AIPMT 1993 Physics Question Paper with Answer and Solution

(B) Let $\overrightarrow{v_b}$ be the velocity of the boat relative to the water and $\overrightarrow{v_r}$ be the velocity of the river water relative to the ground.
The resultant velocity of the boat relative to the ground is given by $\overrightarrow{v_{bg}} = \overrightarrow{v_b} + \overrightarrow{v_r}$.
Since the boat crosses the river perpendicular to the flow,the vectors $\overrightarrow{v_b}$ and $\overrightarrow{v_r}$ are perpendicular to each other.
Therefore,the magnitude of the resultant velocity is $v_{bg} = \sqrt{v_b^2 + v_r^2}$.
Given $v_{bg} = 10 \, km/h$ and $v_b = 8 \, km/h$,we have:
$10 = \sqrt{8^2 + v_r^2}$
$100 = 64 + v_r^2$
$v_r^2 = 100 - 64 = 36$
$v_r = 6 \, km/h$.

(A) The distance travelled by a body in the $n^{th}$ second is given by the formula: $S_n = u + \frac{a}{2}(2n - 1)$.
Since the body starts from rest,the initial velocity $u = 0$.
Therefore,the distance travelled in the $n^{th}$ second is $S_n = \frac{a}{2}(2n - 1)$.
For the $4^{th}$ second $(n = 4)$: $S_4 = \frac{a}{2}(2 \times 4 - 1) = \frac{a}{2}(7) = \frac{7a}{2}$.
For the $3^{rd}$ second $(n = 3)$: $S_3 = \frac{a}{2}(2 \times 3 - 1) = \frac{a}{2}(5) = \frac{5a}{2}$.
The ratio of the distance travelled in the $4^{th}$ second to the $3^{rd}$ second is: $\frac{S_4}{S_3} = \frac{7a/2}{5a/2} = \frac{7}{5}$.

(C) When a block is placed on an inclined plane,the forces acting on it are the gravitational force $(mg)$,the normal force $(N)$,and the static frictional force $(f_s)$.
The component of weight acting down the plane is $mg \sin \alpha$ and the component perpendicular to the plane is $mg \cos \alpha$.
For the block to be at rest,the frictional force $f_s = mg \sin \alpha$ and the normal force $N = mg \cos \alpha$.
The block starts slipping when the angle of inclination reaches the angle of repose,$\theta$. At this point,the static friction reaches its maximum value,$f_{s,max} = \mu_s N$.
Equating the forces at the point of slipping: $mg \sin \theta = \mu_s (mg \cos \theta)$.
Dividing both sides by $mg \cos \theta$,we get $\mu_s = \frac{\sin \theta}{\cos \theta} = \tan \theta$.
Thus,the coefficient of static friction is $\tan \theta$.

(D) The relationship between kinetic energy $(K)$,momentum $(P)$,and mass $(m)$ is given by $K = \frac{P^2}{2m}$.
Since the kinetic energies are the same $(K_1 = K_2)$,we have $\frac{P_1^2}{2m_1} = \frac{P_2^2}{2m_2}$.
Rearranging the terms,we get $\frac{P_1^2}{P_2^2} = \frac{m_1}{m_2}$.
Taking the square root on both sides,the ratio of their momenta is $\frac{P_1}{P_2} = \sqrt{\frac{m_1}{m_2}}$.
Given $m_1 = 1\, g$ and $m_2 = 9\, g$,we substitute these values:
$\frac{P_1}{P_2} = \sqrt{\frac{1}{9}} = \frac{1}{3}$.
Therefore,the ratio of their momentum is $1:3$.

