MHT CET 2008 Physics Question Paper with Answer and Solution

(C) At the top of the hill, the forces acting on the rider are the gravitational force $(mg)$ acting downwards and the normal reaction $(N)$ acting upwards.
The centripetal force required for circular motion is provided by the net force towards the center: $mg - N = \frac{mv^2}{R}$.
For the condition of "weightlessness", the normal reaction $N$ must be zero.
Therefore, $mg = \frac{mv^2}{R}$.
Solving for velocity $v$: $v = \sqrt{Rg}$.
Given $R = 20\, m$ and taking $g = 10\, m/s^2$, we have:
$v = \sqrt{20 \times 10} = \sqrt{200} \approx 14.14\, m/s$.
Thus, the speed is between $14\, m/s$ and $15\, m/s$.

(D) The angular frequencies are given as $\omega_{1} = 100 \, rad \, s^{-1}$ and $\omega_{2} = 1000 \, rad \, s^{-1}$.
Let the common displacement amplitude be $A$.
The maximum acceleration $a_{max}$ for a simple harmonic motion is given by the formula $a_{max} = \omega^2 A$.
For the first motion,$a_{max1} = \omega_{1}^2 A = (100)^2 A$.
For the second motion,$a_{max2} = \omega_{2}^2 A = (1000)^2 A$.
The ratio of their maximum accelerations is $\frac{a_{max1}}{a_{max2}} = \frac{\omega_{1}^2 A}{\omega_{2}^2 A} = \frac{\omega_{1}^2}{\omega_{2}^2}$.
Substituting the values,we get $\frac{a_{max1}}{a_{max2}} = \frac{(100)^2}{(1000)^2} = \frac{10000}{1000000} = \frac{1}{100}$.
Thus,the ratio is $1:10^2$.

(C) The initial velocity of the particle is $\vec{u} = v \cos \theta \hat{i} + v \sin \theta \hat{j}$.
The final velocity of the particle when it lands on the ground is $\vec{v}_f = v \cos \theta \hat{i} - v \sin \theta \hat{j}$.
The initial momentum is $\vec{p}_i = m \vec{u} = m v \cos \theta \hat{i} + m v \sin \theta \hat{j}$.
The final momentum is $\vec{p}_f = m \vec{v}_f = m v \cos \theta \hat{i} - m v \sin \theta \hat{j}$.
The change in momentum is $\Delta \vec{p} = \vec{p}_f - \vec{p}_i = (m v \cos \theta \hat{i} - m v \sin \theta \hat{j}) - (m v \cos \theta \hat{i} + m v \sin \theta \hat{j}) = -2 m v \sin \theta \hat{j}$.
The magnitude of the change in momentum is $|\Delta \vec{p}| = 2 m v \sin \theta$.
Given $\theta = 45^{\circ}$,we have $\sin 45^{\circ} = 1 / \sqrt{2}$.
Therefore,$|\Delta \vec{p}| = 2 m v (1 / \sqrt{2}) = \sqrt{2} m v$.

(C) The gravitational potential energy of an object of mass $m$ at the surface of the earth is given by $U_{1} = -\frac{GMm}{R}$.
At a height $h = R$ from the surface,the distance from the center of the earth is $r = R + h = R + R = 2R$.
The potential energy at this height is $U_{2} = -\frac{GMm}{2R}$.
The gain in potential energy is $\Delta U = U_{2} - U_{1} = -\frac{GMm}{2R} - (-\frac{GMm}{R}) = \frac{GMm}{R} - \frac{GMm}{2R} = \frac{GMm}{2R}$.
Since the acceleration due to gravity at the surface is $g = \frac{GM}{R^{2}}$,we have $GM = gR^{2}$.
Substituting this into the expression for $\Delta U$:
$\Delta U = \frac{1}{2} \frac{(gR^{2})m}{R} = \frac{1}{2} mgR$.

(C) The orbital speed of a satellite is given by $v = \sqrt{\frac{GM}{r}}$,where $G$ is the gravitational constant,$M$ is the mass of the earth,and $r$ is the radius of the orbit.
Since both satellites are in the same orbit,$r$ is the same for both.
Therefore,the orbital speed $v$ is independent of the mass of the satellite.
Thus,$S_{1}$ and $S_{2}$ move with the same speed.
The time period is given by $T = \frac{2\pi r}{v}$,which is also independent of the mass of the satellite.
Potential energy $U = -\frac{GMm}{r}$ and kinetic energy $K = \frac{GMm}{2r}$ both depend on the mass $m$ of the satellite,so they are not equal for $S_{1}$ and $S_{2}$.

