MHT CET 2010 Physics Question Paper with Answer and Solution

(A) The acceleration due to gravity at a height $h$ above the earth's surface is given by the formula:
$g^{\prime} = g \left( \frac{R}{R+h} \right)^{2}$
Given that the height $h = nR$,we substitute this into the equation:
$g^{\prime} = g \left( \frac{R}{R+nR} \right)^{2}$
$g^{\prime} = g \left( \frac{R}{R(1+n)} \right)^{2}$
$g^{\prime} = g \left( \frac{1}{1+n} \right)^{2}$
Now,we find the ratio of the acceleration due to gravity on the surface $(g)$ to that at the altitude $(g^{\prime})$:
$\frac{g}{g^{\prime}} = \frac{g}{g \left( \frac{1}{1+n} \right)^{2}}$
$\frac{g}{g^{\prime}} = (1+n)^{2}$
Therefore,the ratio is $(n+1)^{2}$.

(B) The acceleration due to gravity $g$ on the surface of the Earth is given by the formula: $g = \frac{GM}{R^2}$.
Since mass $M = \text{density} (\rho) \times \text{volume} (V) = \rho \times \frac{4}{3} \pi R^3$,we can substitute this into the formula for $g$:
$g = \frac{G (\rho \cdot \frac{4}{3} \pi R^3)}{R^2} = \frac{4}{3} \pi \rho G R$.
From this expression,it is clear that $g \propto \rho$ when the radius $R$ is kept constant.
Therefore,$\frac{g_2}{g_1} = \frac{\rho_2}{\rho_1}$.
Given that the density is doubled,$\rho_2 = 2\rho_1$.
Thus,$g_2 = 2 \times g_1 = 2 \times 9.8 \ m/s^2 = 19.6 \ m/s^2$.

(B) The pressure $p$ exerted by an ideal gas is given by the kinetic theory of gases as:
$p = \frac{1}{3} \rho \bar{v}^2$
where $\rho$ is the density and $\bar{v}^2$ is the mean square speed.
Since density $\rho = \frac{M}{V}$,we can write:
$p = \frac{1}{3} \frac{M}{V} \bar{v}^2$
Multiplying and dividing by $2$,we get:
$p = \frac{2}{3} \left( \frac{1}{2} \frac{M}{V} \bar{v}^2 \right)$
Here,$\frac{1}{2} M \bar{v}^2$ is the total kinetic energy of the gas molecules.
Thus,$\frac{1}{2} \frac{M}{V} \bar{v}^2$ represents the kinetic energy per unit volume,which is given as $E$.
Therefore,$p = \frac{2}{3} E$.

(C) The average velocity $(v_{av})$ is calculated as the arithmetic mean of the velocities:
$v_{av} = \frac{v_{1} + v_{2} + v_{3} + v_{4}}{N} = \frac{1 + 3 + 5 + 7}{4} = \frac{16}{4} = 4 \ km/s$
The root mean square $(RMS)$ velocity $(v_{rms})$ is calculated as the square root of the mean of the squares of the velocities:
$v_{rms} = \sqrt{\frac{v_{1}^{2} + v_{2}^{2} + v_{3}^{2} + v_{4}^{2}}{N}} = \sqrt{\frac{1^{2} + 3^{2} + 5^{2} + 7^{2}}{4}} = \sqrt{\frac{1 + 9 + 25 + 49}{4}} = \sqrt{\frac{84}{4}} = \sqrt{21} \approx 4.583 \ km/s$
The difference between the $RMS$ velocity and the average velocity is:
$v_{rms} - v_{av} = 4.583 \ km/s - 4 \ km/s = 0.583 \ km/s$

(A) For a particle to just begin to slip on a rotating disc,the required centripetal force is provided by the maximum static friction force $f_{s,max} = \mu m g$.
Thus,$m r \omega^2 = \mu m g$.
Since $\mu$,$m$,and $g$ are constants,we have $r \omega^2 = \text{constant}$,which implies $r \propto \frac{1}{\omega^2}$.
Given $r_1 = 4 \ cm$ at $\omega_1 = \omega$.
When $\omega_2 = 2\omega$,we have $\frac{r_1}{r_2} = \frac{\omega_2^2}{\omega_1^2}$.
Substituting the values: $\frac{4}{r_2} = \frac{(2\omega)^2}{\omega^2} = \frac{4\omega^2}{\omega^2} = 4$.
Therefore,$r_2 = \frac{4}{4} = 1 \ cm$.