(D) According to Kepler's Third Law of Planetary Motion,the square of the time period $T$ of a satellite is directly proportional to the cube of its orbital radius $r$,i.e.,$T^2 \propto r^3$.
This relationship is independent of the mass of the satellite.
Given for satellite $A$: $r_A = r$.
Given for satellite $B$: $r_B = 2r$.
The ratio of their time periods is given by:
$\frac{T_A}{T_B} = \left( \frac{r_A}{r_B} \right)^{3/2} = \left( \frac{r}{2r} \right)^{3/2} = \left( \frac{1}{2} \right)^{3/2} = \frac{1}{2^{3/2}} = \frac{1}{2 \cdot 2^{1/2}} = \frac{1}{2\sqrt{2}}$.
Thus,the ratio is $1:2\sqrt{2}$.

(B) The melting point of ice decreases with an increase in pressure. This is because ice expands upon solidification,meaning its volume increases when it turns from liquid water to solid ice. According to Le Chatelier's principle and the Clausius-Clapeyron relation,for substances that expand upon freezing,an increase in pressure lowers the melting point.

(D) According to Mayer's relation,for any ideal gas,the difference between the molar heat capacity at constant pressure $(C_p)$ and the molar heat capacity at constant volume $(C_v)$ is equal to the universal gas constant $(R)$.
Mathematically,$C_p - C_v = R$.
Since $R$ is a universal constant,it is the same for all ideal gases regardless of their molecular weight or composition.
For hydrogen gas,$C_p - C_v = a = R$.
For oxygen gas,$C_p - C_v = b = R$.
Therefore,$a = b$.

(B) For a diatomic gas molecule,the motion can be described in three-dimensional space.
The number of translational degrees of freedom is always $3$ for any gas molecule,corresponding to motion along the $x, y,$ and $z$ axes.
The number of rotational degrees of freedom for a rigid diatomic molecule is $2$.
Therefore,the number of translational degrees of freedom is $3$.

(B) According to the First Law of Thermodynamics,the heat supplied to a system is equal to the sum of the change in internal energy and the work done by the system.
$\Delta Q = \Delta U + \Delta W$
Given:
Heat added,$\Delta Q = 110\; J$
Change in internal energy,$\Delta U = 40\; J$
Rearranging the formula to find the work done:
$\Delta W = \Delta Q - \Delta U$
$\Delta W = 110\; J - 40\; J = 70\; J$
Therefore,the amount of external work done is $70\; J$.

(B) thermodynamic state function is a property whose value depends only on the current state of the system and not on the path taken to reach that state.
Enthalpy $(H)$,Gibbs energy $(G)$,and Internal energy $(U)$ are all state functions because they depend only on the state variables of the system.
Work done $(W)$ is a path function,meaning its value depends on the specific process or path taken to change the state of the system.
Therefore,work done is not a thermodynamic state function.

(D) The thermodynamic state of a system is defined by state functions such as Pressure $(P)$,Volume $(V)$,Temperature $(T)$,and Internal Energy $(U)$. These depend only on the current state of the system,not on the path taken to reach that state.
Work $(W)$ is a path function,not a state function. It represents the energy transferred between the system and its surroundings due to a process,and its value depends on the specific path taken during the thermodynamic process. Therefore,work does not characterize the thermodynamic state of matter.

(D) According to the Stefan-Boltzmann Law,the power $P$ radiated by a black body is proportional to the fourth power of its absolute temperature $T$,i.e.,$P \propto T^4$.
Since the rate of energy received on Earth is directly proportional to the power radiated by the Sun,we have $E \propto T^4$.
If the temperature is doubled $(T' = 2T)$,the new rate of energy $E'$ will be $E' \propto (2T)^4 = 16T^4$.
Therefore,the rate of energy received increases by a factor of $16$.

