KVPY 2012 Physics Question Paper with Answer and Solution

(D) Let the heat capacity of the bodies be $C(T)$,where $C(T)$ is an increasing function of temperature $T$.
When the two bodies are brought into contact,heat flows from the hotter body $(100^{\circ} C)$ to the colder body $(0^{\circ} C)$ until they reach a common final temperature $T_f$.
According to the principle of calorimetry,the heat lost by the hotter body equals the heat gained by the colder body:
$\int_{T_f}^{100} C(T) dT = \int_{0}^{T_f} C(T) dT$.
Since $C(T)$ increases with temperature,the heat capacity of the hotter body is higher than that of the colder body throughout the process of reaching equilibrium.
Because the hotter body has a higher heat capacity,it requires more energy to change its temperature by a certain amount compared to the colder body.
Therefore,the final equilibrium temperature $T_f$ will be closer to the initial temperature of the body with the higher heat capacity.
Thus,$T_f > 50^{\circ} C$.

(B) The work done by a gas during an expansion process is equal to the area under the $p-V$ curve.
For an ideal gas expanding from volume $V$ to $2V$,the isothermal process is represented by the curve $AB$ and the adiabatic process is represented by the curve $AC$ on the $p-V$ diagram.
Since the adiabatic curve is steeper than the isothermal curve,the area under the isothermal curve $AB$ is greater than the area under the adiabatic curve $AC$ for the same change in volume.
Therefore,the work done in the isothermal process $(W_i)$ is greater than the work done in the adiabatic process $(W_a)$.
Since the gas is expanding,the work done is positive in both cases.
Thus,$W_i > W_a > 0$.

(A) The frequency of vibration $v$ of a liquid drop is governed by surface tension $\sigma$,density $\rho$,and radius $R$. The given ratio is $\frac{v}{\sqrt{\sigma / \rho R^3}}$.
Dimensions of frequency $v = [T^{-1}]$.
Dimensions of surface tension $\sigma = [MT^{-2}]$.
Dimensions of density $\rho = [ML^{-3}]$.
Dimensions of radius $R = [L]$.
Let the ratio be $X = \frac{v}{\sqrt{\sigma / \rho R^3}}$.
Dimensions of $\sqrt{\frac{\sigma}{\rho R^3}} = \sqrt{\frac{MT^{-2}}{ML^{-3} \cdot L^3}} = \sqrt{\frac{MT^{-2}}{M}} = [T^{-1}]$.
Thus,the ratio $\frac{v}{\sqrt{\sigma / \rho R^3}}$ is dimensionless.
We check the dimensions of option $(A)$:
$\frac{k \rho g R^2}{\sigma} = \frac{[ML^{-3}] \cdot [LT^{-2}] \cdot [L^2]}{[MT^{-2}]} = \frac{[ML^0T^{-2}]}{[MT^{-2}]} = [M^0L^0T^0]$.
Since this expression is dimensionless,it is consistent with the dimensional analysis of the given ratio.

(C) Let $A$ be the centre of the central coin. The moment of inertia of the central coin about an axis through $A$ is $I_{central} = \frac{m r^2}{2}$.
For the six surrounding coins,the distance of their centres from $A$ is $2r$. Using the parallel axis theorem,the moment of inertia of each surrounding coin about the axis through $A$ is $I_{surrounding} = \frac{m r^2}{2} + m(2r)^2 = \frac{m r^2}{2} + 4mr^2 = \frac{9}{2} mr^2$.
The total moment of inertia of the system about the axis through $A$ is $I_A = I_{central} + 6 \times I_{surrounding} = \frac{m r^2}{2} + 6 \times \frac{9}{2} mr^2 = \frac{m r^2}{2} + 27 mr^2 = \frac{55}{2} mr^2$.
Now,we need the moment of inertia about point $P$,which is at a distance $d = 2r$ from $A$. Using the parallel axis theorem,$I_P = I_A + M_{total} \times d^2$,where $M_{total} = 7m$.
$I_P = \frac{55}{2} mr^2 + (7m)(2r)^2 = \frac{55}{2} mr^2 + 28 mr^2 = \frac{55 + 56}{2} mr^2 = \frac{111}{2} mr^2$.

(A) By the law of conservation of angular momentum,the angular momentum $L$ of the planet remains constant at all points in its orbit.
At the perihelion $P$ and aphelion $A$,the velocity vector is perpendicular to the position vector,so the angular momentum is given by $L = mvr$.
Since $L_P = L_A$,we have $m v_P r_P = m v_A r_A$,which implies $\frac{v_P}{v_A} = \frac{r_A}{r_P}$.
For an ellipse with semi-major axis $a$ and eccentricity $e$,the distance of the closest point (perihelion) is $r_P = a(1-e)$ and the distance of the farthest point (aphelion) is $r_A = a(1+e)$.
Substituting these values,we get $\frac{v_P}{v_A} = \frac{a(1+e)}{a(1-e)} = \frac{1+e}{1-e}$.

(A) In simple harmonic motion,the velocity of the body is given by $v = \omega \sqrt{a^2 - x^2}$.
The time $dt$ spent by the body in a small interval $dx$ is given by $dt = \frac{dx}{v} = \frac{dx}{\omega \sqrt{a^2 - x^2}}$.
The probability $P(x)dx$ of finding the body in the interval $dx$ is proportional to the time spent in that interval,i.e.,$P(x) \propto \frac{1}{v} = \frac{1}{\omega \sqrt{a^2 - x^2}}$.
As $x \to \pm a$,the velocity $v \to 0$,which means the time spent $dt \to \infty$.
Therefore,the probability of finding the body is highest at the extreme positions $x = \pm a$.
Thus,the correct option is $A$.