(B) An ideal gas is a theoretical gas composed of many randomly moving point particles that are not subject to inter-particle interactions. It obeys the ideal gas law $PV = nRT$ and satisfies all the fundamental assumptions of the kinetic theory of gases,such as negligible molecular volume and elastic collisions.

(A) When the driver applies brakes,the car covers a distance $x$ before coming to rest under the effect of the retarding force $F$. By the work-energy theorem: $\frac{1}{2} m v^{2} = F x$,which gives $x = \frac{m v^{2}}{2 F}$.
When the driver takes a turn,the required centripetal force is provided by the friction force $F$. Thus,$\frac{m v^{2}}{r} = F$,which gives the radius of the turn as $r = \frac{m v^{2}}{F}$.
Comparing the two,we see that $x = \frac{r}{2}$.
This implies that by using the same retarding (friction) force,the car can be brought to rest in a shorter distance by applying brakes compared to the radius required to turn the car. Therefore,braking is more effective.

(D) Using the third equation of rotational motion: $\omega^2 = \omega_0^2 - 2\alpha\theta$.
Initially,the angular velocity becomes $\omega = \frac{\omega_0}{2}$ after $\theta_1 = 36 \times 2\pi$ radians.
Substituting these values: $(\frac{\omega_0}{2})^2 = \omega_0^2 - 2\alpha(36 \times 2\pi)$.
$\frac{\omega_0^2}{4} = \omega_0^2 - 144\pi\alpha$,which gives $144\pi\alpha = \frac{3\omega_0^2}{4}$,so $\alpha = \frac{3\omega_0^2}{576\pi} = \frac{\omega_0^2}{192\pi}$.
Now,for the motion from $\omega = \frac{\omega_0}{2}$ to $\omega = 0$,let the additional rotations be $n$. The angle covered is $\theta_2 = n \times 2\pi$.
Using $0^2 = (\frac{\omega_0}{2})^2 - 2\alpha(n \times 2\pi)$.
$\frac{\omega_0^2}{4} = 2(\frac{\omega_0^2}{192\pi})(n \times 2\pi)$.
$\frac{1}{4} = \frac{4n\pi}{192\pi} = \frac{n}{48}$.
$n = \frac{48}{4} = 12$.
Thus,the fan will make $12$ more rotations before coming to rest.

(D) When the surface area of a liquid is increased,molecules from the interior of the liquid rise to the surface.
As these molecules reach the surface,work is done against the cohesive force.
This work is stored in the molecules in the form of potential energy.
Thus,the potential energy of the molecules lying on the surface is greater than that of the molecules in the interior of the liquid.

(C) The surface energy $(E)$ of a soap film is given by the product of surface tension $(T)$ and the total surface area. Since a soap film has two surfaces,the total area is $2A$.
$E = T \times 2A$
When the area of the frame is reduced by $50 \%$,the new area becomes $A' = A - 0.5A = 0.5A = A/2$.
The new surface energy $(E_1)$ is:
$E_1 = T \times 2(A/2) = T \times A$
The percentage change in surface energy is calculated as:
$\text{Percentage change} = \frac{E - E_1}{E} \times 100$
$\text{Percentage change} = \frac{2TA - TA}{2TA} \times 100 = \frac{TA}{2TA} \times 100 = 50 \%$
Thus,the energy of the soap film changes by $50 \%$.

(D) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{L/g}$,where $L$ is the distance between the point of suspension and the center of mass $(CM)$ of the system.
Initially,the $CM$ of the water-filled sphere is at the geometric center of the sphere.
As water flows out from the bottom hole,the $CM$ of the remaining water shifts downwards,which increases the effective length $L$ of the pendulum,causing the time period $T$ to increase.
As the sphere becomes nearly empty,the $CM$ of the remaining water shifts back upwards towards the center of the sphere.
This causes the effective length $L$ to decrease,which in turn causes the time period $T$ to decrease.
Therefore,the period of oscillation first increases and then decreases.

(D) The time period of a simple pendulum is given by the formula:
$T = 2 \pi \sqrt{\frac{l}{g}}$
Here,$l$ is the length of the pendulum and $g$ is the acceleration due to gravity.
Squaring both sides of the equation,we get:
$T^2 = \frac{4 \pi^2 l}{g}$
This can be written as:
$T^2 = k \cdot l$,where $k = \frac{4 \pi^2}{g}$ is a constant.
This equation is of the form $y^2 = 4ax$,which represents a parabola.
Therefore,the graph between the time period $T$ and the length $l$ is a part of a parabola.