(C) The potential energy $U$ of a particle executing simple harmonic motion at a displacement $y$ is given by $U = \frac{1}{2}m\omega^2 y^2$.
The maximum potential energy $U_{\max}$ occurs at the amplitude $a$,given by $U_{\max} = \frac{1}{2}m\omega^2 a^2$.
According to the problem,the potential energy is one-fourth of its maximum value:
$U = \frac{1}{4} U_{\max}$
Substituting the expressions for $U$ and $U_{\max}$:
$\frac{1}{2}m\omega^2 y^2 = \frac{1}{4} (\frac{1}{2}m\omega^2 a^2)$
Canceling the common terms $\frac{1}{2}m\omega^2$ from both sides:
$y^2 = \frac{1}{4} a^2$
Taking the square root of both sides:
$y = \frac{a}{2}$

(C) The speed of sound in an ideal gas is given by the formula $v = \sqrt{\frac{\gamma RT}{M}}$,where $\gamma$ is the adiabatic index,$R$ is the gas constant,$T$ is the absolute temperature in Kelvin,and $M$ is the molar mass of the gas.
From this relation,we can see that $v \propto \sqrt{T}$.
Let $v_1$ be the speed of sound at temperature $T_1$ and $v_2$ be the speed at $T_2$.
Given $T_1 = 27^\circ C = 27 + 273 = 300 \ K$.
We want the speed to be doubled,so $v_2 = 2v_1$.
Using the proportionality $v_2/v_1 = \sqrt{T_2/T_1}$,we get $2 = \sqrt{T_2/300}$.
Squaring both sides,$4 = T_2/300$,which gives $T_2 = 1200 \ K$.
Converting back to Celsius,$T_2 = 1200 - 273 = 927^\circ C$.

(D) Binary stars are two stars that orbit around a common center of mass. By observing their orbital periods and the separation between them,astronomers can apply Kepler's laws and Newton's law of universal gravitation to determine the masses of the stars. Furthermore,the motion of binary stars provides a direct observational test for Newton's law of gravitation in a celestial context,confirming that the same gravitational laws apply to distant stars as they do to objects on Earth.

(D) Let the side $AB = B$ and $BC = L = 2B$. The mass of the rectangle is $M$.
For an axis passing through the center of mass and parallel to the sides:
$I_{EG} = \frac{MB^2}{12}$
$I_{HF} = \frac{ML^2}{12} = \frac{M(2B)^2}{12} = \frac{4MB^2}{12} = \frac{MB^2}{3}$
For the diagonal $BD$,the moment of inertia is given by $I_{BD} = \frac{M B^2 L^2}{6(B^2 + L^2)}$.
Substituting $L = 2B$:
$I_{BD} = \frac{M B^2 (2B)^2}{6(B^2 + (2B)^2)} = \frac{4MB^4}{6(5B^2)} = \frac{4MB^2}{30} = \frac{2MB^2}{15} \approx 0.133 MB^2$.
Comparing the values:
$I_{EG} = 0.0833 MB^2$
$I_{HF} = 0.333 MB^2$
$I_{BD} = 0.133 MB^2$
Thus,the moment of inertia is minimum about the axis $EG$.

(B) The given equation is $P = P_0 \exp(-\alpha t^2)$.
In any exponential function of the form $e^x$,the exponent $x$ must be dimensionless.
Therefore,the term $\alpha t^2$ must be dimensionless.
This implies that the dimensions of $\alpha$ multiplied by the dimensions of $t^2$ must equal $1$ (dimensionless).
$[\alpha] [t^2] = [M^0 L^0 T^0] = 1$.
Since $[t] = [T]$,we have $[\alpha] [T^2] = 1$.
Thus,$[\alpha] = [T^{-2}]$.

(A) The frequency of the first wave is $f_1 = \frac{1}{T_1} = \frac{1}{4}\; Hz$.
The frequency of the second wave is $f_2 = \frac{1}{T_2} = \frac{1}{3}\; Hz$.
When two waves of different frequencies are combined,they produce beats.
The beat frequency is given by $f_b = |f_2 - f_1| = |\frac{1}{3} - \frac{1}{4}| = \frac{4-3}{12} = \frac{1}{12}\; Hz$.
The time period of the resultant beat phenomenon is $T = \frac{1}{f_b} = \frac{1}{1/12} = 12\; s$.