(A) Molecules in the bulk of a liquid are surrounded by other molecules on all sides,resulting in a net attractive force of zero. However,molecules on the surface are attracted only by the molecules below them,as there are no liquid molecules above the surface. To bring a molecule from the bulk to the surface,work must be done against these inward attractive forces. This work is stored as potential energy. Therefore,molecules on the surface of a liquid in equilibrium possess maximum potential energy compared to those in the bulk.

(D) The relationship between Young's modulus $(Y)$,modulus of rigidity $(\eta)$,and Poisson's ratio $(\sigma)$ is given by the formula:
$Y = 2 \eta (1 + \sigma)$
Therefore,the correct relation is $Y = 2 \eta (1 + \sigma)$.

(D) The formula for Bulk modulus $K$ is given by $K = -\frac{\Delta p}{\Delta V / V}$.
We are given the Bulk modulus $K = 2100 \, MPa = 2100 \times 10^3 \, kPa$.
The fractional change in volume is $\frac{\Delta V}{V} = 0.008 \, \% = \frac{0.008}{100} = 8 \times 10^{-5}$.
To find the increase in pressure $\Delta p$, we rearrange the formula: $\Delta p = K \times \left( \frac{\Delta V}{V} \right)$.
Substituting the values: $\Delta p = (2100 \times 10^3 \, kPa) \times (8 \times 10^{-5})$.
$\Delta p = 2100 \times 8 \times 10^{-2} \, kPa = 21 \times 8 \, kPa = 168 \, kPa$.

(A) The elongation $\Delta L$ is given by the formula $\Delta L = \frac{FL}{AY} = \frac{FL}{\pi r^2 Y}$.
Since the wires are of the same material and stretched by the same load,$F$ and $Y$ are constant.
Therefore,$\Delta L \propto \frac{L}{r^2}$.
Calculating the proportionality factor $\frac{L}{r^2}$ for each case:
For $A$: $\frac{100}{1^2} = 100$.
For $B$: $\frac{200}{3^2} = \frac{200}{9} \approx 22.22$.
For $C$: $\frac{300}{3^2} = \frac{300}{9} \approx 33.33$.
For $D$: $\frac{400}{4^2} = \frac{400}{16} = 25$.
Comparing the values,the wire with dimensions $L=100 \ cm$ and $r=1 \ mm$ has the largest value of $\frac{L}{r^2}$,hence it will elongate the most.

(A) The kinetic energy $(E)$ of a particle of mass $m$ moving with speed $v$ is given by $E = \frac{1}{2} m v^2$.
For uniform circular motion,the centripetal acceleration $(a)$ is given by $a = \frac{v^2}{r}$,which implies $v^2 = a r$.
Substituting $v^2 = a r$ into the kinetic energy equation:
$E = \frac{1}{2} m (a r)$
Rearranging the equation to solve for acceleration $(a)$:
$2 E = m a r$
$a = \frac{2 E}{m r}$

(C) Initial angular velocity $\omega_{0} = 1200 \text{ rpm} = \frac{1200 \times 2\pi}{60} \text{ rad/s} = 40\pi \text{ rad/s}$.
Final angular velocity $\omega = 0 \text{ rad/s}$.
Angular deceleration $\alpha = 4 \text{ rad/s}^{2}$.
Using the kinematic equation $\omega^{2} = \omega_{0}^{2} - 2\alpha\theta$:
$0 = (40\pi)^{2} - 2(4)\theta$
$8\theta = 1600\pi^{2}$
$\theta = 200\pi^{2} \text{ rad}$.
Since the total angle $\theta = 2\pi n$,where $n$ is the number of revolutions:
$2\pi n = 200\pi^{2}$
$n = 100\pi = 100 \times 3.14159 \approx 314 \text{ revolutions}$.

