KCET 2015 Physics Question Paper with Answer and Solution

(B) The acceleration of a body is given by the slope of its velocity-time graph.
Acceleration $a = \frac{dv}{dt} = \tan \theta$,where $\theta$ is the angle the line makes with the time axis.
For body $A$,the angle with the time axis is $25^{\circ}$. Thus,$a_A = \tan 25^{\circ}$.
For body $B$,the angle with the time axis is $50^{\circ}$. Thus,$a_B = \tan 50^{\circ}$.
Therefore,the ratio of the acceleration of $A$ to the acceleration of $B$ is $\frac{a_A}{a_B} = \frac{\tan 25^{\circ}}{\tan 50^{\circ}}$.

(B) Given,mass $m = 0.05 \,kg$.
During the upward motion of the stone,the only force acting on it (ignoring air resistance) is the gravitational force.
The magnitude of the gravitational force is given by $F = m \times g$.
Taking $g = 9.8 \,m/s^2$,we have:
$F = 0.05 \,kg \times 9.8 \,m/s^2 = 0.49 \,N$.
The direction of the gravitational force is always towards the center of the Earth,which is vertically downwards.
Therefore,the net force is $0.49 \,N$ vertically downwards.

(D) Water exhibits anomalous expansion between $0^{\circ} C$ and $4^{\circ} C$.
As water is heated from $0^{\circ} C$ to $4^{\circ} C$,its density increases,which means its volume decreases.
At $4^{\circ} C$,water attains its maximum density and minimum volume.
When water is heated further from $4^{\circ} C$ to $10^{\circ} C$,it expands like a normal liquid,and its volume increases.
Therefore,the volume of water first decreases and then increases.

(B) Given: Beat frequency $= 4 \,Hz$, frequency of fork $B$ $(f_B)$ $= 384 \,Hz$.
The possible frequencies of fork $A$ $(f_A)$ are $f_B \pm 4$, which are $388 \,Hz$ or $380 \,Hz$.
When a prong of a tuning fork is filed, its mass decreases, which increases its frequency ($f_A$ increases).
Case $1$: If $f_A = 380 \,Hz$, filing increases $f_A$ towards $384 \,Hz$, so the beat frequency $(|f_A - f_B|)$ would decrease.
Case $2$: If $f_A = 388 \,Hz$, filing increases $f_A$ away from $384 \,Hz$ (e.g., to $389 \,Hz$), so the beat frequency $(|f_A - f_B|)$ increases.
Since the problem states that the beat frequency increases, the initial frequency of fork $A$ must be $388 \,Hz$.

(C) The moment of inertia of a thin uniform rod of mass $M$ and length $L$ about an axis passing through its centre and perpendicular to its length is $I = \frac{ML^2}{12}$.
When the rod is bent into a ring of radius $R$,the circumference of the ring is equal to the length of the rod,so $L = 2\pi R$,which implies $R = \frac{L}{2\pi}$.
The moment of inertia of a ring about its diameter is $I^{\prime} = \frac{MR^2}{2}$.
Substituting $R = \frac{L}{2\pi}$ into the expression for $I^{\prime}$,we get $I^{\prime} = \frac{M}{2} \left(\frac{L}{2\pi}\right)^2 = \frac{ML^2}{8\pi^2}$.
Now,the ratio $\frac{I}{I^{\prime}}$ is $\frac{\frac{ML^2}{12}}{\frac{ML^2}{8\pi^2}} = \frac{8\pi^2}{12} = \frac{2\pi^2}{3}$.

(B) Mass is a fundamental property of matter and it does not depend on external conditions such as gravity or location.
It represents the amount of matter contained in an object.
Since mass is an intrinsic property,it remains constant regardless of where the object is placed.
Therefore,if the mass of the body on the surface of the Earth is $M$,its mass on the surface of the moon will also be $M$.

(B) Heat required to melt $ 1 \text{ g} $ of ice at $ 0^{\circ} C $ to water at $ 0^{\circ} C $ is $ Q_1 = m L_f = 1 \times 80 = 80 \text{ cal} $.
Heat required to raise the temperature of $ 1 \text{ g} $ of water from $ 0^{\circ} C $ to $ 100^{\circ} C $ is $ Q_2 = m c \Delta T = 1 \times 1 \times 100 = 100 \text{ cal} $.
Total heat required to convert $ 1 \text{ g} $ of ice at $ 0^{\circ} C $ to $ 1 \text{ g} $ of water at $ 100^{\circ} C $ is $ Q_{total} = 80 + 100 = 180 \text{ cal} $.
Heat released by $ 1 \text{ g} $ of steam at $ 100^{\circ} C $ to condense into water at $ 100^{\circ} C $ is $ Q_{steam} = m L_v = 1 \times 540 = 540 \text{ cal} $.
Since $ Q_{steam} > Q_{total} $,the steam has more than enough heat to melt the ice and raise the water to $ 100^{\circ} C $.
Therefore,the mixture will reach thermal equilibrium at $ 100^{\circ} C $ with some steam remaining.

(A) For a stretched string fixed at both ends,the frequency of the $ n $-th overtone is given by $ f_n = (n+1) f_1 $,where $ f_1 $ is the fundamental frequency.
For the second overtone,$ n = 2 $,so the string vibrates in the $ 3^{rd} $ harmonic mode.
In the $ 3^{rd} $ harmonic mode,the string has $ 3 $ loops.
The number of nodes $( N )$ in a standing wave with $ p $ loops is $ p+1 $. Here,$ p = 3 $,so the number of nodes is $ 3+1 = 4 $.
The number of antinodes $( A )$ is equal to the number of loops,which is $ 3 $.
Therefore,there are $ 4 $ nodes and $ 3 $ antinodes.