(C) The dimensions of pressure $P$ are $[ML^{-1}T^{-2}]$.
The dimensions of radius $r$ and distance $x$ are $[L]$.
The dimensions of velocity $v$ are $[LT^{-1}]$.
The dimensions of length $l$ are $[L]$.
Given the formula $\eta = \frac{P(r^2 - x^2)}{4vl}$,we substitute the dimensions:
$\eta = \frac{[ML^{-1}T^{-2}] \cdot [L^2]}{[LT^{-1}] \cdot [L]}$
$\eta = \frac{[MLT^{-2}]}{[L^2T^{-1}]}$
$\eta = [ML^{-1}T^{-1}]$

(C) The time taken for an object to roll down an inclined plane is given by $t = \sqrt{\frac{2h}{g \sin^2 \theta} (1 + \frac{K^2}{R^2})}$.
For a solid sphere,the moment of inertia $I = \frac{2}{5}MR^2$,so $\frac{K^2}{R^2} = \frac{2}{5} = 0.4$.
For a disc and a solid cylinder,the moment of inertia $I = \frac{1}{2}MR^2$,so $\frac{K^2}{R^2} = \frac{1}{2} = 0.5$.
Since the time $t$ is directly proportional to $\sqrt{1 + \frac{K^2}{R^2}}$,the object with the smallest value of $\frac{K^2}{R^2}$ will reach the bottom first.
Comparing the values,$0.4 < 0.5$,therefore the solid sphere reaches the bottom first.

(D) The fundamental frequency $f$ of a stretched string is given by the formula: $f = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$.
Here,$T$ is the tension and $\mu$ is the linear mass density of the string.
Since $T$ and $\mu$ remain constant,the product of frequency and length is constant: $f_1 l_1 = f_2 l_2$.
Given $f_1 = 512 \; Hz$,$l_1 = 0.5 \; m$,and $f_2 = 256 \; Hz$.
Substituting the values: $512 \times 0.5 = 256 \times l_2$.
$256 = 256 \times l_2$.
Therefore,$l_2 = 1 \; m$.

(B) Let $m$ be the mass of the monkey and $g$ be the acceleration due to gravity.
Let $a$ be the acceleration with which the monkey descends.
The force equation for the monkey is given by $mg - T = ma$,where $T$ is the tension in the branch.
To prevent the branch from breaking,the tension $T$ must not exceed the breaking strength of the branch.
The breaking strength is given as $75 \%$ of the weight of the monkey,so $T_{max} = 0.75mg = \frac{3}{4}mg$.
Substituting $T = \frac{3}{4}mg$ into the equation $mg - T = ma$:
$mg - \frac{3}{4}mg = ma$
$\frac{1}{4}mg = ma$
$a = \frac{g}{4}$.
Thus,the minimum acceleration required for the monkey to slide down without breaking the branch is $\frac{g}{4}$.

(B) The electric potential energy $U$ of a system of two point charges $q_1$ and $q_2$ separated by a distance $r$ is given by $U = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r}$.
For two electrons,$q_1 = -e$ and $q_2 = -e$,so the potential energy is $U = \frac{e^2}{4\pi\varepsilon_0 r}$.
When one electron is moved towards the other,the separation distance $r$ decreases.
Since $U \propto \frac{1}{r}$,as the distance $r$ decreases,the potential energy $U$ increases.

(C) The $3\,\Omega$ and $6\,\Omega$ resistors are connected in parallel. Therefore,the potential difference across both is the same.
Let $V_p$ be the potential difference across the parallel combination.
$V_p = I_1 R_1 = I_2 R_2$
$0.8 \times 3 = I_2 \times 6$
$I_2 = \frac{2.4}{6} = 0.4\,A$
The total current $I$ flowing through the $4\,\Omega$ resistor is the sum of the currents through the parallel branches:
$I = I_1 + I_2 = 0.8\,A + 0.4\,A = 1.2\,A$
The potential drop across the $4\,\Omega$ resistor is given by $V = I \times R = 1.2\,A \times 4\,\Omega = 4.8\,V$.

