gain $(A_v)$ of a common emitter transistor amplifier is given by the formula:
$A_v = \beta \times \frac{R_L}{R_i}$
Where:
$\beta$ ($AC$ current gain) = $65$
$R_L$ (Load resistance) = $5400 \text{ }\Omega$
$R_i$ (Input resistance) = $450 \text{ }\Omega$
Substituting the values into the formula:
$A_v = 65 \times \frac{5400}{450}$
$A_v = 65 \times 12$
$A_v = 780$
Therefore, the voltage gain is $780$.

(D) photodiode is a special type of $p-n$ junction diode that is designed to operate under reverse bias conditions.
When light (photons) with energy greater than the bandgap energy of the semiconductor falls on the junction, it generates electron-hole pairs.
Under the influence of the reverse bias electric field, these charge carriers are swept across the junction, creating a photocurrent.
The magnitude of this photocurrent is directly proportional to the intensity of the incident light.
Therefore, a photodiode is always operated in reverse bias to detect light signals efficiently.

(C) An $LED$ (Light Emitting Diode) is a heavily doped $p-n$ junction diode that emits spontaneous radiation when forward biased.
When the $LED$ is forward biased, electrons from the $n$-region and holes from the $p$-region move towards the junction and recombine.
During this recombination, the energy released is in the form of photons (light).
The energy of the emitted photons is approximately equal to or slightly less than the band gap energy $(E_g)$ of the semiconductor material.
Therefore, the statement that it emits light when reverse biased is incorrect.

(D) An $LED$ (Light Emitting Diode) is a semiconductor device that emits light when an electric current flows through it.
$1$. LEDs are known for their long operational lifetime when manufactured correctly.
$2$. The light produced by LEDs is monochromatic or specific in wavelength, and the colours do not fade over time.
$3$. LEDs are highly energy-efficient compared to traditional incandescent or fluorescent bulbs.
$4$. The brightness of an $LED$ can be easily controlled by varying the amount of current flowing through it (using techniques like Pulse Width Modulation or simple current limiting).
Therefore, the statement that the brightness of light emitted by an $LED$ cannot be controlled is '$WRONG$'.

(C) For a solar cell to be efficient, the material must be able to absorb a significant portion of the solar spectrum.
Most of the solar radiation reaching the Earth's surface lies in the visible and near-infrared range.
$A$ band gap in the range of $1.0 \text{ eV}$ to $1.8 \text{ eV}$ is ideal because it allows the material to absorb photons with energies corresponding to the peak intensity of the solar spectrum.
Silicon, with a band gap of approximately $1.1 \text{ eV}$, is the most commonly used material for solar cells due to this property.

(A) $Zener$ diode is a specially designed $pn$ junction diode that is heavily doped to operate in the reverse breakdown region.
In forward bias, it behaves like an ordinary $pn$ junction diode, conducting current once the threshold voltage is reached.
In reverse bias, once the voltage across the $Zener$ diode reaches its $Zener$ breakdown voltage $(V_Z)$, the current increases significantly while the voltage across it remains nearly constant.
This property allows the $Zener$ diode to function as a voltage regulator in the reverse bias region.

(C) In a $p-n$ junction diode, the depletion region width increases with an increase in reverse bias voltage.
When the reverse bias voltage is increased to a critical value known as the breakdown voltage, the electric field across the junction becomes very strong.
This strong electric field causes a large number of charge carriers to be generated (either through Zener breakdown or Avalanche breakdown), leading to a sudden increase in the reverse current.
Therefore, breakdown occurs when the reverse bias is increased to this critical value.

(C) An ideal diode in forward bias acts as a short circuit (zero resistance), and in reverse bias, it acts as an open circuit (infinite resistance).
Case $1$: Diode $D$ is forward biased.
The upper branch has a resistance of $30 \ \Omega$ in series with the diode (which acts as $0 \ \Omega$). The lower branch has a resistance of $30 \ \Omega$. These two branches are in parallel.
$R_1 = (30 \ \Omega \parallel 30 \ \Omega) = \frac{30 \times 30}{30 + 30} = \frac{900}{60} = 15 \ \Omega$.
Case $2$: Diode $D$ is reverse biased.
The upper branch acts as an open circuit (infinite resistance). Thus, only the lower branch with $30 \ \Omega$ resistance is connected between $A$ and $B$.
$R_2 = 30 \ \Omega$.
Ratio $R_1/R_2 = 15/30 = 1/2$.