(B) The potential energy of an oscillating particle (Simple Harmonic Motion) is given by $U = \frac{1}{2} k x^{2}$.
We know that the restoring force acting on the particle is $F = -k x$.
From the potential energy equation,we can write:
$2 U = k x^{2}$
Substitute $k = -\frac{F}{x}$ into the equation:
$2 U = -\left( \frac{F}{x} \right) x^{2}$
$2 U = -F x$
Rearranging the terms,we get:
$\frac{2 U}{F} = -x$
Therefore:
$\frac{2 U}{F} + x = 0$

(C) The acceleration of a particle performing $SHM$ is given by $a = -\omega^{2} x$,where $x = A \sin(\omega t + \phi)$.
To find the average acceleration over one complete oscillation (time period $T$),we integrate the acceleration over the interval $[0, T]$ and divide by $T$.
$\text{Average acceleration} = \frac{1}{T} \int_{0}^{T} a(t) dt = \frac{1}{T} \int_{0}^{T} -\omega^{2} A \sin(\omega t + \phi) dt$.
Since the integral of a sine function over one full period is zero,the average acceleration is $0$.

(B) The maximum velocity of a particle in $SHM$ is given by $v = a \omega$,where $a$ is the amplitude of displacement and $\omega$ is the angular frequency.
From this,we can find the angular frequency: $\omega = \frac{v}{a}$.
The maximum acceleration (amplitude of acceleration) is given by $A_{max} = \omega^2 a$.
Substituting the value of $\omega$ into the acceleration formula:
$A_{max} = \left(\frac{v}{a}\right)^2 \times a = \frac{v^2}{a^2} \times a = \frac{v^2}{a}$.

(B) The moment of inertia of a thin uniform rod of length $L$ and mass $M$ about an axis passing through its center of mass $(CM)$ and perpendicular to the rod is $I_{CM} = \frac{M L^2}{12}$.
The distance of the given axis from the center of mass is $x = \frac{L}{2} - \frac{L}{3} = \frac{L}{6}$.
Using the parallel axis theorem,$I = I_{CM} + M x^2$.
Substituting the values,we get $I = \frac{M L^2}{12} + M \left( \frac{L}{6} \right)^2$.
$I = \frac{M L^2}{12} + \frac{M L^2}{36}$.
Taking the least common multiple,$I = \frac{3 M L^2 + M L^2}{36} = \frac{4 M L^2}{36} = \frac{M L^2}{9}$.

(A) The moment of inertia of a disc of mass $M$ and radius $R$ about its diameter is given by $I = \frac{1}{4} M R^2$.
From this,we get $M R^2 = 4 I$.
To find the moment of inertia about an axis perpendicular to the plane and passing through the rim,we use the parallel axis theorem.
First,the moment of inertia about an axis perpendicular to the plane and passing through the center (center of mass) is $I_{cm} = \frac{1}{2} M R^2$.
By the parallel axis theorem,$I_{rim} = I_{cm} + M R^2$.
Substituting $I_{cm} = \frac{1}{2} M R^2$,we get $I_{rim} = \frac{1}{2} M R^2 + M R^2 = \frac{3}{2} M R^2$.
Substituting $M R^2 = 4 I$ into the expression,we get $I_{rim} = \frac{3}{2} (4 I) = 6 I$.

(C) $1$. Option $A$ is correct: $\tau = I \alpha$,which is the rotational analogue of Newton's second law $(F = ma)$.
$2$. Option $B$ is correct: $\tau = p \times B$ (or $\mu \times B$),representing the torque on a dipole in a magnetic field.
$3$. Option $C$ is incorrect: The correct relation is $I = \frac{\tau}{\alpha}$. The given relation $I = \tau \times \alpha$ is dimensionally and physically wrong.
$4$. Option $D$ is incorrect: The correct relation for angular momentum is $L = I \omega$. Linear momentum is $p = mv$. The relation $p = I \omega$ is physically incorrect.
Note: Since the question asks for the incorrect relation and both $C$ and $D$ are incorrect,$C$ is the most fundamental error in rotational dynamics definitions.

