AIPMT 1991 Physics Question Paper with Answer and Solution

(B) The expression given is a scalar triple product of the form $\overrightarrow{A} \cdot (\overrightarrow{B} \times \overrightarrow{A})$.
By the properties of the cross product,the vector $\overrightarrow{C} = (\overrightarrow{B} \times \overrightarrow{A})$ is perpendicular to both $\overrightarrow{A}$ and $\overrightarrow{B}$.
Since $\overrightarrow{C}$ is perpendicular to $\overrightarrow{A}$,the dot product of $\overrightarrow{A}$ and $\overrightarrow{C}$ must be zero.
Mathematically,$\overrightarrow{A} \cdot (\overrightarrow{B} \times \overrightarrow{A}) = 0$ because the vector $(\overrightarrow{B} \times \overrightarrow{A})$ lies in a plane perpendicular to $\overrightarrow{A}$.

(A) The force per unit length between two parallel current-carrying wires is given by the formula: $F/l = \frac{\mu_0 I_1 I_2}{2\pi r}$.
Rearranging for $\mu_0$,we get: $\mu_0 = \frac{2\pi r F}{l I_1 I_2}$.
The dimensional formula for force $F$ is $[M^1 L^1 T^{-2}]$.
The dimensional formula for length $l$ is $[L^1]$.
The dimensional formula for current $I$ is $[A^1]$.
The dimensional formula for distance $r$ is $[L^1]$.
Substituting these into the formula: $[\mu_0] = \frac{[L^1] [M^1 L^1 T^{-2}]}{[L^1] [A^1] [A^1]}$.
Simplifying the expression: $[\mu_0] = [M^1 L^1 T^{-2} A^{-2}]$.

(A) Let the total length of the chain be $L$ and its total mass be $M$. The mass per unit length is $\lambda = M/L$.
Let $x$ be the length of the chain hanging over the edge. Then the length of the chain on the table is $(L - x)$.
The mass of the hanging part is $m_h = \lambda x$ and the mass of the part on the table is $m_t = \lambda (L - x)$.
The force pulling the chain down is the weight of the hanging part: $F_g = m_h g = \lambda x g$.
The maximum static frictional force acting on the part on the table is $f_{max} = \mu N = \mu m_t g = \mu \lambda (L - x) g$.
For the chain to be on the verge of sliding,the pulling force must equal the maximum frictional force: $\lambda x g = \mu \lambda (L - x) g$.
Dividing both sides by $\lambda g$,we get $x = \mu (L - x)$.
Substituting $\mu = 0.25$: $x = 0.25(L - x) \implies x = 0.25L - 0.25x \implies 1.25x = 0.25L$.
Thus,the fraction $x/L = 0.25 / 1.25 = 1/5 = 0.20$.
Converting to percentage,the fraction is $20\%$.

(A) For a satellite in a circular orbit,the gravitational force provides the necessary centripetal force: $\frac{GMm}{r^2} = \frac{Mv^2}{r}$.
This implies that the potential energy $U = -\frac{GMm}{r} = -Mv^2$.
The kinetic energy $K = \frac{1}{2}Mv^2$.
The total energy $E$ is the sum of kinetic and potential energy: $E = K + U = \frac{1}{2}Mv^2 - Mv^2 = -\frac{1}{2}Mv^2$.

(C) According to Dalton's Law of Partial Pressures,the total pressure exerted by a mixture of non-reacting gases in a fixed volume is equal to the sum of the partial pressures of the individual gases.
Since all three gases are mixed and placed into a single container of the same volume $V$,the total pressure $P$ is the sum of the individual pressures.
Therefore,$P = P_1 + P_2 + P_3$.

(D) The pressure $P$ exerted by an ideal gas is given by the kinetic theory of gases as $P = \frac{1}{3} \rho v_{rms}^2$,where $\rho$ is the density of the gas and $v_{rms}$ is the root mean square velocity.
We know that the average kinetic energy per unit volume $E$ is given by $E = \frac{1}{2} \rho v_{rms}^2$.
From this,we can write $\rho v_{rms}^2 = 2E$.
Substituting this into the pressure equation: $P = \frac{1}{3} (2E) = \frac{2}{3}E$.
Therefore,the correct relation is $P = \frac{2}{3}E$.