(B) $1$. Analyze the circuit: The diodes $D_1$ and $D_2$ are forward-biased, while $D_3$ is reverse-biased.
$2$. Since $D_3$ is reverse-biased, it acts as an open circuit (infinite resistance), so no current flows through the branch containing $D_3$.
$3$. The circuit simplifies to two parallel branches connected in series with the $100 \text{ } \Omega$ resistor.
$4$. Branch $1$ (containing $D_1$): Total resistance $R_1 = R_{D1} + 150 \text{ } \Omega = 50 \text{ } \Omega + 150 \text{ } \Omega = 200 \text{ } \Omega$.
$5$. Branch $2$ (containing $D_2$): Total resistance $R_2 = R_{D2} + 50 \text{ } \Omega = 50 \text{ } \Omega + 50 \text{ } \Omega = 100 \text{ } \Omega$.
$6$. The equivalent resistance of the parallel combination of $R_1$ and $R_2$ is $R_p = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{200 \times 100}{200 + 100} = \frac{20000}{300} = \frac{200}{3} \text{ } \Omega$.
$7$. The total resistance of the circuit is $R_{total} = R_p + 100 \text{ } \Omega = \frac{200}{3} + 100 = \frac{500}{3} \text{ } \Omega$.
$8$. The total current $I$ flowing from the battery is $I = \frac{V}{R_{total}} = \frac{5}{500/3} = \frac{15}{500} = 0.03 \text{ A} = 30 \text{ mA}$.

(D) $Zener$ diode is a special type of diode designed to operate in the breakdown region.
When used as a voltage regulator, the $Zener$ diode is connected in reverse bias across the load.
This configuration ensures that the voltage across the load remains constant at the $Zener$ breakdown voltage, even if the input voltage or load current changes.
Thus, the correct connection is reverse bias and in parallel with the load.

(D) diode is forward biased if the potential at the $p$-side (anode) is higher than the potential at the $n$-side (cathode). Otherwise, it is reverse biased.
In Fig. $(1)$:
The $p$-side of diode $D_1$ is at $-5 \ V$ and the $n$-side is connected to $-3 \ V$ through resistor $R_1$. Since $-5 \ V < -3 \ V$, the potential at the $p$-side is lower than the potential at the $n$-side. Therefore, diode $D_1$ is reverse biased.
In Fig. $(2)$:
The $p$-side of diode $D_2$ is at $0 \ V$ and the $n$-side is connected to $-4 \ V$ through resistor $R_2$. Since $0 \ V > -4 \ V$, the potential at the $p$-side is higher than the potential at the $n$-side. Therefore, diode $D_2$ is forward biased.
Thus, Fig. $(1)$ is reverse biased and Fig. $(2)$ is forward biased.

(B) The given circuit consists of a $5 \text{ V}$ $DC$ source, a silicon diode, and a resistor of $200 \text{ } \Omega$ connected in series.
Since the diode is forward-biased, it will conduct current when the applied voltage exceeds its knee voltage.
The effective voltage across the resistor $(V_R)$ is given by:
$V_R = V_{\text{source}} - V_{\text{knee}}$
$V_R = 5 \text{ V} - 0.7 \text{ V} = 4.3 \text{ V}$
Using Ohm's law, the current $(I)$ flowing through the circuit is:
$I = \frac{V_R}{R}$
$I = \frac{4.3 \text{ V}}{200 \text{ } \Omega} = 0.0215 \text{ A}$
To convert the current into milliamperes (mA):
$I = 0.0215 \times 1000 \text{ mA} = 21.5 \text{ mA}$
Therefore, the reading in the milliammeter is $21.5 \text{ mA}$.