(B) Let the radii of the thin spherical shell and the solid sphere be $R_1$ and $R_2$,respectively.
The moment of inertia of a thin spherical shell about its diameter is given by:
$I_{\text{shell}} = \frac{2}{3} MR_1^2$ ... $(i)$
The moment of inertia of a solid sphere about its diameter is given by:
$I_{\text{sphere}} = \frac{2}{5} MR_2^2$ ... (ii)
Given that the masses $(M)$ and the moments of inertia $(I)$ for both bodies are equal,we equate $(i)$ and (ii):
$\frac{2}{3} MR_1^2 = \frac{2}{5} MR_2^2$
Dividing both sides by $M$ and simplifying:
$\frac{R_1^2}{R_2^2} = \frac{3}{5}$
Taking the square root on both sides:
$\frac{R_1}{R_2} = \frac{\sqrt{3}}{\sqrt{5}}$
Thus,the ratio of their radii is $\sqrt{3}: \sqrt{5}$.

(C) An open window acts as a perfectly black body because any radiation entering it is absorbed by the interior of the room and does not escape.
For a perfectly black body,the absorption coefficient is defined as $1$.

(B) During clear nights,objects on the surface of the Earth radiate heat,causing the temperature to fall. Hence,option $(A)$ is incorrect.
According to Stefan-Boltzmann law,the total energy radiated by a black body per unit time per unit area is $E \propto T^{4}$. Hence,option $(C)$ is incorrect.
The energy radiated per second (power) is given by $P = A \sigma \varepsilon T^{4} = 4 \pi r^{2} \sigma \varepsilon T^{4}$.
For the two spheres,the ratio of power radiated is:
$\frac{P_{1}}{P_{2}} = \left(\frac{r_{1}}{r_{2}}\right)^{2} \left(\frac{T_{1}}{T_{2}}\right)^{4} = \left(\frac{1}{4}\right)^{2} \left(\frac{4000}{2000}\right)^{4} = \frac{1}{16} \times (2)^{4} = \frac{16}{16} = 1$.
Since $P_{1} = P_{2}$,option $(D)$ is incorrect.
Newton's law of cooling is an approximate form of Stefan's law of radiation and is valid for small temperature differences,typically observed in natural convection. Thus,option $(B)$ is correct.

(C) In a transverse wave,the particles of the medium vibrate about their mean positions in a direction perpendicular to the direction of wave propagation.
Since the wave propagates along the $X$-axis and the particles oscillate along the $Y$-axis,the direction of particle velocity is perpendicular to the direction of wave velocity.
Therefore,the angle between the particle velocity and the wave velocity is $90^{\circ}$ or $\frac{\pi}{2}$ radians.

(A) Given the displacement of a particle in the wave is $y = A \sin (\omega t - k x)$.
The particle velocity $v_p$ is defined as the rate of change of displacement with respect to time:
$v_p = \frac{dy}{dt}$
Differentiating the displacement equation with respect to $t$:
$v_p = \frac{d}{dt} [A \sin (\omega t - k x)] = A \omega \cos (\omega t - k x)$
For the maximum particle velocity,the cosine term must be at its maximum value,which is $1$:
$v_{p, \text{max}} = A \omega (1) = A \omega$
Thus,the maximum particle velocity is $A \omega$.

(D) The apparent frequency $n'$ heard by an observer moving towards a stationary source is given by the Doppler effect formula: $n' = n \left( \frac{v + v_o}{v} \right)$,where $v$ is the velocity of sound and $v_o$ is the velocity of the observer.
Given that $v_o = \frac{v}{5}$,we substitute this into the formula:
$n' = n \left( \frac{v + v/5}{v} \right) = n \left( \frac{6v/5}{v} \right) = 1.2n$.
The fractional change in frequency is $\frac{n' - n}{n} = \frac{1.2n - n}{n} = 0.2$.
To find the percentage increase,we multiply by $100$: $0.2 \times 100 \% = 20 \%$.

(D) The potential energy of a stretched spring is given by the formula $U = \frac{1}{2} k x^2$,where $k$ is the spring constant and $x$ is the elongation.
For the first case,$x_1 = 2 \ cm$,so $U = \frac{1}{2} k (2)^2 = 2k$ ... $(i)$.
For the second case,$x_2 = 10 \ cm$,so the new potential energy $U'$ is $U' = \frac{1}{2} k (10)^2 = 50k$ ... $(ii)$.
Dividing equation $(ii)$ by equation $(i)$,we get $\frac{U'}{U} = \frac{50k}{2k} = 25$.
Therefore,$U' = 25U$.