(D) The efficiency $\eta$ of a Carnot engine is given by the formula: $\eta = \left(1 - \frac{T_{2}}{T_{1}}\right) \times 100 \%$.
Given: $T_{1} = 500 \ K$ (source temperature) and $T_{2} = 300 \ K$ (sink temperature).
Substituting the values into the formula:
$\eta = \left(1 - \frac{300}{500}\right) \times 100 \%$.
$\eta = \left(1 - 0.6\right) \times 100 \%$.
$\eta = 0.4 \times 100 \% = 40 \%$.
Thus,the efficiency of the Carnot engine is $40 \%$.

(B) The dimensions of Planck's constant $(h)$ are $[M L^{2} T^{-1}]$.
The dimensions of moment of inertia $(I)$ are $[M L^{2}]$.
The ratio of the dimensions of Planck's constant to the moment of inertia is given by $\frac{[h]}{[I]} = \frac{[M L^{2} T^{-1}]}{[M L^{2}]} = [T^{-1}]$.
The dimension of frequency is $[T^{-1}]$.
The dimension of velocity is $[L T^{-1}]$.
The dimension of angular momentum is $[M L^{2} T^{-1}]$.
The dimension of time is $[T]$.
Therefore,the ratio of the dimensions of Planck's constant and the moment of inertia has the dimensions of frequency.

(A) For the second-hand,the time period $T_s = 60 \text{ s}$.
So,the angular speed $\omega_s = \frac{2\pi}{T_s} = \frac{2\pi}{60} \text{ rad/s}$.
For the hour-hand,the time period $T_h = 12 \text{ hours} = 12 \times 60 \times 60 \text{ s} = 43200 \text{ s}$.
So,the angular speed $\omega_h = \frac{2\pi}{T_h} = \frac{2\pi}{43200} \text{ rad/s}$.
The ratio of the angular speed of the second-hand to the hour-hand is $\frac{\omega_s}{\omega_h} = \frac{2\pi / 60}{2\pi / 43200} = \frac{43200}{60} = 720$.
Thus,the ratio is $720: 1$.

(C) Given: mass $m = 4 \,kg$, momentum $p = 6 \,Ns$.
The relationship between kinetic energy $K$ and momentum $p$ is given by $K = \frac{p^2}{2m}$.
Substituting the given values into the formula:
$K = \frac{6^2}{2 \times 4} = \frac{36}{8} = 4.5 \,J$.
Therefore, the kinetic energy of the body is $4.5 \,J$.

(C) Given: Horizontal range $(R)$ $= 2 \times$ Maximum height $(H)$.
Formula for range: $R = \frac{v^{2} \sin 2\theta}{g}$.
Formula for maximum height: $H = \frac{v^{2} \sin^{2} \theta}{2g}$.
According to the problem: $\frac{v^{2} \sin 2\theta}{g} = 2 \times \frac{v^{2} \sin^{2} \theta}{2g}$.
Simplifying: $\sin 2\theta = \sin^{2} \theta$.
Using the identity $\sin 2\theta = 2 \sin \theta \cos \theta$: $2 \sin \theta \cos \theta = \sin^{2} \theta$.
Dividing by $\sin \theta$ (assuming $\theta \neq 0$): $2 \cos \theta = \sin \theta$,which means $\tan \theta = 2$.
From $\tan \theta = 2$,we have $\sin \theta = \frac{2}{\sqrt{5}}$ and $\cos \theta = \frac{1}{\sqrt{5}}$.
Substituting these into the range formula: $R = \frac{v^{2} (2 \sin \theta \cos \theta)}{g} = \frac{2v^{2}}{g} \times \frac{2}{\sqrt{5}} \times \frac{1}{\sqrt{5}} = \frac{4v^{2}}{5g}$.

(B) The ratio of hydraulic stress to the corresponding volumetric strain is defined as the Bulk modulus $(B)$.
Mathematically,$B = -\frac{\Delta P}{\Delta V / V}$,where $\Delta P$ is the change in pressure and $\frac{\Delta V}{V}$ is the volumetric strain.

(A) The kinetic energy $(KE)$ of a particle executing $SHM$ is given by:
$KE = \frac{1}{2} m \omega^{2} (A^{2} - y^{2})$
Given the distance $y = \frac{A}{2}$,where $A$ is the amplitude:
$KE = \frac{1}{2} m \omega^{2} (A^{2} - (\frac{A}{2})^{2}) = \frac{1}{2} m \omega^{2} (A^{2} - \frac{A^{2}}{4}) = \frac{1}{2} m \omega^{2} (\frac{3A^{2}}{4})$
The potential energy $(PE)$ of a particle executing $SHM$ is given by:
$PE = \frac{1}{2} m \omega^{2} y^{2}$
Substituting $y = \frac{A}{2}$:
$PE = \frac{1}{2} m \omega^{2} (\frac{A}{2})^{2} = \frac{1}{2} m \omega^{2} (\frac{A^{2}}{4})$
The ratio of kinetic energy to potential energy is:
$\frac{KE}{PE} = \frac{\frac{1}{2} m \omega^{2} (\frac{3A^{2}}{4})}{\frac{1}{2} m \omega^{2} (\frac{A^{2}}{4})} = \frac{3/4}{1/4} = 3$
Therefore,the ratio is $3:1$.