(C) The velocity $v$ of a body executing $S.H.M.$ at a displacement $y$ is given by the formula: $v = \omega \sqrt{a^2 - y^2}$,where $a$ is the amplitude and $\omega$ is the angular frequency.
For $y_1 = 4 \, cm$,$v_1 = 10 \, cm/sec$: $10 = \omega \sqrt{a^2 - 4^2} \implies 100 = \omega^2(a^2 - 16)$ --- $(1)$
For $y_2 = 5 \, cm$,$v_2 = 8 \, cm/sec$: $8 = \omega \sqrt{a^2 - 5^2} \implies 64 = \omega^2(a^2 - 25)$ --- $(2)$
Subtracting equation $(2)$ from equation $(1)$:
$100 - 64 = \omega^2(a^2 - 16 - a^2 + 25)$
$36 = \omega^2(9)$
$\omega^2 = 4 \implies \omega = 2 \, rad/sec$.
The time period $T$ is given by $T = \frac{2\pi}{\omega}$.
$T = \frac{2\pi}{2} = \pi \, sec$.

(D) For a simple harmonic oscillator,the kinetic energy $T$ at displacement $X$ is given by $T = \frac{1}{2} m \omega^2 (a^2 - X^2)$.
The potential energy $V$ at displacement $X$ is given by $V = \frac{1}{2} m \omega^2 X^2$.
To find the ratio of kinetic energy $T$ to potential energy $V$,we divide the two expressions:
$\frac{T}{V} = \frac{\frac{1}{2} m \omega^2 (a^2 - X^2)}{\frac{1}{2} m \omega^2 X^2}$.
Canceling the common terms $\frac{1}{2} m \omega^2$,we get:
$\frac{T}{V} = \frac{a^2 - X^2}{X^2}$.

(D) When a trolley moves with a horizontal acceleration $a$,a pseudo force $ma$ acts on the bob of the pendulum in the direction opposite to the acceleration of the trolley.
The effective acceleration due to gravity $g'$ is the vector sum of the acceleration due to gravity $g$ (acting downwards) and the pseudo acceleration $a$ (acting horizontally).
Since these two vectors are perpendicular to each other,the magnitude of the effective acceleration $g'$ is given by:
$g' = \sqrt{g^2 + a^2}$
Thus,the correct option is $D$.

(A) The intensity $I$ of a wave is directly proportional to the square of its amplitude $a$,i.e.,$I \propto a^2$.
Given the ratio of intensities is $\frac{I_1}{I_2} = \frac{4}{1}$.
Since $\frac{I_1}{I_2} = \left( \frac{a_1}{a_2} \right)^2$,we have $\left( \frac{a_1}{a_2} \right)^2 = \frac{4}{1}$.
Taking the square root on both sides,we get $\frac{a_1}{a_2} = \sqrt{\frac{4}{1}} = \frac{2}{1}$.
Therefore,the ratio of the amplitudes is $2:1$.

(C) Density $\rho$ is given by the formula $\rho = \frac{m}{V}$.
For a product or quotient,the relative error is the sum of the relative errors of the individual measurements: $\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + \frac{\Delta V}{V}$.
Given: $m = 22.42 \; g$,$\Delta m = 0.01 \; g$,$V = 4.7 \; cc$,$\Delta V = 0.1 \; cc$.
Substituting the values:
$\frac{\Delta \rho}{\rho} = \frac{0.01}{22.42} + \frac{0.1}{4.7}$.
$\frac{\Delta \rho}{\rho} \approx 0.000446 + 0.021276 \approx 0.021722$.
To find the percentage error,multiply by $100 \%$:
Percentage error $= 0.021722 \times 100 \% \approx 2.17 \%$.
Rounding to the nearest significant value provided in the options,the maximum percentage error is approximately $2 \%$.

(B) Let the total distance be $d = 200 \; m$. The first half distance is $d_1 = 100 \; m$ and the second half distance is $d_2 = 100 \; m$.
The time taken for the first half is $t_1 = \frac{d_1}{v_1} = \frac{100}{40} \; h$.
The time taken for the second half is $t_2 = \frac{d_2}{v} = \frac{100}{v} \; h$.
The average speed $v_{avg}$ is given by the formula: $v_{avg} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{d}{t_1 + t_2}$.
Substituting the given values: $48 = \frac{200}{\frac{100}{40} + \frac{100}{v}}$.
Dividing the numerator and denominator by $100$: $48 = \frac{2}{\frac{1}{40} + \frac{1}{v}}$.
Rearranging the terms: $\frac{1}{40} + \frac{1}{v} = \frac{2}{48} = \frac{1}{24}$.
Solving for $v$: $\frac{1}{v} = \frac{1}{24} - \frac{1}{40} = \frac{5 - 3}{120} = \frac{2}{120} = \frac{1}{60}$.
Therefore,$v = 60 \; km/h$.