(A) An opaque body does not transmit any radiation.
By definition,the coefficient of transmission $(t)$ is the ratio of the transmitted energy to the incident energy.
Since no radiation passes through an opaque body,the transmitted energy is $0$.
Therefore,the coefficient of transmission for an opaque body is $0$.

(C) The energy radiated by a body is given by Stefan-Boltzmann Law: $Q = A \varepsilon \sigma T^{4} t$.
Since the time $t$,emissivity $\varepsilon$,and Stefan-Boltzmann constant $\sigma$ are the same for both,the energy radiated $Q$ is proportional to $A T^{4}$.
Since $A = 4 \pi r^{2}$,we have $Q \propto r^{2} T^{4}$.
The temperatures must be in Kelvin: $T_{1} = 127 + 273 = 400 \ K$ and $T_{2} = 527 + 273 = 800 \ K$.
The ratio of energy radiated is:
$\frac{Q_{1}}{Q_{2}} = \left(\frac{r_{1}}{r_{2}}\right)^{2} \left(\frac{T_{1}}{T_{2}}\right)^{4}$
$\frac{Q_{1}}{Q_{2}} = \left(\frac{8}{2}\right)^{2} \left(\frac{400}{800}\right)^{4}$
$\frac{Q_{1}}{Q_{2}} = (4)^{2} \times \left(\frac{1}{2}\right)^{4}$
$\frac{Q_{1}}{Q_{2}} = 16 \times \frac{1}{16} = 1$.

(C) The surface tension of a liquid decreases with an increase in temperature.
As the temperature rises,the kinetic energy of the molecules increases,which weakens the intermolecular cohesive forces responsible for surface tension.
The surface tension of a liquid becomes zero at its boiling point and vanishes at the critical temperature.
At the critical temperature,the intermolecular forces for liquids and gases become equal,and the liquid can expand without any restriction.
For small temperature differences,the variation in surface tension with temperature is linear and is given by the relation: $T_{t} = T_{0}(1 - \alpha t)$,where $T_{t}$ and $T_{0}$ are the surface tensions at $t^{\circ}C$ and $0^{\circ}C$ respectively,and $\alpha$ is the temperature coefficient of surface tension.

(C) The correct option is $C$.
From the relation $E = h \nu$,where $E$ is energy,$h$ is Planck's constant,and $\nu$ is frequency.
$h = \frac{E}{\nu}$.
The dimension of energy $E$ is $[ML^2 T^{-2}]$.
The dimension of frequency $\nu$ is $[T^{-1}]$.
Therefore,the dimension of $h = \frac{[ML^2 T^{-2}]}{[T^{-1}]} = [ML^2 T^{-1}]$.
Now,let us check the dimensions of the product of force,displacement,and time:
Dimension of force $F = [MLT^{-2}]$.
Dimension of displacement $d = [L]$.
Dimension of time $t = [T]$.
Product dimension $= [MLT^{-2}] \times [L] \times [T] = [ML^2 T^{-1}]$.
This matches the dimension of Planck's constant.

(B) The velocity of a transverse wave in a stretched string is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Since $\mu = \rho A = \rho (\pi R^2)$,where $\rho$ is the density of the material and $A$ is the cross-sectional area,we have:
$v = \sqrt{\frac{T}{\rho \pi R^2}} = \frac{1}{R} \sqrt{\frac{T}{\rho \pi}}$.
From this expression,it is clear that $v \propto \frac{1}{R}$.
Since $R_{1} > R_{2}$,the velocity $v_{1}$ in the thicker wire will be less than the velocity $v_{2}$ in the thinner wire $(v_{1} < v_{2})$.
Therefore,the transverse wave travels faster in the thinner wire.