(C) In the given circuit,the positive terminal of the $12 \ V$ battery is connected such that diode $D_2$ is forward-biased and diode $D_1$ is reverse-biased.
Since $D_1$ is reverse-biased,it acts as an open circuit and no current flows through the $3 \ \Omega$ resistor branch.
Since $D_2$ is forward-biased and ideal,it acts as a short circuit. The current flows through the $4 \ \Omega$ resistor and the $2 \ \Omega$ resistor branch.
The total resistance of the circuit is $R = 4 \ \Omega + 2 \ \Omega = 6 \ \Omega$.
Using Ohm's law,the current $i$ in the circuit is given by:
$i = \frac{V}{R} = \frac{12 \ V}{6 \ \Omega} = 2 \ A$.

(C) According to Curie's Law,the magnetic susceptibility $\chi$ of a paramagnetic material is inversely proportional to its absolute temperature $T$,i.e.,$\chi \propto \frac{1}{T}$.
Therefore,$\frac{\chi_1}{\chi_2} = \frac{T_2}{T_1}$.
Given:
$T_1 = -73^{\circ} C = 273 - 73 = 200 \ K$
$T_2 = -173^{\circ} C = 273 - 173 = 100 \ K$
$\chi_1 = 0.0075$
Substituting the values:
$\chi_2 = \chi_1 \times \frac{T_1}{T_2} = 0.0075 \times \frac{200}{100} = 0.0075 \times 2 = 0.0150$.
Thus,the correct option is $C$.

(D) Given: Magnification $m = -3$ (since the image is real,it must be inverted). Radius of curvature $R = -30 \ cm$. Focal length $f = R/2 = -15 \ cm$.
Using the magnification formula: $m = -v/u \Rightarrow -3 = -v/u \Rightarrow v = 3u$.
Using the mirror formula: $1/f = 1/v + 1/u$.
Substituting the values: $1/(-15) = 1/(3u) + 1/u$.
$1/(-15) = (1 + 3)/(3u) = 4/(3u)$.
$-3u = 60 \Rightarrow u = -20 \ cm$.
The negative sign indicates that the person should stand $20 \ cm$ in front of the concave mirror.

(D) The energy difference between two orbits in a hydrogen atom is given by the Rydberg formula for wavenumber: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
Since the frequency $f$ is related to wavelength $\lambda$ by $f = \frac{C}{\lambda}$,we can write $\frac{f}{C} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
Rearranging for frequency,we get $f = RC \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
Given $n_1 = 2$ and $n_2 = 3$,we substitute these values into the equation:
$f = RC \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = RC \left( \frac{1}{4} - \frac{1}{9} \right)$.
Calculating the fraction: $\frac{1}{4} - \frac{1}{9} = \frac{9-4}{36} = \frac{5}{36}$.
Thus,the frequency is $f = \frac{5RC}{36}$.

(B) The speed of light in free space is given by $c = \frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}$.
The speed of light in a medium is given by $v = \frac{1}{\sqrt{\mu \varepsilon}}$.
The refractive index $n$ of a medium is defined as the ratio of the speed of light in vacuum to the speed of light in the medium:
$n = \frac{c}{v} = \frac{\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}}{\frac{1}{\sqrt{\mu \varepsilon}}} = \frac{\sqrt{\mu \varepsilon}}{\sqrt{\mu_{0} \varepsilon_{0}}} = \sqrt{\frac{\mu \varepsilon}{\mu_{0} \varepsilon_{0}}}$.
Thus,the correct option is $B$.

(C) truth table defines the output of a logic gate for all possible input combinations.
For the given table:
- When $A=0, B=0$,output $Y=1$.
- When $A=0, B=1$,output $Y=1$.
- When $A=1, B=0$,output $Y=1$.
- When $A=1, B=1$,output $Y=0$.
This behavior corresponds to the $NAND$ gate,which is the inverse of the $AND$ gate. The Boolean expression for a $NAND$ gate is $Y = \overline{A \cdot B}$.

(B) Space waves are used for line-of-sight $(LOS)$ communication.
Line-of-sight is a type of communication that can transmit and receive data only when transmitter and receiver stations are in view of each other with no obstacle between them.
This mode of propagation is typically used for high-frequency signals like $VHF$,$UHF$,and microwave signals.

(C) The energy of a photon emitted during a transition from $n_2$ to $n_1$ is given by $E = hc / \lambda$. The energy is least when the transition is between the closest energy levels.
For the Lyman series, the transitions occur to $n_1 = 1$. The least energetic photon corresponds to the transition from $n_2 = 2$ to $n_1 = 1$.
The energy difference is $\Delta E = 13.6 \text{ eV} \times (1/n_1^2 - 1/n_2^2) = 13.6 \times (1/1^2 - 1/2^2) = 13.6 \times (3/4) = 10.2 \text{ eV}$.
Using the relation $\lambda = hc / \Delta E$:
$\lambda = 1240 \text{ eV nm} / 10.2 \text{ eV} \approx 121.57 \text{ nm} \approx 122 \text{ nm}$.

(A) Given: Initial temperature $T_{1} = 100^{\circ} C$,initial resistance $R_{1} = 100 \ \Omega$,final resistance $R_{2} = 200 \ \Omega$,and temperature coefficient $\alpha = 0.005 \ ^{\circ} C^{-1}$.
Using the formula for temperature dependence of resistance: $R_{2} = R_{1}[1 + \alpha(T_{2} - T_{1})]$.
Rearranging for $T_{2}$: $T_{2} - T_{1} = \frac{R_{2} - R_{1}}{\alpha R_{1}}$.
Substituting the values: $T_{2} - 100 = \frac{200 - 100}{0.005 \times 100}$.
$T_{2} - 100 = \frac{100}{0.5} = 200$.
$T_{2} = 200 + 100 = 300^{\circ} C$.