(B) Translational kinetic energy is given by $K_t = \frac{1}{2}mv^2$.
Rotational kinetic energy is given by $K_r = \frac{1}{2}I\omega^2$.
For a solid sphere,the moment of inertia is $I = \frac{2}{5}mr^2$ and the relation between linear and angular velocity is $v = r\omega$,so $\omega = \frac{v}{r}$.
Substituting these into the rotational kinetic energy formula:
$K_r = \frac{1}{2} \left( \frac{2}{5}mr^2 \right) \left( \frac{v}{r} \right)^2 = \frac{1}{5}mv^2$.
The total kinetic energy is $K_{total} = K_t + K_r = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \left( \frac{5+2}{10} \right)mv^2 = \frac{7}{10}mv^2$.
The ratio $K_t : (K_t + K_r)$ is $\frac{\frac{1}{2}mv^2}{\frac{7}{10}mv^2} = \frac{1}{2} \times \frac{10}{7} = \frac{5}{7}$.

(D) In an elastic collision between two bodies of equal mass,the velocities of the bodies are interchanged after the collision.
Given: Initial velocity of ball $A$ $(u_A)$ = $0.5 \, m s^{-1}$ and initial velocity of ball $B$ $(u_B)$ = $-0.3 \, m s^{-1}$.
Since the masses are identical and the collision is elastic,the final velocity of ball $A$ $(v_A)$ will be equal to $u_B$,and the final velocity of ball $B$ $(v_B)$ will be equal to $u_A$.
Therefore,$v_A = -0.3 \, m s^{-1}$ and $v_B = 0.5 \, m s^{-1}$.
The question asks for the velocities of $B$ and $A$ respectively,which are $v_B$ and $v_A$.
Thus,the velocities are $0.5 \, m s^{-1}$ and $-0.3 \, m s^{-1}$.

(D) For a closed organ pipe,the frequencies of the harmonics are given by $f_n = (2n - 1)f_1$,where $n = 1, 2, 3, ...$ is the harmonic number.
The fundamental frequency $(n=1)$ is the first harmonic.
The first overtone is the third harmonic $(n=2)$.
The second overtone is the fifth harmonic $(n=3)$.
The third overtone is the seventh harmonic $(n=4)$.
For the $n$-th harmonic in a closed pipe,the number of nodes is $n$ and the number of antinodes is $n$.
Since the third overtone corresponds to the $4$-th harmonic $(n=4)$,the air column in the pipe will have $4$ nodes and $4$ antinodes.

(C) When milk is churned,it undergoes circular motion. The cream particles,being lighter than the milk,experience a smaller centripetal force required to maintain circular motion. Consequently,they are pushed away from the outer edge towards the center of the container due to the centrifugal force,which is a pseudo-force acting in a rotating frame of reference.

(A) Let the speed of each train be $u$ and the frequency of the whistle be $\nu$. The speed of sound is $v = 340 \; m/s$.
Since the trains are moving towards each other,the apparent frequency $\nu'$ is given by the Doppler effect formula:
$\nu' = \nu \left( \frac{v + u}{v - u} \right)$
Given that the pitch changes by a factor of $9/8$,we have $\nu' = \frac{9}{8} \nu$.
Substituting the values:
$\frac{9}{8} \nu = \nu \left( \frac{340 + u}{340 - u} \right)$
$\frac{9}{8} = \frac{340 + u}{340 - u}$
Cross-multiplying gives:
$9(340 - u) = 8(340 + u)$
$3060 - 9u = 2720 + 8u$
$3060 - 2720 = 8u + 9u$
$340 = 17u$
$u = \frac{340}{17} = 20 \; m/s$.
Thus,the speed of each train is $20 \; m/s$.

(B) The equation of the line is $Y = X + 4$,which can be written as $X - Y + 4 = 0$.
The perpendicular distance $d$ from the origin $(0, 0)$ to this line is given by the formula $d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$.
Substituting the values,$d = \frac{|1(0) - 1(0) + 4|}{\sqrt{1^2 + (-1)^2}} = \frac{4}{\sqrt{2}}$.
The angular momentum $L$ about the origin is given by $L = mvd$.
Substituting the given values $m = 5$,$v = 3\sqrt{2}$,and $d = \frac{4}{\sqrt{2}}$:
$L = 5 \times 3\sqrt{2} \times \frac{4}{\sqrt{2}} = 5 \times 3 \times 4 = 60$ units.