(B) The time period of a magnet in a vibration magnetometer is given by $T = 2\pi \sqrt{\frac{I}{MB}}$,where $I$ is the moment of inertia and $M$ is the magnetic moment.
Initially,$T = 2 \, s = 2\pi \sqrt{\frac{I}{MB}}$.
When the magnet is cut into three equal parts along its length,for each part,the mass $m' = m/3$ and length $l' = l/3$.
The moment of inertia of each part about the center is $I' = \frac{1}{12} m' (l')^2 = \frac{1}{12} (m/3) (l/3)^2 = \frac{1}{27} I$.
The magnetic moment of each part is $M' = M/3$.
When three such parts are placed on each other,the total moment of inertia $I_s = 3 \times I' = 3 \times (I/27) = I/9$.
The total magnetic moment $M_s = 3 \times M' = 3 \times (M/3) = M$.
The new time period $T_s = 2\pi \sqrt{\frac{I_s}{M_s B}} = 2\pi \sqrt{\frac{I/9}{MB}} = \frac{1}{3} \times 2\pi \sqrt{\frac{I}{MB}} = \frac{T}{3}$.
Substituting $T = 2 \, s$,we get $T_s = 2/3 \, s$.

(A) Let the shunt resistance be $S$.
Given:
Total current to be measured,$I = 750\, A$
Full-scale deflection current of the ammeter,$I_g = 100\, A$
Resistance of the ammeter,$R_G = 13\, \Omega$
When a shunt $S$ is connected in parallel with the ammeter,the potential difference across the ammeter and the shunt must be equal:
$I_g R_G = (I - I_g) S$
Substituting the given values:
$100 \times 13 = (750 - 100) \times S$
$1300 = 650 \times S$
Solving for $S$:
$S = \frac{1300}{650} = 2\, \Omega$
Thus,the value of the shunt resistance is $2\, \Omega$.

(D) The energy of the photon required to excite an electron across the band gap is given by $E = h\nu$.
For the minimum frequency $\nu$,the energy of the photon must be equal to the band gap energy $E_g$.
Given $E_g = 2.0\, eV$.
We know that $1\, eV = 1.6 \times 10^{-19}\, J$,so $E_g = 2.0 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-19}\, J$.
Using the relation $E = h\nu$,we have $\nu = \frac{E_g}{h}$.
Substituting the values $h = 6.63 \times 10^{-34}\, J\cdot s$:
$\nu = \frac{3.2 \times 10^{-19}}{6.63 \times 10^{-34}} \approx 0.482 \times 10^{15}\, Hz$.
$\nu \approx 4.82 \times 10^{14}\, Hz$.
Rounding this value,we get $\nu \approx 5 \times 10^{14}\, Hz$.

(A) The current in an $LCR$ series circuit is given by $i = \frac{V}{\sqrt{R^2 + (X_L - X_C)^2}}$,where $V$ is the $rms$ voltage,$R$ is the resistance,$X_L = \omega L$ is the inductive reactance,and $X_C = \frac{1}{\omega C}$ is the capacitive reactance.
For the current to be maximum,the impedance must be minimum,which occurs when $X_L = X_C$.
This condition is known as resonance,where $\omega L = \frac{1}{\omega C}$.
Rearranging for $L$,we get $L = \frac{1}{\omega^2 C}$.
Given $\omega = 1000 \ s^{-1}$ and $C = 10 \ \mu F = 10 \times 10^{-6} \ F$.
Substituting these values: $L = \frac{1}{(1000)^2 \times 10 \times 10^{-6}} = \frac{1}{10^6 \times 10^{-5}} = \frac{1}{10} = 0.1 \ H$.
Converting to millihenry: $0.1 \ H = 100 \ mH$.