(A) Given: $P=2 \Omega, Q=2 \Omega, R=2 \Omega, S=3 \Omega$.
The condition for a balanced Wheatstone bridge is $\frac{P}{Q} = \frac{R}{S_{eq}}$,where $S_{eq}$ is the equivalent resistance of $S$ shunted with $X$.
The equivalent resistance $S_{eq}$ when $S$ is shunted with $X$ is given by $S_{eq} = \frac{S \cdot X}{S+X}$.
Substituting the values into the balance equation: $\frac{2}{2} = \frac{2}{\left(\frac{3X}{3+X}\right)}$.
This simplifies to $1 = \frac{2(3+X)}{3X}$.
Therefore,$3X = 6 + 2X$,which gives $X = 6 \Omega$.
Thus,$S$ must be shunted with a resistance of $6 \Omega$ to balance the bridge.

(C) From the circuit diagram,the voltmeter is connected in parallel to the combination of the ammeter and the resistor $R$.
Let the resistance of the ammeter be $R_A$ and the resistance of the voltmeter be $R_V$.
The voltmeter reading $V = 6 \text{ V}$ is the potential difference across the series combination of the ammeter and the resistor $R$.
The ammeter reading $I = 3 \text{ A}$ is the current flowing through the ammeter and the resistor $R$.
According to Ohm's law for the series combination,the total resistance $R_{total} = R + R_A = \frac{V}{I} = \frac{6 \text{ V}}{3 \text{ A}} = 2 \Omega$.
Since the ammeter has some finite resistance $R_A > 0$,it follows that $R = 2 \Omega - R_A$.
Therefore,$R < 2 \Omega$.

(B) cyclotron is a particle accelerator that uses a magnetic field to keep charged particles in a circular path and an alternating electric field to increase their kinetic energy.
It is primarily used to accelerate positively charged particles such as protons,deuterons,and alpha particles.
Neutrons cannot be accelerated by a cyclotron because they are electrically neutral and do not experience the Lorentz force $(F = q(v \times B))$ required for circular motion or the electric force $(F = qE)$ required for acceleration.

(A) Given: Output power $(P_{out})$ = $100 \,W$, Input voltage $(V_{in})$ = $220 \,V$, Input current $(I_{in})$ = $0.5 \,A$.
The input power $(P_{in})$ is calculated as $P_{in} = V_{in} \times I_{in} = 220 \,V \times 0.5 \,A = 110 \,W$.
The efficiency $(\eta)$ of a transformer is defined as the ratio of output power to input power: $\eta = \frac{P_{out}}{P_{in}} \times 100$.
Substituting the values: $\eta = \frac{100 \,W}{110 \,W} \times 100 = \frac{10}{11} \times 100 \approx 90.91 \%$.
Rounding to the nearest provided option, the efficiency is $90 \%$.

(D) Given: Resistance of galvanometer $G = 50 \Omega$,full-scale deflection current $I_g = 5 \times 10^{-4} \text{ A}$,and target voltage $V = 3 \text{ V}$.
To convert a galvanometer into a voltmeter,a high resistance $R$ must be connected in series with it.
The formula for the total resistance is $V = I_g(R + G)$.
Rearranging for $R$: $R = \frac{V}{I_g} - G$.
Substituting the values: $R = \frac{3}{5 \times 10^{-4}} - 50$.
$R = 0.6 \times 10^4 - 50 = 6000 - 50 = 5950 \Omega$.
Thus,the required resistance is $5950 \Omega$.

(A) The core of an electromagnet is designed to concentrate the magnetic flux produced by the current-carrying coil.
To achieve this efficiently,the material must have high magnetic permeability,allowing it to be easily magnetized.
Additionally,it must have low retentivity so that the magnetism disappears quickly when the electric current is switched off.
Therefore,soft iron is commonly used for this purpose because it possesses high permeability and low retentivity.

(A) According to the law of conservation of linear momentum,for a nucleus initially at rest,the total momentum after splitting must be zero.
$m_1 v_1 = m_2 v_2 \implies \frac{v_1}{v_2} = \frac{m_2}{m_1}$
Since the density of nuclear matter is constant,the mass $m$ of a nucleus is proportional to its volume,which is proportional to the cube of its radius $R$.
$m \propto R^3 \implies \frac{m_2}{m_1} = \left(\frac{R_2}{R_1}\right)^3$
Given the ratio of radii $\frac{R_1}{R_2} = \frac{1}{2}$,we have $\frac{R_2}{R_1} = \frac{2}{1}$.
Substituting this into the velocity ratio equation:
$\frac{v_1}{v_2} = \left(\frac{2}{1}\right)^3 = \frac{8}{1}$.
Thus,the ratio of their velocities is $8:1$.

(B) Given, half-life of the radioactive substance, $ T_{1/2} = 20 \text{ minutes} $.
At $ 50\% $ decay, the remaining amount $ N_1 = 50\% $ of $ N_0 = 0.5 N_0 $. This corresponds to $ 1 $ half-life, so $ t_1 = 20 \text{ minutes} $.
At $ 87.5\% $ decay, the remaining amount $ N_2 = (100 - 87.5)\% $ of $ N_0 = 12.5\% $ of $ N_0 = 0.125 N_0 = (1/8) N_0 = (1/2)^3 N_0 $.
This corresponds to $ 3 $ half-lives, so $ t_2 = 3 \times T_{1/2} = 3 \times 20 = 60 \text{ minutes} $.
The time taken between $ 50\% $ decay and $ 87.5\% $ decay is $ \Delta t = t_2 - t_1 = 60 - 20 = 40 \text{ minutes} $.