(A) In a sine wave,the displacement of a particle is given by $y = A \sin(kx - \omega t)$.
The particle velocity is $v_p = \frac{\partial y}{\partial t} = -A\omega \cos(kx - \omega t)$.
The speed of the particle is $|v_p| = |A\omega \cos(kx - \omega t)|$.
Two particles have the same speed if their displacement magnitudes are equal,i.e.,$|y_1| = |y_2|$.
For a sine wave,points with the same speed are those that have the same magnitude of displacement from the mean position.
Specifically,the points at the crest $(A)$ and the trough $(B)$ both have zero velocity (speed = $0$). The distance between a crest and the adjacent trough is $\frac{\lambda}{2}$.
Alternatively,any two points symmetric about the mean position or the extremes will have the same speed. The minimum distance between two such points is $\frac{\lambda}{2}$.

(A) The given equation of the wave is $y=A \sin (100 \pi t-3 x)$.
Comparing this with the standard wave equation $y=A \sin (\omega t-kx)$,we get the wave number $k=3 \ m^{-1}$.
The relationship between phase difference $(\Delta \phi)$ and path difference $(\Delta x)$ is given by $\Delta \phi = k \cdot \Delta x$.
Given the phase difference $\Delta \phi = \frac{\pi}{3}$.
Substituting the values,we have $\frac{\pi}{3} = 3 \cdot \Delta x$.
Solving for the distance $\Delta x$,we get $\Delta x = \frac{\pi}{3 \times 3} = \frac{\pi}{9} \ m$.

(A) The fundamental frequency of a pipe closed at one end is $n_{1} = \frac{v}{4 l_{1}}$,which gives $l_{1} = \frac{v}{4 n_{1}}$.
The fundamental frequency of a pipe open at both ends is $n_{2} = \frac{v}{2 l_{2}}$,which gives $l_{2} = \frac{v}{2 n_{2}}$.
When both pipes are joined end to end,the new pipe is closed at one end and open at the other,with a total length $L = l_{1} + l_{2}$.
The fundamental frequency $n$ of this new closed pipe is given by $n = \frac{v}{4 L} = \frac{v}{4 (l_{1} + l_{2})}$.
Substituting the values of $l_{1}$ and $l_{2}$: $\frac{1}{4 n} = \frac{1}{4 n_{1}} + \frac{1}{2 n_{2}}$.
Multiplying by $4$: $\frac{1}{n} = \frac{1}{n_{1}} + \frac{2}{n_{2}} = \frac{n_{2} + 2 n_{1}}{n_{1} n_{2}}$.
Therefore,$n = \frac{n_{1} n_{2}}{n_{2} + 2 n_{1}}$.

(A) In the fundamental mode of a pipe closed at one end,the length of the pipe $l$ corresponds to one-fourth of the wavelength,i.e.,$l = \frac{\lambda}{4}$,which implies $\lambda = 4l$.
Given that the time taken by the wave to travel the length of the pipe $l$ is $t = 0.01 \ s$,the speed of sound $v$ is given by $v = \frac{l}{t}$.
The fundamental frequency $n$ is given by $n = \frac{v}{\lambda}$.
Substituting the values,we get $n = \frac{l/t}{4l} = \frac{1}{4t}$.
Substituting $t = 0.01 \ s$,we get $n = \frac{1}{4 \times 0.01} = \frac{1}{0.04} = 25 \ Hz$.

(D) The average power of an $L-C-R$ circuit is given by the formula:
$P_{\text{av}} = V_{\text{rms}} \cdot I_{\text{rms}} \cos \phi$
Where:
$V_{\text{rms}}$ is the root mean square value of the electromotive force (emf),
$I_{\text{rms}}$ is the root mean square value of the current,
$\phi$ is the phase difference between the voltage and the current.
Therefore,the average power depends on the current,the emf,and the phase difference.

(B) The time period $T$ of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula: $T = \frac{4 \varepsilon_{0}^{2} n^{3} h^{3}}{m Z^{2} e^{4}}$.
Since the orbital frequency $f$ is the reciprocal of the time period $(f = 1/T)$, we have:
$f \propto \frac{1}{n^{3}}$.
Therefore, the orbital frequency is proportional to $n^{-3}$.