(C) The instantaneous power $p$ in an $AC$ circuit is given by the product of instantaneous $emf$ and current:
$p = e \cdot i = (E_{0} \sin \omega t) \cdot (I_{0} \sin (\omega t - \phi))$
Using the trigonometric identity $\sin A \sin B = \frac{1}{2} [\cos(A-B) - \cos(A+B)]$,we get:
$p = \frac{E_{0} I_{0}}{2} [\cos \phi - \cos(2\omega t - \phi)]$
The average power $P_{av}$ over one complete cycle $T$ is the average of $p$ over time $T$:
$P_{av} = \frac{1}{T} \int_{0}^{T} p \, dt = \frac{E_{0} I_{0}}{2T} \int_{0}^{T} [\cos \phi - \cos(2\omega t - \phi)] \, dt$
Since the average of $\cos(2\omega t - \phi)$ over a full cycle is $0$,the expression simplifies to:
$P_{av} = \frac{E_{0} I_{0}}{2} \cos \phi$
Here,$\cos \phi$ is known as the power factor of the $AC$ circuit.

(C) The efficiency of a transformer is defined as the ratio of output power to input power.
$\eta = \frac{\text{Output Power}}{\text{Input Power}}$
Given:
Output power $(P_{out})$ = $100 \,W$
Input voltage $(V_p)$ = $220 \,V$
Input current $(I_p)$ = $0.5 \,A$
Input power $(P_{in})$ = $V_p \times I_p = 220 \,V \times 0.5 \,A = 110 \,W$
Efficiency $(\eta)$ = $\frac{100 \,W}{110 \,W} \approx 0.909$
$\eta \approx 90.9 \% \approx 90 \%$
Thus, the efficiency is approximately $90 \%$.

(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \text{ eV}$.
For the ground state $(n=1)$,$E_1 = -13.6 \text{ eV}$.
The first excited state corresponds to $n=2$.
Thus,the energy of the electron in the first excited state is $E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4 \text{ eV}$.
The excitation energy required to move the electron from the ground state to the first excited state is $\Delta E = E_2 - E_1$.
$\Delta E = -3.4 \text{ eV} - (-13.6 \text{ eV}) = 10.2 \text{ eV}$.

(A) We know that in the steady state,the capacitor behaves like an open circuit,meaning no current flows through it.
Therefore,the potential difference across the capacitor is equal to the battery voltage,which is $45 \text{ V}$.
The final charge $q$ on the capacitor is given by the formula:
$q = C \times V$
Given:
Capacitance $C = 20 \mu F = 20 \times 10^{-6} \text{ F}$
Voltage $V = 45 \text{ V}$
Substituting the values:
$q = 20 \times 10^{-6} \times 45$
$q = 900 \times 10^{-6} \text{ C}$
$q = 9 \times 10^{-4} \text{ C}$

(D) The energy stored in a parallel plate capacitor is given by the formula $U = \frac{1}{2} C V^{2}$.
Here,the capacitance $C = \frac{\varepsilon_{0} A}{d}$.
The potential difference $V$ between the plates is related to the electric field $E$ and separation $d$ by $V = E d$.
Substituting these values into the energy formula:
$U = \frac{1}{2} \left( \frac{\varepsilon_{0} A}{d} \right) (E d)^{2}$
$U = \frac{1}{2} \left( \frac{\varepsilon_{0} A}{d} \right) (E^{2} d^{2})$
$U = \frac{1}{2} \varepsilon_{0} E^{2} A d$.

(A) In a potentiometer experiment to find the internal resistance $r$ of a cell,let $E$ be the emf of the cell and $V$ be the terminal potential difference across the external resistance $R$. The balancing lengths are $l_1 = 110 \ cm$ (open circuit) and $l_2 = 100 \ cm$ (closed circuit with $R = 10 \ \Omega$).
We know that $E \propto l_1$ and $V \propto l_2$.
Therefore,$\frac{E}{V} = \frac{l_1}{l_2} = \frac{110}{100} = 1.1$.
Also,the relationship between emf,terminal voltage,and internal resistance is given by $\frac{E}{V} = \frac{R+r}{R} = 1 + \frac{r}{R}$.
Equating the two expressions: $1 + \frac{r}{R} = 1.1$.
$\frac{r}{R} = 1.1 - 1 = 0.1$.
Given $R = 10 \ \Omega$,we have $r = 0.1 \times 10 \ \Omega = 1.0 \ \Omega$.

(D) The magnetic force $F$ on a moving charge $q$ in a magnetic field $B$ is given by the Lorentz force formula: $F = q(v \times B)$.
Protons,cathode rays (electrons),and alpha particles are all charged particles. Therefore,they experience a magnetic force when moving through a magnetic field and are deflected.
Neutrons are electrically neutral particles,meaning their charge $q = 0$.
Since the charge is zero,the magnetic force $F = 0 \times (v \times B) = 0$.
Thus,neutrons cannot be deflected by a magnetic field.