(B) The person has a near point of $ 75 \ cm $. To read at a normal distance of $ 25 \ cm $,the lens must form a virtual image of an object placed at $ 25 \ cm $ at the person's near point of $ 75 \ cm $.
Given: Object distance $ u = -25 \ cm $,Image distance $ v = -75 \ cm $.
Using the lens formula: $ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} $
$ \frac{1}{f} = \frac{1}{-75} - \frac{1}{-25} $
$ \frac{1}{f} = -\frac{1}{75} + \frac{3}{75} = \frac{2}{75} $
$ f = \frac{75}{2} = 37.5 \ cm $
Thus,the focal length of the reading glass is $ 37.5 \ cm $.

(C) Given: Wavelength $\lambda = 500 \ nm = 500 \times 10^{-9} \ m$,Distance $D = 0.5 \ m$,Slit separation $d = 0.5 \ mm = 0.5 \times 10^{-3} \ m$.
Position of $n^{th}$ maxima is $x_n = \frac{n \lambda D}{d}$. For $n=3$,$x_3 = \frac{3 \lambda D}{d}$.
Position of $m^{th}$ minima on the other side is $x'_m = \frac{(2m-1) \lambda D}{2d}$. For $m=2$,$x'_2 = \frac{(2 \times 2 - 1) \lambda D}{2d} = \frac{3 \lambda D}{2d}$.
The total distance between them is $x = x_3 + x'_2 = \frac{3 \lambda D}{d} + \frac{3 \lambda D}{2d} = \frac{9 \lambda D}{2d}$.
Substituting the values: $x = \frac{9 \times 500 \times 10^{-9} \times 0.5}{2 \times 0.5 \times 10^{-3}} = \frac{4500 \times 10^{-9}}{2 \times 10^{-3}} = 2250 \times 10^{-6} \ m = 2.25 \ mm$.

(B) The distance of closest approach $(d)$ is the distance at which the kinetic energy of the $\alpha$-particle is completely converted into electrostatic potential energy.
$d = \frac{1}{4 \pi \varepsilon_{0}} \frac{(Z_1 e)(Z_2 e)}{K}$
Here,$Z_1 = 2$ (for $\alpha$-particle),$Z_2 = 79$ (for gold nucleus),and $K = 5 \text{ MeV} = 5 \times 10^6 \times 1.6 \times 10^{-19} \text{ J} = 8 \times 10^{-13} \text{ J}$.
Substituting the values:
$d = (9 \times 10^9) \times \frac{(2 \times 1.6 \times 10^{-19}) \times (79 \times 1.6 \times 10^{-19})}{8 \times 10^{-13}}$
$d = \frac{9 \times 10^9 \times 2 \times 79 \times 2.56 \times 10^{-38}}{8 \times 10^{-13}}$
$d \approx 4.55 \times 10^{-14} \text{ m} = 4.55 \times 10^{-12} \text{ cm}$.
Thus,the order of magnitude is $10^{-12} \text{ cm}$.

(C) When $n$ identical cells,each of emf $E$ and internal resistance $r$,are connected in series,the total emf is $nE$ and the total internal resistance is $nr$.
If $m$ cells are connected with reversed polarity,the effective emf $E'$ is given by the formula:
$E' = (n - 2m)E$
Here,the total number of cells $n = 4$ and the number of wrongly connected cells $m = 1$.
Substituting these values:
$E' = (4 - 2 \times 1)E = (4 - 2)E = 2E$
The internal resistance of cells in series is additive regardless of their polarity. Therefore,the effective internal resistance $r_{eq}$ remains:
$r_{eq} = n \times r = 4r$
Thus,the equivalent emf is $2E$ and the effective internal resistance is $4r$.

(C) For a charged spherical shell of radius $R = 10 \,cm$ carrying charge $q$:
$1$. Inside the shell $(r < R)$, the electric potential is constant and equal to the potential at the surface: $V = \frac{1}{4 \pi \varepsilon_{0}} \frac{q}{R}$.
$2$. Outside the shell $(r > R)$, the potential varies inversely with distance: $V = \frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r}$.
Given distances are $r_{1} = 5 \,cm$, $r_{2} = 10 \,cm$, and $r_{3} = 15 \,cm$.
Since $r_{1} < R$ and $r_{2} = R$, we have $V_{1} = V_{2} = \frac{1}{4 \pi \varepsilon_{0}} \frac{q}{10}$.
For $r_{3} = 15 \,cm$, $V_{3} = \frac{1}{4 \pi \varepsilon_{0}} \frac{q}{15}$.
Comparing the values, since $15 > 10$, it follows that $\frac{q}{15} < \frac{q}{10}$, therefore $V_{3} < V_{1} = V_{2}$.
Thus, the correct relation is $V_{1} = V_{2} > V_{3}$.

(D) Given that the parallel plate capacitor is charged and then isolated. Therefore,the charge $Q$ remains constant.
Capacitance of a parallel plate capacitor is given by $C = \frac{\varepsilon_{0} A}{d}$,which implies $C \propto \frac{1}{d}$.
If the separation $d$ between the plates is increased,the capacitance $C$ decreases.
From the relation $Q = CV$,we have $V = \frac{Q}{C}$. Since $Q$ is constant and $C$ decreases,the potential $V$ must increase.
Thus,on increasing the plate separation,the charge remains constant,the potential increases,and the capacitance decreases.