(A) The linear momentum $p$ of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $p = mv = \frac{mZe^2}{2 \epsilon_0 nh}$. Since $Z=1$ for hydrogen,$p \propto \frac{1}{n}$.
The angular momentum $L$ of an electron in the $n^{th}$ orbit is given by Bohr's quantization condition: $L = \frac{nh}{2\pi}$. Thus,$L \propto n$.
The product of linear momentum and angular momentum is $p \times L \propto \left(\frac{1}{n}\right) \times n = n^0$.
Comparing this with $n^x$,we get $x = 0$.

(C) The capacitance $C$ of a parallel plate capacitor is given by the formula:
$C = \frac{k \varepsilon_{0} A}{d}$
where $k$ is the dielectric constant,$\varepsilon_{0}$ is the permittivity of free space,$A$ is the area of the plates,and $d$ is the distance between the plates.
From the formula,we can see that $C \propto A$.
Therefore,the capacity of the capacitor increases if the area of the plate is increased.

(A) The energy stored in a capacitor is given by $U = \frac{q^2}{2C}$.
Given: $C = 48 \mu F = 48 \times 10^{-6} \ F$,$q_1 = 0.1 \ C$,$q_2 = 0.5 \ C$.
The change in energy $\Delta U$ is given by:
$\Delta U = U_2 - U_1 = \frac{q_2^2}{2C} - \frac{q_1^2}{2C} = \frac{1}{2C} (q_2^2 - q_1^2)$
Substituting the values:
$\Delta U = \frac{1}{2 \times 48 \times 10^{-6}} ((0.5)^2 - (0.1)^2)$
$\Delta U = \frac{1}{96 \times 10^{-6}} (0.25 - 0.01)$
$\Delta U = \frac{0.24}{96 \times 10^{-6}}$
$\Delta U = \frac{0.24 \times 10^6}{96} = \frac{240000}{96} = 2500 \ J$.

(B) Let the full-scale deflection current of the galvanometer be $I_g$.
When a resistance of $100 \Omega$ is connected in series,the total resistance is $(100 + R)$. The voltage range $V$ is given by $V = I_g(100 + R)$ --- $(i)$
When a resistance of $1000 \Omega$ is connected in series,the new range is $2V$. The total resistance is $(1000 + R)$. Thus,$2V = I_g(1000 + R)$ --- (ii)
Dividing equation (ii) by equation $(i)$:
$\frac{2V}{V} = \frac{I_g(1000 + R)}{I_g(100 + R)}$
$2 = \frac{1000 + R}{100 + R}$
$2(100 + R) = 1000 + R$
$200 + 2R = 1000 + R$
$2R - R = 1000 - 200$
$R = 800 \Omega$

(D) potentiometer works on the principle of the null deflection method.
In the balanced condition,the potential difference across the cell is measured without drawing any current from the circuit.
Since no current flows through the secondary circuit at the balance point,the potentiometer measures the true electromotive force $(EMF)$ or potential difference.
In contrast,a voltmeter has a finite resistance and draws some current from the circuit,which leads to a voltage drop across the internal resistance of the source,resulting in an inaccurate reading.

(B) According to Bohr's quantization postulate,the angular momentum of an electron in a stationary orbit is given by:
$mvr = \frac{nh}{2\pi}$
Rearranging this equation,we get:
$2\pi r = \frac{nh}{mv}$
Since the de-Broglie wavelength is defined as $\lambda = \frac{h}{mv}$,we can substitute this into the equation:
$2\pi r = n\lambda$
For the ground state of the hydrogen atom,the principal quantum number is $n = 1$.
Substituting $n = 1$ into the equation,we get:
$\lambda = 2\pi r$

(B) Consider a small element of length $dr$ at a distance $r$ from the fixed end of the rod.
As the rod rotates with angular velocity $\omega$,the linear velocity of this element is $v = r\omega$.
The motional emf $de$ induced across this small element is given by $de = B v dr = B (r\omega) dr$.
To find the total emf $e$ induced across the entire length of the rod,we integrate this expression from $r = 0$ to $r = l$:
$e = \int_{0}^{l} B \omega r dr = B \omega \left[ \frac{r^2}{2} \right]_{0}^{l} = \frac{1}{2} B l^2 \omega$.