(A) In a $p-n$ junction, the depletion region is formed by the diffusion of charge carriers across the junction.
When the junction is forward biased, the external electric field opposes the internal electric field, which pushes the charge carriers towards the junction, thereby decreasing the width of the depletion region.
When the junction is reverse biased, the external electric field supports the internal electric field, which pulls the charge carriers away from the junction, thereby increasing the width of the depletion region.
Therefore, the depletion region increases if the junction is reverse biased.

(B) semiconductor in its extremely pure form is known as an intrinsic semiconductor.
In an intrinsic semiconductor, the number of free electrons is equal to the number of holes.
The conductivity of such a material is solely due to the thermal excitation of electrons, which results in the breaking of covalent bonds.
Therefore, when conductivity is only due to the breaking of covalent bonds, it is called an intrinsic semiconductor.

(D) Given: Intrinsic carrier concentration $n_i = 10^{16} \text{ m}^{-3}$.
Donor concentration (phosphorus atoms) $n_D = 10^{21} \text{ m}^{-3}$.
Since phosphorus is a pentavalent impurity, it acts as a donor, making the crystal an $n$-type semiconductor.
In an $n$-type semiconductor, the electron concentration $n_e \approx n_D = 10^{21} \text{ m}^{-3}$.
According to the law of mass action for semiconductors, $n_e \cdot n_h = n_i^2$, where $n_h$ is the hole concentration.
Substituting the values: $10^{21} \cdot n_h = (10^{16})^2$.
$10^{21} \cdot n_h = 10^{32}$.
$n_h = 10^{32} / 10^{21} = 10^{11} \text{ m}^{-3}$.
Therefore, the new hole concentration is $10^{11} \text{ per m}^3$.

(B) In a semiconductor, the number of charge carriers (electrons and holes) increases exponentially with an increase in temperature $(T)$.
This increase in the number of charge carriers dominates over the effect of increased scattering of charge carriers.
As a result, the resistivity $(\rho)$ of a semiconductor decreases as the temperature increases.
The relationship is given by $\rho = \rho_0 e^{E_g / 2kT}$, where $E_g$ is the energy band gap, $k$ is the Boltzmann constant, and $T$ is the absolute temperature.
This represents an exponential decay curve.
Therefore, graph $B$ correctly represents the temperature dependence of resistivity for a semiconductor.

(B) The process of adding impurity atoms to a pure semiconductor is known as doping.
Adding impurities increases the number of charge carriers (electrons or holes) in the semiconductor.
Since the conductivity $(\sigma)$ of a semiconductor is directly proportional to the number of charge carriers $(n)$, i.e., $\sigma = ne\mu$, where $e$ is the charge and $\mu$ is the mobility, an increase in the number of charge carriers leads to an increase in conductivity.
Resistivity $(\rho)$ is the reciprocal of conductivity $(\rho = 1/\sigma)$.
Therefore, as the conductivity increases due to doping, the resistivity of the semiconductor decreases.

(C) Insulators are materials that do not allow the flow of electric current through them easily.
This is because they have a very large energy band gap between the valence band and the conduction band, which prevents electrons from moving to the conduction band.
Consequently, their electrical conductivity is extremely small, typically in the range of $10^{-20}$ to $10^{-10} \ S/m$.

(B) $1$. Germanium $(Ge)$ is a group $14$ element. Indium $(In)$ is a group $13$ element. Doping $Ge$ with $In$ creates a p-type semiconductor. Thus, $X$ is p-type.
$2$. Arsenic $(As)$ is a group $15$ element. Doping $Ge$ with $As$ creates an n-type semiconductor. Thus, $Y$ is n-type.
$3$. The junction formed is a p-n junction.
$4$. In the circuit diagram, the p-side $(X)$ is connected to the negative terminal of the battery and the n-side $(Y)$ is connected to the positive terminal of the battery.
$5$. When the p-side is connected to the negative terminal and the n-side is connected to the positive terminal, the p-n junction is reverse biased.

(C) In an intrinsic semiconductor, the material is pure and contains no impurities.
When thermal energy is supplied, electrons are excited from the valence band to the conduction band.
This process creates a free electron in the conduction band and a corresponding hole in the valence band.
Since each electron-hole pair is generated simultaneously, the number of free electrons per unit volume $(n_e)$ must be equal to the number of holes per unit volume $(n_h)$.
Therefore, $n_e = n_h$.