(A) Newton's first law of motion states that an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force.
This law defines force as that which changes the state of rest or uniform motion of an object.
It implies that force is an external agent that can exist independently of the object's motion,establishing the concept of the physical independence of force.
Therefore,the first law of motion is the basis for the physical independence of force.

(D) The magnitude of the sum of two vectors is given by $|\vec{A}+\vec{B}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}| \cos \theta$.
Similarly,the magnitude of the difference of two vectors is given by $|\vec{A}-\vec{B}|^2 = |\vec{A}|^2 + |\vec{B}|^2 - 2|\vec{A}||\vec{B}| \cos \theta$.
Given that $|\vec{A}+\vec{B}| = |\vec{A}-\vec{B}|$,we square both sides to get $|\vec{A}+\vec{B}|^2 = |\vec{A}-\vec{B}|^2$.
Substituting the expressions,we have $|\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}| \cos \theta = |\vec{A}|^2 + |\vec{B}|^2 - 2|\vec{A}||\vec{B}| \cos \theta$.
Canceling the common terms $|\vec{A}|^2$ and $|\vec{B}|^2$ from both sides,we get $2|\vec{A}||\vec{B}| \cos \theta = -2|\vec{A}||\vec{B}| \cos \theta$.
Rearranging the terms,we get $4|\vec{A}||\vec{B}| \cos \theta = 0$.
Since the vectors are non-zero,$|\vec{A}| \neq 0$ and $|\vec{B}| \neq 0$,therefore $\cos \theta = 0$.
This implies $\theta = 90^{\circ}$.

(C) The radius $r$ of a circular path of a charged particle moving in a uniform magnetic field is given by the formula $r = \frac{mv}{qB}$,where $m$ is the mass,$v$ is the speed,$q$ is the charge,and $B$ is the magnetic field strength.
Since $m$,$q$,and $B$ are constant,the radius is directly proportional to the speed,i.e.,$r \propto v$.
Given the initial radius $r_1 = 2\, cm$ and the initial speed $v_1 = v$.
If the speed is doubled,$v_2 = 2v$.
Therefore,the new radius $r_2$ is given by $\frac{r_2}{r_1} = \frac{v_2}{v_1} = \frac{2v}{v} = 2$.
Thus,$r_2 = 2 \times r_1 = 2 \times 2\, cm = 4\, cm$.

(D) The radius of a circular path of a charged particle in a magnetic field is given by $r = \frac{\sqrt{2mK}}{qB}$,where $m$ is the mass,$K$ is the kinetic energy,and $q$ is the charge.
From this,we can write $K = \frac{r^2 q^2 B^2}{2m}$.
Since $r$ and $B$ are the same for both particles,$K \propto \frac{q^2}{m}$.
For a deuteron $(d)$,$q_d = e$ and $m_d = 2m_p$. For a proton $(p)$,$q_p = e$ and $m_p = m_p$.
Taking the ratio: $\frac{K_p}{K_d} = \left( \frac{q_p}{q_d} \right)^2 \times \left( \frac{m_d}{m_p} \right) = \left( \frac{e}{e} \right)^2 \times \left( \frac{2m_p}{m_p} \right) = 1^2 \times 2 = 2$.
Therefore,$K_p = 2 \times K_d = 2 \times 50 \, keV = 100 \, keV$.

(A) Faraday's laws of electromagnetic induction describe the process where a changing magnetic flux induces an electromotive force $(EMF)$.
This process involves the conversion of mechanical work done to move a conductor or change the magnetic field into electrical energy.
Since energy is neither created nor destroyed,this phenomenon is a direct consequence of the law of conservation of energy.

(A) The induced emf is given by Faraday's law: $|e| = N \frac{|\Delta \Phi|}{\Delta t}$.
Here,$N = 50$,$A = 100 \, cm^2 = 100 \times 10^{-4} \, m^2 = 10^{-2} \, m^2$,$B_1 = 2 \times 10^{-2} \, T$,$B_2 = 0 \, T$,and $e = 0.1 \, V$.
The change in magnetic flux is $\Delta \Phi = A(B_2 - B_1) = 10^{-2} \times (0 - 2 \times 10^{-2}) = -2 \times 10^{-4} \, Wb$.
The magnitude of induced emf is $|e| = N \frac{|\Delta \Phi|}{t}$.
Substituting the values: $0.1 = 50 \times \frac{2 \times 10^{-4}}{t}$.
$t = \frac{50 \times 2 \times 10^{-4}}{0.1} = \frac{100 \times 10^{-4}}{0.1} = \frac{10^{-2}}{10^{-1}} = 0.1 \, s$.