(C) photon is a quantum of electromagnetic radiation that carries energy and momentum.
It has zero rest mass and zero electric charge.
The energy of a photon is given by $E = h\nu$,where $h$ is Planck's constant and $\nu$ is the frequency.
The momentum of a photon is given by $p = E/c = h/\lambda$.
Photons travel at the speed of light $c$ in a vacuum.
Since a photon is electrically neutral,it does not possess any electric charge.
Therefore,charge is not a property of photons.

(C) The velocity of electromagnetic radiation in free space is given by the speed of light $(c)$.
According to Maxwell's equations,the speed of electromagnetic waves in a vacuum is related to the permeability of free space $(\mu_{0})$ and the permittivity of free space $(\varepsilon_{0})$ by the formula:
$c = \frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}$
Thus,the correct expression for the velocity is $\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}$.

(B) When an ion of charge $q$ and mass $m$ is accelerated through a potential difference $V$,its kinetic energy is given by: $E = qV = \frac{1}{2}mv^2$. From this,the velocity $v$ is $v = \sqrt{\frac{2qV}{m}}$.
When this ion enters a magnetic field $B$ perpendicular to its motion,it follows a circular path of radius $R$ due to the Lorentz force acting as the centripetal force: $qvB = \frac{mv^2}{R}$.
Substituting the expression for $v$ into the force equation: $qvB = \frac{m}{R} \left(\frac{2qV}{m}\right) = \frac{2qV}{R}$.
Simplifying for the charge-to-mass ratio: $qB = \frac{2qV}{vR} \implies B = \frac{mv}{qR} \implies R = \frac{mv}{qB}$.
Substituting $v = \sqrt{\frac{2qV}{m}}$ into the radius equation: $R = \frac{m}{qB} \sqrt{\frac{2qV}{m}} = \frac{1}{B} \sqrt{\frac{2mV}{q}}$.
Squaring both sides: $R^2 = \frac{2mV}{qB^2}$.
Rearranging to find the ratio $\frac{q}{m}$: $\frac{q}{m} = \frac{2V}{R^2 B^2}$.
Since $V$ and $B$ are constant,$\frac{q}{m} \propto \frac{1}{R^2}$.

(C) When a magnetic field is perpendicular to the motion of a charged particle,the magnetic force provides the necessary centripetal force.
$F_{c} = F_{m}$
$\frac{m v^{2}}{R} = B q v$
From this,the radius of the circular path is given by:
$R = \frac{m v}{B q}$
The time period $T$ of the circular motion is the time taken to complete one full circumference:
$T = \frac{2 \pi R}{v}$
Substituting the expression for $R$:
$T = \frac{2 \pi}{v} \left( \frac{m v}{B q} \right)$
$T = \frac{2 \pi m}{B q}$
Since $T$ depends only on the mass $m$,charge $q$,and magnetic field $B$,it is independent of both the radius $R$ and the speed $v$.

(A) Nickel exhibits ferromagnetism due to a quantum physical effect known as exchange coupling,where the electron spins of one atom interact with those of neighboring atoms.
This interaction results in the alignment of the magnetic dipole moments of the atoms,overcoming the randomizing tendency of thermal collisions.
This persistent alignment is responsible for the permanent magnetism in ferromagnetic materials.
When the temperature of a ferromagnetic material is raised above a specific critical value,known as the Curie temperature $(T_C)$,the exchange coupling is no longer effective.
Consequently,the material transitions from being ferromagnetic to paramagnetic.
In the paramagnetic state,the dipoles still tend to align with an external magnetic field,but this alignment is much weaker,and thermal agitation can easily disrupt it.

(B) The nuclear binding energy $(BE)$ is defined as the energy equivalent of the mass defect $(\Delta m)$.
The mass defect is the difference between the sum of the masses of individual nucleons and the actual mass of the nucleus.
For the oxygen isotope ${ }_{8}^{17}O$, the number of protons $(Z)$ is $8$ and the number of neutrons $(N)$ is $A - Z = 17 - 8 = 9$.
The mass of the nucleons is $(8 M_{p} + 9 M_{n})$.
The mass defect $\Delta m = (8 M_{p} + 9 M_{n} - M_{O})$.
However, binding energy is typically expressed as the energy released when nucleons form a nucleus, which is $BE = (8 M_{p} + 9 M_{n} - M_{O}) c^{2}$.
Note: The provided options suggest the magnitude of the mass defect calculation. Given the standard definition $BE = \Delta m c^2$, the correct expression for the binding energy is $(8 M_{p} + 9 M_{n} - M_{O}) c^{2}$. Since the options provided are in the form $(M_{O} - 8 M_{p} - 9 M_{n}) c^{2}$, this represents the negative of the binding energy (or the mass defect with a sign convention often used in specific contexts). Based on the structure of the options, $B$ is the intended answer.