(D) When three resistors of $4\,\Omega$ are connected in an equilateral triangle,let the corners be $A$,$B$,and $C$.
To find the effective resistance between any two corners (e.g.,$A$ and $B$),we observe the circuit configuration.
The resistor connected directly between $A$ and $B$ is in parallel with the series combination of the other two resistors (connected between $A-C$ and $C-B$).
Resistance of the series branch = $4\,\Omega + 4\,\Omega = 8\,\Omega$.
Now,this $8\,\Omega$ branch is in parallel with the $4\,\Omega$ resistor connected directly across $A$ and $B$.
The equivalent resistance $R_{eq}$ is given by:
$\frac{1}{R_{eq}} = \frac{1}{4} + \frac{1}{8} = \frac{2+1}{8} = \frac{3}{8}$.
Therefore,$R_{eq} = \frac{8}{3}\,\Omega$.

(A) The magnetic force on a moving charge is given by the Lorentz force formula: $\overrightarrow{F_m} = q(\overrightarrow{v} \times \overrightarrow{B})$.
Here,the velocity vector $\overrightarrow{v}$ is in the $+X$-direction,and the magnetic field $\overrightarrow{B}$ is in the $-X$-direction.
The angle $\theta$ between $\overrightarrow{v}$ and $\overrightarrow{B}$ is $180^\circ$.
Since the cross product $\overrightarrow{v} \times \overrightarrow{B} = vB \sin(180^\circ) = 0$,the magnetic force $\overrightarrow{F_m}$ is zero.
Therefore,the charge will continue to move in the $X$-direction with constant velocity and remain unaffected.

(A) current-carrying coil placed in a uniform magnetic field experiences a magnetic torque given by $\vec{\tau} = \vec{M} \times \vec{B}$,where $\vec{M}$ is the magnetic moment and $\vec{B}$ is the magnetic field.
According to Faraday's law of electromagnetic induction,an $E.M.F.$ is induced only when there is a change in magnetic flux linked with the coil $(\varepsilon = -d\phi/dt)$.
Since the magnetic field is uniform and the coil is not specified to be moving or changing its orientation relative to the field,there is no change in magnetic flux. Therefore,no $E.M.F.$ is induced.
Thus,only torque is formed. The correct option is $A$.

(D) The induced $e.m.f.$ $(e)$ in a coil due to self-inductance is given by the formula: $e = L \cdot \left| \frac{di}{dt} \right|$.
Given:
$e = 5 \,V$
Change in current,$di = 3 \,A - 2 \,A = 1 \,A$
Time interval,$dt = 1 \,ms = 1 \times 10^{-3} \,s$
Substituting the values into the formula:
$5 = L \cdot \frac{1 \,A}{1 \times 10^{-3} \,s}$
$5 = L \cdot 10^3$
$L = \frac{5}{10^3} \,H = 5 \times 10^{-3} \,H = 5 \,mH$.
Therefore,the correct option is $D$.

(A) The energy of a photon is given by $E = h\nu$,where $h$ is Planck's constant and $\nu$ is the frequency.
According to Einstein's mass-energy equivalence,$E = mc^2$,where $m$ is the relativistic mass and $c$ is the speed of light.
Equating the two expressions for energy: $mc^2 = h\nu$.
Since the momentum $p$ of a photon is defined as $p = mc$,we can substitute $m = \frac{p}{c}$ into the equation: $p = \frac{h\nu}{c}$.
Using the relation $\nu = \frac{c}{\lambda}$,we substitute for $\nu$: $p = \frac{h(c/\lambda)}{c} = \frac{h}{\lambda}$.
Therefore,the momentum of a photon is $\frac{h}{\lambda}$.

(D) According to the photoelectric effect,the number of photo-electrons emitted per second is directly proportional to the number of incident photons per second.
Since the intensity of light is defined as the energy incident per unit area per unit time,it is directly proportional to the number of photons incident on the metal surface.
Therefore,increasing the intensity of the incident light increases the number of incident photons,which in turn increases the number of photo-electrons emitted per second.
Thus,the correct option is $D$.