(C) The formula for self-inductance $L$ is given by $L = \frac{N\phi}{i}$,where $N$ is the number of turns,$\phi$ is the magnetic flux,and $i$ is the current.
Given values are $N = 100$,$\phi = 5 \times 10^{-5} \ Wb$,and $i = 2 \ A$.
Substituting these values into the formula:
$L = \frac{100 \times 5 \times 10^{-5}}{2}$
$L = \frac{500 \times 10^{-5}}{2}$
$L = 250 \times 10^{-5} \ H$
$L = 2.5 \times 10^{-3} \ H$

(D) The ozone layer in the Earth's stratosphere absorbs the majority of the Sun's harmful ultraviolet $(UV)$ radiation. This absorption is crucial as it protects living organisms on Earth from the damaging effects of high-energy $UV$ rays,such as skin cancer and cataracts.

(A) According to Gauss's Law,the electric field $E$ at a distance $r$ from the center of a uniformly charged spherical shell of radius $R$ (where $r > R$) is given by $E = \frac{q}{4\pi\varepsilon_{0}r^{2}}$.
Since the surface charge density $\sigma = \frac{q}{4\pi R^{2}}$,we have $q = \sigma(4\pi R^{2})$.
Substituting the value of $q$ in the electric field formula:
$E = \frac{\sigma(4\pi R^{2})}{4\pi\varepsilon_{0}r^{2}} = \frac{\sigma R^{2}}{\varepsilon_{0}r^{2}}$.

(B) Let the length of the wire be $l$.
For the circular loop,the circumference is $2 \pi r = l$,so the radius $r = \frac{l}{2 \pi}$.
The magnetic dipole moment of the circular loop is $M_{1} = i A_{1} = i \pi r^{2}$.
Substituting $r$,we get $M_{1} = i \pi \left(\frac{l}{2 \pi}\right)^{2} = \frac{i l^{2}}{4 \pi}$.
For the square loop,the perimeter is $4 a = l$,so the side length $a = \frac{l}{4}$.
The magnetic dipole moment of the square loop is $M_{2} = i A_{2} = i a^{2}$.
Substituting $a$,we get $M_{2} = i \left(\frac{l}{4}\right)^{2} = \frac{i l^{2}}{16}$.
The ratio of the dipole moments is $\frac{M_{1}}{M_{2}} = \frac{i l^{2} / 4 \pi}{i l^{2} / 16} = \frac{16}{4 \pi} = \frac{4}{\pi}$.

(A) toroid is a hollow circular ring on which a large number of turns of a metallic wire are closely wound.
It can be thought of as a solenoid that has been bent into a circular shape to form a closed loop.
Therefore,a toroid is a ring-shaped closed solenoid.

(A) The radius of a circular path for a charged particle in a perpendicular magnetic field is given by $r = \frac{mv}{qB}$.
For particle $a$,the radius is $r_{a} = \frac{m_{a} v_{a}}{q B}$.
For particle $b$,the radius is $r_{b} = \frac{m_{b} v_{b}}{q B}$.
Given that $r_{a} > r_{b}$,we substitute the expressions:
$\frac{m_{a} v_{a}}{q B} > \frac{m_{b} v_{b}}{q B}$.
Since the charge $q$ and magnetic field $B$ are the same for both particles,we can cancel them out.
Therefore,$m_{a} v_{a} > m_{b} v_{b}$.

(B) When a magnet is cut along its length,the pole strength remains the same,but when it is cut perpendicular to its length,the pole strength is halved.
In this case,the magnet is cut into four equal parts by cutting it once along its length (parallel to the axis) and once perpendicular to its length.
$1$. Cutting along the length (parallel to the axis) divides the pole strength $m$ into $m/2$ and $m/2$.
$2$. Cutting perpendicular to the length divides the magnet into two parts,but the pole strength of each cross-section remains $m/2$.
Therefore,for each of the four resulting parts,the pole strength $m^{\prime}$ is $m/2$.

(B) The resolving power of a telescope is given by the formula: $RP = \frac{D}{1.22 \lambda}$,where $D$ is the aperture of the telescope and $\lambda$ is the wavelength of light used.
From this relation,it is clear that the resolving power is directly proportional to the aperture $(RP \propto D)$.
Therefore,if the aperture of the telescope is decreased,the resolving power will also decrease.