(A) The self-inductance $L$ of a solenoid is given by the formula $L = \frac{\mu_0 N^2 A}{l}$,where $N$ is the number of turns,$A$ is the cross-sectional area,and $l$ is the length of the coil.
Since $A$ and $l$ remain constant,we have $L \propto N^2$.
Given that the number of turns is doubled,$N_2 = 2N_1$.
Therefore,the new self-inductance $L_2$ is given by $L_2 = L_1 \left( \frac{N_2}{N_1} \right)^2 = L_1 \left( \frac{2N_1}{N_1} \right)^2 = 4L_1$.
Thus,the self-inductance becomes four times the original value.

(C) The energy $U$ stored in an inductor is given by the formula $U = \frac{1}{2} L I^2$.
Given:
Inductance $L = 100\, mH = 100 \times 10^{-3}\, H = 0.1\, H$.
Current $I = 1\, A$.
Substituting the values into the formula:
$U = \frac{1}{2} \times 0.1 \times (1)^2$
$U = 0.05\, J$.
Therefore,the energy stored in the magnetic field is $0.05\, J$.

(B) The energy of a photon is given by the relation $E = h\nu$,where $h$ is Planck's constant and $\nu$ is the frequency.
Given energy $E = 1 \;MeV = 1 \times 10^6 \;eV = 10^6 \times 1.6 \times 10^{-19} \;J = 1.6 \times 10^{-13} \;J$.
Planck's constant $h \approx 6.63 \times 10^{-34} \;J \cdot s$.
Using the formula $\nu = \frac{E}{h}$:
$\nu = \frac{1.6 \times 10^{-13} \;J}{6.63 \times 10^{-34} \;J \cdot s} \approx 0.241 \times 10^{21} \;Hz = 2.41 \times 10^{20} \;Hz$.
Thus,the frequency is approximately $2.4 \times 10^{20} \;Hz$.

(D) According to Einstein's photoelectric equation: $E = W_0 + K_{\max}$.
First,calculate the energy of the incident photon: $E = \frac{hc}{\lambda} \approx \frac{12400 \; eV \cdot \mathring{A}}{3000 \; \mathring{A}} \approx 4.13 \; eV$.
Given the work function $W_0 = 1 \; eV$,the maximum kinetic energy is $K_{\max} = E - W_0 = 4.13 \; eV - 1 \; eV = 3.13 \; eV$.
Convert $K_{\max}$ to Joules: $K_{\max} = 3.13 \times 1.6 \times 10^{-19} \; J \approx 5 \times 10^{-19} \; J$.
Using $K_{\max} = \frac{1}{2} m v^2$,where $m = 9.1 \times 10^{-31} \; kg$:
$v = \sqrt{\frac{2 K_{\max}}{m}} = \sqrt{\frac{2 \times 5 \times 10^{-19}}{9.1 \times 10^{-31}}} \approx \sqrt{1.1 \times 10^{12}} \approx 1.05 \times 10^6 \; m/s$.
Thus,the velocity is approximately $1 \times 10^6 \; m/s$.

(D) The formula for the remaining fraction of a radioactive sample is given by $\frac{N}{N_0} = (1/2)^n$,where $n$ is the number of half-lives.
Given,half-life $T_{1/2} = 1600$ years and total time $t = 6400$ years.
The number of half-lives $n = \frac{t}{T_{1/2}} = \frac{6400}{1600} = 4$.
Therefore,the fraction remaining is $\frac{N}{N_0} = (1/2)^4 = \frac{1}{16}$.

(B) When a $P-N$ junction is formed,there is a high concentration of electrons in the $N$-region and a high concentration of holes in the $P$-region.
Due to this concentration gradient,electrons diffuse from the $N$-side to the $P$-side,and holes diffuse from the $P$-side to the $N$-side.
As these charge carriers cross the junction,they recombine near the interface,leaving behind immobile ionized impurity atoms.
This region,depleted of mobile charge carriers,is known as the depletion layer.