(C) The magnetic field at the centre of a circular coil is given by $B = \frac{\mu_0 I}{2a}$,where $a$ is the radius.
Given $a = 2\pi \text{ cm} = 2\pi \times 10^{-2} \text{ m}$.
For the first coil with current $I_1 = 3 \text{ A}$:
$B_1 = \frac{\mu_0 \times 3}{2 \times 2\pi \times 10^{-2}} = \frac{4\pi \times 10^{-7} \times 3}{4\pi \times 10^{-2}} = 3 \times 10^{-5} \text{ T}$.
For the second coil with current $I_2 = 4 \text{ A}$:
$B_2 = \frac{\mu_0 \times 4}{2 \times 2\pi \times 10^{-2}} = \frac{4\pi \times 10^{-7} \times 4}{4\pi \times 10^{-2}} = 4 \times 10^{-5} \text{ T}$.
Since the coils are placed at right angles,the resultant magnetic field is $B = \sqrt{B_1^2 + B_2^2}$.
$B = \sqrt{(3 \times 10^{-5})^2 + (4 \times 10^{-5})^2} = \sqrt{9 \times 10^{-10} + 16 \times 10^{-10}} = \sqrt{25 \times 10^{-10}} = 5 \times 10^{-5} \text{ T}$.

(C) Given: Magnetic field $ B = 10^{-4} \,Wb m^{-2} $.
Specific charge of the proton $ \frac{q}{m} = 10^{11} \,C kg^{-1} $.
Velocity of the proton $ v = 10^{9} \,ms^{-1} $.
When a charged particle enters a magnetic field normally, it follows a circular path with radius $ r $ given by the formula:
$ r = \frac{mv}{qB} $.
We can rewrite this as $ r = \frac{v}{(\frac{q}{m}) \times B} $.
Substituting the given values:
$ r = \frac{10^{9}}{10^{11} \times 10^{-4}} $.
$ r = \frac{10^{9}}{10^{7}} $.
$ r = 10^{2} \,m = 100 \,m $.
Thus, the radius of the circle described is $ 100 \,m $.

(A) In an $LCR$ circuit,resonance occurs when the inductive reactance $(X_L = \omega L)$ is equal to the capacitive reactance $(X_C = 1/(\omega C))$.
At this condition,the net reactance $X = X_L - X_C = 0$.
The impedance of the circuit is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
At resonance,$Z = R$,which is the minimum possible impedance.
Since the net reactance is zero,the phase angle $\phi = \tan^{-1}((X_L - X_C)/R) = 0$.
Therefore,the current and voltage are in phase at resonance.

(C) magnetic dipole of magnetic moment $M$ and moment of inertia $I$ placed in a uniform magnetic field $B$ experiences a restoring torque $\tau = -MB \sin \theta$.
For small oscillations, $\sin \theta \approx \theta$, so $\tau = -MB \theta$.
Comparing this with the equation for angular $SHM$, $\tau = -C \theta$, where $C = MB$.
The time period $T$ of an angular $SHM$ is given by $T = 2 \pi \sqrt{\frac{I}{C}}$.
Substituting $C = MB$, we get $T = 2 \pi \sqrt{\frac{I}{MB}}$.

(A) The force per unit length $f$ between two parallel wires carrying currents $I_1$ and $I_2$ separated by a distance $d$ is given by the formula:
$f = \frac{\mu_0}{4\pi} \frac{2 I_1 I_2}{d}$
Given values are $I_1 = 1 \text{ A}$,$I_2 = 3 \text{ A}$,$d = 1 \text{ m}$,and $\frac{\mu_0}{4\pi} = 10^{-7} \text{ T m A}^{-1}$.
Substituting these values into the formula:
$f = 10^{-7} \times \frac{2 \times 1 \times 3}{1} = 6 \times 10^{-7} \text{ N m}^{-1}$
Since the currents flow in opposite directions,the force between the wires is repulsive.
Therefore,the force per unit length is $6 \times 10^{-7} \text{ N m}^{-1}$ and it is repulsive.

(D) The average power dissipated in an $AC$ circuit is given by the formula $ P_{av} = VI \cos \phi $,where $ V $ is the $RMS$ voltage,$ I $ is the $RMS$ current,and $ \phi $ is the phase angle between voltage and current.
In a pure inductor,the current lags behind the voltage by a phase angle of $ \phi = \pi / 2 $.
Substituting this into the power formula: $ P_{av} = VI \cos(\pi / 2) $.
Since $ \cos(\pi / 2) = 0 $,we get $ P_{av} = VI \times 0 = 0 $.
Therefore,the average power dissipated in a pure inductor is zero.

(C) Given: Wingspan $l = 40 \text{ m}$,speed $v = 1080 \text{ km h}^{-1}$.
Convert speed to $SI$ units: $v = 1080 \times \frac{5}{18} \text{ m s}^{-1} = 300 \text{ m s}^{-1}$.
Vertical component of Earth's magnetic field $B = 1.75 \times 10^{-5} \text{ T}$.
The motional emf induced across the wingspan is given by $E = B l v$.
Substituting the values: $E = (1.75 \times 10^{-5} \text{ T}) \times (40 \text{ m}) \times (300 \text{ m s}^{-1})$.
$E = 1.75 \times 10^{-5} \times 12000 = 1.75 \times 0.12 = 0.21 \text{ V}$.
Thus,the emf developed between the tips of the wings is $0.21 \text{ V}$.