(D) The speed of light in a medium with refractive index $\mu$ is given by $v = \frac{c}{\mu}$,where $c$ is the speed of light in a vacuum.
Time taken $(T)$ to travel a distance $t$ is given by the formula $T = \frac{\text{distance}}{\text{speed}}$.
Substituting the values,we get $T = \frac{t}{v} = \frac{t}{(c/\mu)}$.
Therefore,$T = \frac{\mu t}{c}$.

(B) To form an $AND$ gate using $NAND$ gates,we first pass the inputs $A$ and $B$ through a $NAND$ gate to get $\overline{A \cdot B}$.
Then,we pass this output through another $NAND$ gate configured as a $NOT$ gate (by shorting its inputs together).
Let the output of the first $NAND$ gate be $X = \overline{A \cdot B}$.
The second $NAND$ gate acts as a $NOT$ gate,so its output $Y = \overline{X \cdot X} = \overline{X} = \overline{\overline{A \cdot B}} = A \cdot B$.
Thus,two $NAND$ gates are required to form an $AND$ gate.

(A) When a $p-n$ junction diode is forward biased,the potential barrier is reduced,allowing electrons from the $n$-region and holes from the $p$-region to cross the junction.
At the junction,these charge carriers recombine,and the energy released during this process is emitted in the form of photons.
In diodes made of specific semiconductor materials like gallium arsenide or indium phosphide,this energy corresponds to the visible light spectrum.
Such a device is known as a Light Emitting Diode $(LED)$.

(A) The relationship between the neutral temperature $(T_{n})$,the temperature of the cold junction $(T_{c})$,and the inversion temperature $(T_{i})$ is given by the formula:
$T_{n} = \frac{T_{c} + T_{i}}{2}$
Given:
$T_{i} = 600^{\circ} C$
$T_{n} = 320^{\circ} C$
Substituting the values into the formula:
$320^{\circ} = \frac{T_{c} + 600^{\circ}}{2}$
$640^{\circ} = T_{c} + 600^{\circ}$
$T_{c} = 640^{\circ} - 600^{\circ}$
$T_{c} = 40^{\circ} C$
Therefore,the temperature of the cold junction is $40^{\circ} C$.

(C) In an interference experiment,the resultant intensity $I$ is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$,where $\phi$ is the phase difference between the two interfering waves.
The intensity is minimum when $\cos \phi = -1$.
This occurs when the phase difference $\phi$ is an odd multiple of $\pi$.
Therefore,the condition for minimum intensity is $\phi = (2n-1) \pi$,where $n = 1, 2, 3, \ldots$.

(A) According to Brewster's law,when unpolarised light is incident on a transparent medium at the polarising angle $i$,the reflected ray is completely plane-polarised.
In this condition,the reflected ray and the refracted ray are perpendicular to each other,meaning the angle between them is $90^{\circ}$.
From the geometry of the reflection and refraction at the interface,the sum of the angle of incidence $i$,the angle between the reflected and refracted rays $(90^{\circ})$,and the angle of refraction $r$ must equal $180^{\circ}$ (as they form a straight line along the interface).
Therefore,$i + 90^{\circ} + r = 180^{\circ}$.
Solving for $r$,we get $r = 180^{\circ} - 90^{\circ} - i = 90^{\circ} - i$.

(A) The fringe width in air is given by $\beta = \frac{\lambda D}{d} = 2 \,mm$.
When the apparatus is immersed in a medium of refractive index $\mu$,the wavelength of light changes to $\lambda' = \frac{\lambda}{\mu}$.
Consequently,the new fringe width $\beta'$ becomes $\beta' = \frac{\lambda' D}{d} = \frac{\lambda D}{\mu d} = \frac{\beta}{\mu}$.
Given $\beta = 2 \,mm$ and $\mu = 1.33$,we have $\beta' = \frac{2}{1.33} \approx 1.5 \,mm$.
The change in fringe width is $\Delta \beta = \beta - \beta' = 2 \,mm - 1.5 \,mm = 0.5 \,mm$.