(D) We know that the neutrino is the particle which has almost zero mass and exactly zero charge.
Positron has the charge $+e$ and mass $m_{e}$ (equal to electron mass).
Electron is a particle which has $-e$ charge and $9.1 \times 10^{-31} \ kg$ mass.
Neutron is a particle which has zero charge and $1838 \ m_{e}$ mass.
Therefore,the required answer is neutrino.

(D) The nuclear density $\rho$ is defined as the ratio of the mass of the nucleus to its volume.
Since the mass of a nucleus is approximately $M = A \cdot m_p$ (where $A$ is the mass number and $m_p$ is the mass of a nucleon) and the volume is $V = \frac{4}{3} \pi R^3$,where $R = R_0 A^{1/3}$.
Substituting the expression for $R$,we get $V = \frac{4}{3} \pi (R_0 A^{1/3})^3 = \frac{4}{3} \pi R_0^3 A$.
Thus,$\rho = \frac{A \cdot m_p}{\frac{4}{3} \pi R_0^3 A} = \frac{m_p}{\frac{4}{3} \pi R_0^3}$.
This expression shows that the nuclear density is independent of the mass number $A$.
Therefore,the ratio of the nuclear densities of any two nuclei is always $1: 1$.

(B) Beta decay can involve the emission of either electrons or positrons.
The electrons or positrons emitted in a $\beta$-decay do not exist inside the nucleus.
They are only created at the time of emission,just as photons are created when an atom makes a transition from a higher to a lower energy state.
In negative $\beta$-decay,a neutron in the nucleus is transformed into a proton,an electron,and an antineutrino.
Hence,in the radioactive decay process,the negatively charged emitted $\beta$-particles are the electrons produced as a result of the decay of neutrons present inside the nucleus.

(D) The primary advantages of optical fibres include:
$1$. They possess a very high bandwidth,allowing for the transmission of a large amount of data.
$2$. They have a high data transmission capacity,significantly exceeding that of copper wires or radio waves.
$3$. They are practically free from electromagnetic $(EM)$ interference and crosstalk,which are common issues in ordinary cables and microwave links.

(C) Metals reflect incident light due to the oscillations of free electrons under the influence of the electric field of the incident light wave.
Electrical conductivity of metals decreases with an increase in temperature because the increased random thermal motion of ions increases the scattering of free electrons,thereby increasing resistance.
Therefore,the bonding in such a solid is metallic.

(A) In the given energy band diagram,we observe the presence of an energy gap $E_g$ between the valence band $(E_V)$ and the conduction band $(E_C)$. This indicates that the material is a semiconductor.
Looking closely at the diagram,we see that there are more holes (represented by open circles) in the valence band than electrons (represented by filled circles) in the conduction band. Specifically,the presence of acceptor energy levels just above the valence band is characteristic of a $p$-type semiconductor,where holes are the majority charge carriers.
Therefore,the material is a $p$-type semiconductor.

(D) Given that, the change in emitter current is $\Delta I_{E} = 8.0 \,mA$.
The change in collector current is $\Delta I_{C} = 7.9 \,mA$.
We know that the current gain $\alpha$ is defined as the ratio of change in collector current to the change in emitter current:
$\alpha = \frac{\Delta I_{C}}{\Delta I_{E}} = \frac{7.9}{8.0} = 0.9875 \approx 0.99$.
We also know the relationship between $\alpha$ and $\beta$ is given by $\beta = \frac{\alpha}{1 - \alpha}$.
Alternatively, using the currents directly: $\beta = \frac{\Delta I_{C}}{\Delta I_{B}}$.
Since $\Delta I_{E} = \Delta I_{C} + \Delta I_{B}$, we have $\Delta I_{B} = \Delta I_{E} - \Delta I_{C} = 8.0 \,mA - 7.9 \,mA = 0.1 \,mA$.
Therefore, $\beta = \frac{7.9 \,mA}{0.1 \,mA} = 79$.
Thus, the values are $\alpha = 0.99$ and $\beta = 79$.