(B) The work function $W_0$ of a photoelectric emitter is given by the formula $W_0 = \frac{hc}{\lambda_0}$,where $h$ is Planck's constant,$c$ is the speed of light,and $\lambda_0$ is the threshold wavelength.
Assuming the given wavelengths are the threshold wavelengths for the respective emitters,we have $\lambda_{01} = 300 \; nm$ and $\lambda_{02} = 600 \; nm$.
The ratio of the work functions is $\frac{W_{01}}{W_{02}} = \frac{hc / \lambda_{01}}{hc / \lambda_{02}} = \frac{\lambda_{02}}{\lambda_{01}}$.
Substituting the values: $\frac{W_{01}}{W_{02}} = \frac{600 \; nm}{300 \; nm} = \frac{2}{1}$.
Thus,the ratio is $2:1$.

(B) The number of spectral lines emitted when an electron transitions from an excited state with principal quantum number $n$ to the ground state is given by the formula:
$N = \frac{n(n - 1)}{2}$
Given that the electron is excited to the state with principal quantum number $n = 4$,we substitute this value into the formula:
$N = \frac{4(4 - 1)}{2}$
$N = \frac{4 \times 3}{2}$
$N = \frac{12}{2} = 6$
Therefore,the total number of spectral lines in the emission spectra is $6$.

(B) The binding energy per nucleon $(B.E./A)$ is a measure of the stability of a nucleus.
According to the binding energy curve,the value of $B.E./A$ increases with the mass number $A$ for light nuclei,reaches a maximum value of approximately $8.8 \text{ MeV}$ for iron $(_{26}^{56}Fe)$,and then gradually decreases for heavier nuclei.
Therefore,the binding energy per nucleon is maximum for $_{26}^{56}Fe$.

(B) Power $P = 5 \, W = 5 \, J/s$.
Energy released per fission $E = 200 \, MeV = 200 \times 10^6 \times 1.6 \times 10^{-19} \, J = 3.2 \times 10^{-11} \, J$.
Fission rate $R = \frac{P}{E} = \frac{5}{3.2 \times 10^{-11}} \, s^{-1}$.
$R = 1.5625 \times 10^{11} \, s^{-1} \approx 1.56 \times 10^{11} \, s^{-1}$.

(B) The given reaction sequence is: $_Z{X^A} \to _{Z+1}{Y^A} \to _{Z-1}{K^{A-4}} \to _{Z-1}{K^{A-4}}$.
$1$. In the first step,$_Z{X^A} \to _{Z+1}{Y^A}$,the atomic number increases by $1$ while the mass number remains the same. This corresponds to the emission of a $\beta^-$-particle $(_{-1}e^0)$.
$2$. In the second step,$_{Z+1}{Y^A} \to _{Z-1}{K^{A-4}}$,the atomic number decreases by $2$ and the mass number decreases by $4$. This corresponds to the emission of an $\alpha$-particle $(_{2}He^4)$.
$3$. In the third step,$_{Z-1}{K^{A-4}} \to _{Z-1}{K^{A-4}}$,there is no change in atomic number or mass number,which indicates the emission of a $\gamma$-ray (electromagnetic radiation).
Therefore,the sequence of emissions is $\beta, \alpha, \gamma$.

(B) In a transistor,the $Emitter$ is the section that is heavily doped. The purpose of heavy doping is to provide a large number of majority charge carriers,which are then injected into the $Base$ region. The $Base$ is lightly doped and thin,while the $Collector$ is moderately doped. Therefore,the correct option is $B$.

(C) The velocity of light in a medium is given by $v = \frac{c}{n}$,where $c$ is the speed of light in vacuum $(3 \times 10^8 \, m/s)$ and $n$ is the refractive index.
Given $n = 1.5$,the velocity of light in the window is $v = \frac{3 \times 10^8}{1.5} = 2 \times 10^8 \, m/s$.
The thickness of the window is $d = 4 \, mm = 4 \times 10^{-3} \, m$.
The time taken $t$ is given by $t = \frac{d}{v}$.
Substituting the values,$t = \frac{4 \times 10^{-3}}{2 \times 10^8} = 2 \times 10^{-11} \, s$.