(C) An $n$-type semiconductor is formed by doping an intrinsic semiconductor (like $Si$ or $Ge$) with pentavalent impurity atoms (such as $P$, $As$, or $Sb$).
These pentavalent atoms provide extra electrons to the conduction band.
Therefore, in an $n$-type semiconductor, electrons are the majority charge carriers and holes are the minority charge carriers.

(C) In an $n$-type semiconductor, pentavalent impurity atoms (like phosphorus, arsenic, etc.) are added to the intrinsic semiconductor.
These impurity atoms are called donor atoms.
Each donor atom provides an extra electron.
These donated electrons occupy discrete energy levels known as donor energy levels $(E_D)$.
These donor energy levels are located within the band gap, just below the conduction band edge $(E_C)$.
Because they are very close to the conduction band, these electrons can easily be thermally excited into the conduction band at room temperature, contributing to electrical conductivity.

(A) In insulators, the energy band gap between the valence band and the conduction band is very large (typically $> 3 \ eV$).
Due to this large energy gap, electrons cannot easily jump from the valence band to the conduction band even at room temperature.
Consequently, the conduction band remains empty because no electrons are available to conduct electricity.

(A) The given circuit consists of four gates: $G_1$ and $G_2$ are $NOT$ gates, $G_3$ is a $NOR$ gate, and $G_4$ is a $NOT$ gate.
Let the inputs be $A$ and $B$.
The outputs of $G_1$ and $G_2$ are $\bar{A}$ and $\bar{B}$ respectively.
These are fed into the $NOR$ gate $G_3$. The output of $G_3$ is $\overline{\bar{A} + \bar{B}}$.
By De Morgan's Law, $\overline{\bar{A} + \bar{B}} = \overline{\bar{A}} \cdot \overline{\bar{B}} = A \cdot B$.
This output is then passed through the $NOT$ gate $G_4$.
The final output $Y$ is $\overline{A \cdot B}$, which is the Boolean expression for a $NAND$ gate.

(C) $NAND$ gate is a combination of an $AND$ gate followed by a $NOT$ gate.
For an $AND$ gate, the output is $A \cdot B$.
Applying the $NOT$ operation to this result gives the output of the $NAND$ gate.
Therefore, the Boolean expression for a $NAND$ gate is $Y = \overline{A \cdot B}$.

(B) The circuit consists of a $NOR$ gate followed by a $NOT$ gate, which together form an $OR$ gate for the inputs $A$ and $B$. Let this output be $Y_1$. Thus, $Y_1 = A + B$.
The inputs $C$ and $D$ are connected to an $AND$ gate, so $Y_2 = C \cdot D$.
The outputs $Y_1$ and $Y_2$ are then fed into a $NOR$ gate to produce $Y_3$. Thus, $Y_3 = \overline{Y_1 + Y_2}$.
Given inputs: $A = 0, B = 1, C = 0, D = 1$.
Calculating $Y_1$: $Y_1 = A + B = 0 + 1 = 1$.
Calculating $Y_2$: $Y_2 = C \cdot D = 0 \cdot 1 = 0$.
Calculating $Y_3$: $Y_3 = \overline{Y_1 + Y_2} = \overline{1 + 0} = \overline{1} = 0$.
Therefore, the new outputs are $(1, 0, 0)$.

(A) When a $p-n$ junction diode is forward biased, the positive terminal of the external battery is connected to the $p$-type region and the negative terminal to the $n$-type region.
This external electric field opposes the internal electric field of the depletion region.
As a result, the majority charge carriers are pushed towards the junction, which reduces the width of the depletion layer.
Consequently, the potential barrier height decreases, allowing current to flow through the diode.

(A) Brewster's law states that the tangent of the Brewster's angle $(i_B)$ is equal to the refractive index $(\mu)$ of the medium: $\tan(i_B) = \mu$.
Since the refractive index $(\mu)$ of a medium depends on the wavelength $(\lambda)$ of light (due to dispersion), and different colours of light have different wavelengths, the refractive index varies for different colours.
Consequently, the Brewster's angle $(i_B = \arctan(\mu))$ is different for light of different colours.