(D) The distance of the $n^{th}$ dark fringe from the central fringe is given by the formula:
$x_n = \frac{(2n - 1) \lambda D}{2d}$
For the second dark fringe $(n = 2)$:
$x_2 = \frac{(2 \times 2 - 1) \lambda D}{2d} = \frac{3 \lambda D}{2d}$
Given: $x_2 = 1 \, mm = 1 \times 10^{-3} \, m$,$D = 1 \, m$,and $d = 0.90 \, mm = 0.9 \times 10^{-3} \, m$.
Substituting the values:
$1 \times 10^{-3} = \frac{3 \times \lambda \times 1}{2 \times 0.9 \times 10^{-3}}$
$1 \times 10^{-3} = \frac{3 \lambda}{1.8 \times 10^{-3}}$
$3 \lambda = 1.8 \times 10^{-6}$
$\lambda = 0.6 \times 10^{-6} \, m = 6 \times 10^{-7} \, m = 6 \times 10^{-5} \, cm$.

(C) The nuclear force is charge-independent.
This means that the strong nuclear force acting between two nucleons is independent of whether they are protons or neutrons.
Therefore,the magnitude of the nuclear force between two protons $(F_{pp})$,two neutrons $(F_{nn})$,and a proton and a neutron $(F_{pn})$ is approximately the same.
Thus,$F_{pp} = F_{pn} = F_{nn}$.

(A) The ground state of a hydrogen atom corresponds to the principal quantum number $n = 1$.
The second excited state corresponds to the principal quantum number $n = 3$.
The energy of an electron in the $n$-th orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \; eV$.
For the second excited state $(n = 3)$,the energy is $E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} = -1.51 \; eV$.
The energy required to ionize the atom is the energy needed to bring the electron from the current state to the state where $E = 0$.
Therefore,the ionization energy $E_{ion} = 0 - E_3 = 0 - (-1.51) = 1.51 \; eV$.

(C) According to the theory of special relativity,the relativistic mass $m$ of an object moving with velocity $v$ is given by the formula:
$m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$
Given that the speed $v = 0.8 c$,we substitute this value into the equation:
$m = \frac{m_0}{\sqrt{1 - \frac{(0.8c)^2}{c^2}}}$
$m = \frac{m_0}{\sqrt{1 - 0.64}}$
$m = \frac{m_0}{\sqrt{0.36}}$
$m = \frac{m_0}{0.6}$
$m = \frac{m_0}{6/10} = \frac{10}{6} m_0 = \frac{5}{3} m_0$
Therefore,the mass of the electron at this speed is $\frac{5}{3} m_0$.

(B) For a transistor to operate as an amplifier,it must be in the active region.
In the active region,the emitter-base junction is forward-biased,which allows charge carriers to flow from the emitter into the base.
The base-collector junction is reverse-biased,which allows the collector to collect the majority of the charge carriers injected from the emitter.
Therefore,the correct condition is that the emitter-base junction is forward-biased and the base-collector junction is reverse-biased.

(A) The wavelength of light in a medium is given by the relation $\lambda_{m} = \frac{\lambda_{a}}{\mu}$,where $\lambda_{a}$ is the wavelength in air and $\mu$ is the refractive index of the medium.
Given: $\lambda_{a} = 5460 \mathring{A}$ and $\mu = 1.5$.
Substituting the values,we get $\lambda_{g} = \frac{5460}{1.5} = 3640 \mathring{A}$.
Thus,the wavelength of light in glass is $3640 \mathring{A}$.

(A) The truth table shows that the output $Y$ is $1$ if either input $A$ or input $B$ (or both) is $1$. If both inputs are $0$,the output is $0$.
This behavior corresponds to the Boolean expression $Y = A + B$,which is the characteristic operation of an $OR$ gate.

$A$	$B$	$Y = A + B$
$0$	$0$	$0$
$0$	$1$	$1$
$1$	$0$	$1$
$1$	$1$	$1$

(A) The notation ${ }_{Z} X^{A}$ represents an atom where $Z$ is the atomic number and $A$ is the mass number.
For ${ }_{11} Na^{23}$,the atomic number $Z = 11$ and the mass number $A = 23$.
The number of protons in the nucleus is equal to the atomic number $Z$,so protons = $11$.
The number of neutrons is given by $N = A - Z = 23 - 11 = 12$.
The nucleus of an atom contains only protons and neutrons; it does not contain electrons.
Therefore,the number of electrons in the nucleus is $0$.
Thus,the number of protons,neutrons,and electrons in the nucleus are $11, 12, 0$ respectively.