(B) Given: Mutual inductance $ M = 0.005 \ H $,current $ i = i_{m} \sin \omega t $,peak current $ i_{m} = 10 \ A $,and angular frequency $ \omega = 100 \pi \ rad \ s^{-1} $.
The induced emf $ \varepsilon $ in the second coil is given by $ \varepsilon = M \frac{di}{dt} $.
Substituting the expression for current: $ \varepsilon = M \frac{d}{dt} (i_{m} \sin \omega t) = M i_{m} \omega \cos \omega t $.
The maximum value of the induced emf $ \varepsilon_{\max} $ occurs when $ \cos \omega t = 1 $.
Therefore,$ \varepsilon_{\max} = M \omega i_{m} $.
Substituting the values: $ \varepsilon_{\max} = 0.005 \times 100 \pi \times 10 = 5 \pi \ V $.
Thus,the maximum value of the emf induced in the second coil is $ 5 \pi \ V $.

(D) Diffraction is the phenomenon of the bending or spreading of light around the corners of an obstacle or through an aperture.
For significant diffraction to occur,the size of the obstacle or the aperture must be comparable to the wavelength $( \lambda )$ of the incident light.
If the size of the obstacle is much larger than the wavelength,the light travels in a straight line (rectilinear propagation),and diffraction is negligible.
Therefore,the condition for observing diffraction is that the size of the obstacle should be of the order of the wavelength of light.

(D) The potential energy $U$ of a system of point charges is given by the sum of the potential energies of all pairs of charges:
$U = \frac{1}{4 \pi \epsilon_0} \left( \frac{q_1 q_2}{r_{12}} + \frac{q_2 q_3}{r_{23}} + \frac{q_3 q_1}{r_{31}} \right)$
Given $q_1 = 3 \times 10^{-9} \text{ C}$,$q_2 = 6 \times 10^{-9} \text{ C}$,$q_3 = 9 \times 10^{-9} \text{ C}$,and $r = 0.1 \text{ m}$.
$U = \frac{9 \times 10^9}{0.1} \left( (3 \times 6) + (6 \times 9) + (9 \times 3) \right) \times 10^{-18} \text{ J}$
$U = 9 \times 10^{11} \times (18 + 54 + 27) \times 10^{-18} \text{ J}$
$U = 9 \times 10^{11} \times 99 \times 10^{-18} \text{ J}$
$U = 891 \times 10^{-7} \text{ J} = 8.91 \times 10^{-6} \text{ J}$

(A) The input characteristics of a transistor in common emitter $(CE)$ configuration describe the relationship between the input current and the input voltage.
Specifically,it is the curve obtained by plotting the base current $(I_{B})$ against the base-emitter voltage $(V_{BE})$ while keeping the collector-emitter voltage $(V_{CE})$ constant.
This graph shows how the base current changes as the input voltage increases for different fixed values of the output voltage.

(A) In amplitude modulation,the amplitude of the carrier wave is varied in accordance with the instantaneous value of the modulating signal.
Mathematically,the modulated wave consists of the carrier frequency $(f_c)$ and two sideband frequencies: the upper sideband $(f_c + f_m)$ and the lower sideband $(f_c - f_m)$,where $f_m$ is the frequency of the modulating signal.
Therefore,amplitude modulation consists of one carrier frequency and two sideband frequencies.

(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of emitted electrons is given by $K_{max} = E - \Phi$,where $E$ is the energy of the incident photon and $\Phi$ is the work function of the metal.
For the first photon with energy $E_1 = 1 \text{ eV}$:
$K_1 = 1 \text{ eV} - 0.5 \text{ eV} = 0.5 \text{ eV}$.
For the second photon with energy $E_2 = 2.5 \text{ eV}$:
$K_2 = 2.5 \text{ eV} - 0.5 \text{ eV} = 2.0 \text{ eV}$.
Since $K_{max} = \frac{1}{2}mv^2$,the ratio of the kinetic energies is:
$\frac{K_1}{K_2} = \frac{0.5}{2.0} = \frac{1}{4}$.
Substituting the expression for kinetic energy:
$\frac{\frac{1}{2}mv_1^2}{\frac{1}{2}mv_2^2} = \frac{1}{4} \Rightarrow \frac{v_1^2}{v_2^2} = \frac{1}{4}$.
Taking the square root on both sides:
$\frac{v_1}{v_2} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Thus,the ratio of the maximum speeds is $1:2$.

(C) Given, kinetic energy of electron $K = 120 eV$.
The de-Broglie wavelength $\lambda$ is related to the accelerating potential $V$ (where $K = eV$) as:
$\lambda = \frac{1.227 \text{ nm}}{\sqrt{V}}$
Substituting $V = 120 V$:
$\lambda = \frac{1.227 \times 10^{-9} \text{ m}}{\sqrt{120}}$
$\lambda = \frac{1.227 \times 10^{-9}}{10.954}$
$\lambda \approx 0.112 \times 10^{-9} \text{ m}$
$\lambda = 112 \times 10^{-12} \text{ m} = 112 \text{ pm}$.