(B) line emission spectrum is produced by excited atoms in a low-pressure gas.
Among the given options,a neon street sign contains neon gas at low pressure.
When an electric discharge passes through the gas,the neon atoms are excited and emit light at specific,discrete wavelengths,resulting in a line emission spectrum.
In contrast,sources like an electric fire or the Sun produce continuous spectra due to thermal radiation from solids or high-pressure gases.

(D) The fringe width $\beta$ in an interference experiment is given by $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength of light,$D$ is the distance between the slits and the screen,and $d$ is the distance between the slits.
When the chamber is filled with air,the wavelength of light is $\lambda_{air} = \frac{\lambda_0}{\mu_{air}}$,where $\lambda_0$ is the wavelength in vacuum and $\mu_{air} \approx 1.0003$.
When the chamber is evacuated,the refractive index becomes $\mu_{vac} = 1$.
Since $\mu_{vac} < \mu_{air}$,the wavelength of light in the vacuum becomes larger than in air $(\lambda_{vac} > \lambda_{air})$.
As $\beta \propto \lambda$,the fringe width $\beta$ increases when the chamber is evacuated.

(C) For the first minimum in a single slit diffraction pattern,the condition is given by $d \sin \theta = n \lambda$,where $n = 1$ for the first minimum.
Given: Wavelength $\lambda = 5000 \; \mathring{A} = 5000 \times 10^{-10} \; m = 5 \times 10^{-7} \; m$.
Slit width $d = 0.001 \; mm = 10^{-3} \; mm = 10^{-6} \; m$.
Substituting the values into the formula: $\sin \theta = \frac{\lambda}{d} = \frac{5 \times 10^{-7}}{10^{-6}} = 0.5$.
Therefore,$\theta = \sin^{-1}(0.5) = 30^o$.

(B) The given decay sequence is: ${}_Z^AX \to {}_{Z + 1}^AY \to {}_{Z - 1}^{A - 4}K \to {}_{Z - 1}^{A - 4}K$.
$1$. In the first step,${}_Z^AX \to {}_{Z + 1}^AY$: The atomic number increases by $1$ while the mass number remains the same. This corresponds to the emission of a $\beta^-$ particle.
$2$. In the second step,${}_{Z + 1}^AY \to {}_{Z - 1}^{A - 4}K$: The atomic number decreases by $2$ and the mass number decreases by $4$. This corresponds to the emission of an $\alpha$ particle.
$3$. In the third step,${}_{Z - 1}^{A - 4}K \to {}_{Z - 1}^{A - 4}K$: There is no change in the atomic number or mass number,which indicates the emission of a $\gamma$ ray (de-excitation of the nucleus).
Therefore,the sequence of particles emitted is $\beta, \alpha, \gamma$.

(D) The self-inductance $L$ of a coil is defined as $L = \frac{N \phi}{i}$,where $N$ is the number of turns,$\phi$ is the magnetic flux through each turn,and $i$ is the current.
For a solenoid,the magnetic flux $\phi$ is given by $\phi = B A$,where $B$ is the magnetic field and $A$ is the cross-sectional area.
The magnetic field $B$ inside a solenoid is $B = \mu_0 n i = \frac{\mu_0 N i}{l}$,where $n$ is the number of turns per unit length $(n = N/l)$ and $l$ is the length of the solenoid.
Substituting $B$ into the flux equation: $\phi = \left( \frac{\mu_0 N i}{l} \right) A$.
Now,substituting $\phi$ into the expression for $L$: $L = \frac{N}{i} \left( \frac{\mu_0 N i A}{l} \right) = \frac{\mu_0 N^2 A}{l}$.
Thus,the self-inductance $L$ is proportional to $N^2$.