(C) $NAND$ gate is defined as an $AND$ gate followed by a $NOT$ gate. The Boolean expression for a $NAND$ gate with inputs $A$ and $B$ is $Y = \overline{A \cdot B}$.
When the two inputs are shorted,$A = B$. Substituting this into the expression,we get $Y = \overline{A \cdot A} = \overline{A}$.
This is the Boolean expression for a $NOT$ gate.
The truth table for this configuration is:

$A$	$Y = \overline{A \cdot A}$
$0$	$1$
$1$	$0$

Since the output is the inverse of the input,the resulting gate is a $NOT$ gate.

(C) In a $p$-type semiconductor,trivalent impurity atoms are added to an intrinsic semiconductor,which creates an abundance of holes.
Therefore,holes act as the majority charge carriers,and thermally generated electrons act as the minority charge carriers.
Thus,in a $p$-type semiconductor,there are holes in larger numbers and electrons in smaller numbers.

(D) The neutral temperature $(T_n)$ of a thermocouple is a characteristic property of the materials used to form the thermocouple. It depends only on the nature of the metals used and is independent of the temperature of the cold junction $(T_c)$ and the inversion temperature $(T_i)$. Therefore,if the temperature of the cold junction is lowered,the neutral temperature remains the same.

(A) When unpolarized light of intensity $I_{0}$ is incident on a polarizing sheet, the intensity of the transmitted light is given by $I_{t} = \frac{I_{0}}{2}$.
Since the total incident intensity is $I_{0}$ and the transmitted intensity is $\frac{I_{0}}{2}$, the intensity of the light that does not get transmitted is the difference between the incident and transmitted intensities.
Intensity of untransmitted light = $I_{0} - I_{t} = I_{0} - \frac{I_{0}}{2} = \frac{I_{0}}{2}$.

(B) In an interference experiment,the distance between two consecutive bright fringes (maxima) or two consecutive dark fringes (minima) is known as the fringe width,denoted by $\beta$.
According to the theory of Young's double-slit experiment,the fringe width is given by the formula:
$\beta = \frac{D \lambda}{d}$
where $D$ is the distance between the screen and the slits,$\lambda$ is the wavelength of the light used,and $d$ is the distance between the two slits.

(D) The resultant intensity of two periodic waves is given by the formula:
$I = I_{1} + I_{2} + 2\sqrt{I_{1}I_{2}}\cos\delta$
where $\delta$ is the phase difference between the waves.
For maximum intensity,$\cos\delta = 1$,so:
$I_{\max} = I_{1} + I_{2} + 2\sqrt{I_{1}I_{2}} = (\sqrt{I_{1}} + \sqrt{I_{2}})^{2}$
For minimum intensity,$\cos\delta = -1$,so:
$I_{\min} = I_{1} + I_{2} - 2\sqrt{I_{1}I_{2}} = (\sqrt{I_{1}} - \sqrt{I_{2}})^{2}$
The sum of the maximum and minimum intensities is:
$I_{\max} + I_{\min} = (\sqrt{I_{1}} + \sqrt{I_{2}})^{2} + (\sqrt{I_{1}} - \sqrt{I_{2}})^{2}$
Expanding the squares:
$I_{\max} + I_{\min} = (I_{1} + I_{2} + 2\sqrt{I_{1}I_{2}}) + (I_{1} + I_{2} - 2\sqrt{I_{1}I_{2}})$
$I_{\max} + I_{\min} = 2(I_{1} + I_{2})$

(C) Resistance of galvanometer,$R_g = 50 \Omega$.
Emf of battery,$V = 3 \text{ V}$.
Resistance connected in series,$R_s = 2950 \Omega$.
Total resistance,$R' = R_g + R_s = 50 + 2950 = 3000 \Omega$.
Therefore,the initial current,$I = \frac{V}{R'} = \frac{3}{3000} = 10^{-3} \text{ A}$.
If the deflection is reduced to $20$ divisions from $30$ divisions,the new current $I'$ will be $I' = I \times \frac{20}{30} = 10^{-3} \times \frac{2}{3} = \frac{2}{3} \times 10^{-3} \text{ A}$.
Let $R_E$ be the new total resistance of the circuit.
Using Ohm's law,$V = I' R_E \Rightarrow R_E = \frac{V}{I'} = \frac{3}{\frac{2}{3} \times 10^{-3}} = \frac{9}{2} \times 10^3 = 4500 \Omega$.
Since $R_E = R_g + R_{new}$,the new series resistance required is $R_{new} = R_E - R_g = 4500 - 50 = 4450 \Omega$.