(C) The initial force between the two charges is given by Coulomb's Law: $F = k \frac{q_1 q_2}{d^2}$.
Substituting the initial values: $F = k \frac{(6 \mu C)(9 \mu C)}{d^2} = k \frac{54 \mu C^2}{d^2} \dots (1)$.
When a charge of $-3 \mu C$ is added to both spheres,the new charges are:
$q_1' = 6 \mu C - 3 \mu C = 3 \mu C$
$q_2' = 9 \mu C - 3 \mu C = 6 \mu C$.
The new force $F'$ is: $F' = k \frac{q_1' q_2'}{d^2} = k \frac{(3 \mu C)(6 \mu C)}{d^2} = k \frac{18 \mu C^2}{d^2} \dots (2)$.
Dividing equation $(2)$ by equation $(1)$:
$\frac{F'}{F} = \frac{18}{54} = \frac{1}{3}$.
Therefore,the new force is $F' = F/3$.

(B) Statement $A$ is correct: The tangent drawn to a line of force at any point gives the direction of the electric field at that point.
Statement $B$ is incorrect: Electric field lines originate from positive charges and terminate at negative charges. They do not form closed loops,unlike magnetic field lines.
Statement $C$ is correct: The force on a charge $q$ in an electric field $E$ is given by $F = qE$. If $q$ is negative,the force $F$ is in the direction opposite to $E$.
Statement $D$ is correct: Two electric field lines can never intersect,because if they did,there would be two directions for the electric field at the point of intersection,which is physically impossible.

(C) The electric dipole moment $\vec{p}$ is a vector directed from the negative charge $(-q)$ to the positive charge $(+q)$.
At any point on the equatorial plane of an electric dipole,the resultant electric field $\vec{E}$ is directed parallel to the dipole axis but in the opposite direction to the dipole moment vector $\vec{p}$.
Since the electric field vector $\vec{E}$ and the dipole moment vector $\vec{p}$ are anti-parallel,the angle between them is $180^{\circ}$.

(C) The energy gap $E_g$ of a semiconductor is related to the wavelength $\lambda$ of the emitted light by the formula: $E_g = \frac{hc}{\lambda}$.
Given $E_g = 1.9 \ eV$. Converting this to Joules: $E_g = 1.9 \times 1.6 \times 10^{-19} \ J$.
Using the values $h = 6.6 \times 10^{-34} \ J \cdot s$ and $c = 3 \times 10^8 \ m/s$:
$\lambda = \frac{hc}{E_g} = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{1.9 \times 1.6 \times 10^{-19}} \ m$.
$\lambda = \frac{19.8 \times 10^{-26}}{3.04 \times 10^{-19}} \ m \approx 6.513 \times 10^{-7} \ m$.
Rounding to the nearest given option,the wavelength is $6.5 \times 10^{-7} \ m$.

(B) Let the original nucleus be $ _{Z}^{A}X $.
When an $ \alpha $-particle $( _{2}^{4}He )$ is emitted,the mass number decreases by $ 4 $ and the atomic number decreases by $ 2 $. The new nucleus is $ _{Z-2}^{A-4}Y $.
When a $ \beta $-particle $( _{-1}^{0}e )$ is emitted,the atomic number increases by $ 1 $ and the mass number remains unchanged.
If we emit one $ \alpha $ and two $ \beta $ particles:
$1$. After $ \alpha $ emission: $ _{Z-2}^{A-4}Y $
$2$. After first $ \beta $ emission: $ _{Z-1}^{A-4}Y' $
$3$. After second $ \beta $ emission: $ _{Z}^{A-4}Y'' $
Since the atomic number $ Z $ is the same as the original nucleus,it is an isotope of the original nucleus.

(B) Given,the polarizing angle of glass,$i_{p} = 57^{\circ}$.
We know that when light is incident at the polarizing angle,the reflected ray and the refracted ray are at right angles to each other.
According to Brewster's law,the relationship between the angle of incidence $(i_{p})$ and the angle of refraction $(r)$ is given by $i_{p} + r = 90^{\circ}$.
Therefore,$r = 90^{\circ} - i_{p}$.
Substituting the given value,$r = 90^{\circ} - 57^{\circ} = 33^{\circ}$.
Thus,the angle of refraction is $33^{\circ}$.

(D) The two cells are connected in opposition,so the net electromotive force (emf) of the circuit is $E_{net} = E_{1} - E_{2}$.
The total resistance of the circuit is the sum of the internal resistances and the external resistance,which is $R_{total} = r_{1} + r_{2} + R$.
According to Ohm's law,the current $I$ flowing through the circuit is given by $I = \frac{E_{net}}{R_{total}} = \frac{E_{1} - E_{2}}{r_{1} + r_{2} + R}$.
The terminal potential difference $V$ across the external resistance $R$ is given by $V = I \times R$.
Substituting the value of $I$,we get $V = \frac{E_{1} - E_{2}}{r_{1} + r_{2} + R} \times R$.

(B) According to Ohm's law,$V = IR$,where $V$ is the potential difference,$I$ is the current,and $R$ is the resistance.
Since the resistors are connected in parallel,the potential difference $V$ across each resistor is the same.
Therefore,$I = V/R$,which implies $I \propto 1/R$.
Let the currents through the $2 \Omega, 3 \Omega$,and $4 \Omega$ resistors be $I_1, I_2$,and $I_3$ respectively.
Then,$I_1 : I_2 : I_3 = \frac{1}{2} : \frac{1}{3} : \frac{1}{4}$.
To simplify this ratio,multiply each term by the least common multiple $(LCM)$ of $2, 3$,and $4$,which is $12$.
$I_1 : I_2 : I_3 = (\frac{1}{2} \times 12) : (\frac{1}{3} \times 12) : (\frac{1}{4} \times 12) = 6 : 4 : 3$.