(A) The force acting on the particle is given by $F = -\alpha x^3 - \beta x^4$.
Since $F = -\frac{dU}{dx}$,we have $\frac{dU}{dx} = \alpha x^3 + \beta x^4$.
At $x = 0$,$\frac{dU}{dx} = 0$,which implies the particle is in equilibrium.
To determine the stability,we examine the derivatives of the potential energy $U$ at $x = 0$:
$\frac{d^2U}{dx^2} = 3\alpha x^2 + 4\beta x^3$,which is $0$ at $x = 0$.
$\frac{d^3U}{dx^3} = 6\alpha x + 12\beta x^2$,which is $0$ at $x = 0$.
$\frac{d^4U}{dx^4} = 6\alpha + 24\beta x$,which is $6\alpha$ at $x = 0$.
Since $\alpha > 0$,the first non-zero derivative at $x = 0$ is of an even order ($4^{th}$ order) and is positive. This indicates that $U$ has a local minimum at $x = 0$.
Therefore,the particle is in stable equilibrium.

(D) The potential energy of the particle is given by $V(x) = -\alpha x + \beta \sin(x / \gamma)$.
For the equation to be dimensionally consistent,the dimensions of each term must be the same as the dimensions of potential energy $[V] = [ML^2T^{-2}]$.
$1$. For the term $\alpha x$: $[\alpha][x] = [V] \implies [\alpha][L] = [ML^2T^{-2}] \implies [\alpha] = [MLT^{-2}]$.
$2$. For the term $\beta \sin(x / \gamma)$: The argument of the sine function must be dimensionless,so $[x / \gamma] = [M^0L^0T^0] \implies [L] / [\gamma] = 1 \implies [\gamma] = [L]$.
$3$. Also,$[\beta] = [V] = [ML^2T^{-2}]$.
Now,check the dimensions of the given options:
For option $(d)$: $\frac{[\alpha][\gamma]}{[\beta]} = \frac{[MLT^{-2}][L]}{[ML^2T^{-2}]} = \frac{[ML^2T^{-2}]}{[ML^2T^{-2}]} = [M^0L^0T^0]$.
Since this combination is dimensionless,the correct option is $(d)$.

(D) Let the velocity of the ball of mass $m$ just before the collision be $u$. By conservation of energy, $\frac{1}{2}mu^2 = mgh$, so $u = \sqrt{2gh}$.
In an elastic collision, both kinetic energy and linear momentum are conserved.
Let $v_1$ be the velocity of the ball (in the opposite direction) and $v_2$ be the velocity of the block after the collision.
Conservation of momentum: $mu = 2mv_2 - mv_1 \Rightarrow u = 2v_2 - v_1 \quad \dots (i)$
Conservation of kinetic energy: $\frac{1}{2}mu^2 = \frac{1}{2}mv_1^2 + \frac{1}{2}(2m)v_2^2 \Rightarrow u^2 = v_1^2 + 2v_2^2 \quad \dots (ii)$
From $(i)$, $2v_2 = u + v_1$. Substituting this into $(ii)$:
$u^2 = v_1^2 + 2\left(\frac{u+v_1}{2}\right)^2 = v_1^2 + \frac{(u+v_1)^2}{2}$
$2u^2 = 2v_1^2 + u^2 + v_1^2 + 2uv_1$
$u^2 - 2uv_1 - 3v_1^2 = 0$
$(u - 3v_1)(u + v_1) = 0$
Since $v_1$ is the speed in the opposite direction, $v_1 = \frac{u}{3}$.
The new height $h'$ reached by the ball is given by $mgh' = \frac{1}{2}mv_1^2$.
$h' = \frac{v_1^2}{2g} = \frac{(u/3)^2}{2g} = \frac{u^2}{18g} = \frac{2gh}{18g} = \frac{h}{9}$.

(A) When a particle falls through a fluid,it is acted upon by three forces: gravitational force $(mg)$ acting downwards,buoyant force $(F_B)$ acting upwards,and a resistive force $(F_R)$ acting upwards.
The net force on the particle is $F_{net} = mg - F_B - F_R$.
Given that the resistive force is proportional to the square of its speed,we have $F_R = kv^2$,where $k$ is a constant.
Initially,at $t = 0$,the speed $v = 0$,so the resistive force $F_R = 0$. The acceleration is maximum,and the speed starts increasing.
As the speed $v$ increases,the resistive force $F_R = kv^2$ also increases. Consequently,the net force $F_{net} = mg - F_B - kv^2$ decreases,which means the acceleration $(a = F_{net}/m)$ decreases.
Eventually,the resistive force increases to a point where the net force becomes zero $(mg - F_B - kv^2 = 0)$. At this point,the acceleration becomes zero,and the particle attains a constant terminal velocity.
The graph of speed $v$ versus time $t$ should show an initial increase in speed with a decreasing slope,eventually approaching a constant value (asymptote). Graph $(a)$ correctly represents this behavior.

(A) In steady state,the rate of heat flow through both rods is equal.
Let $T$ be the temperature at the junction.
$\left(\frac{kA(T_1 - T)}{l}\right)_{\text{steel}} = \left(\frac{kA(T - T_2)}{l}\right)_{\text{copper}}$
Since the area $A$ is the same for both rods,it cancels out:
$\frac{50(100 - T)}{0.1} = \frac{400(T - 0)}{0.2}$
$500(100 - T) = 2000(T)$
$100 - T = 4T$
$5T = 100$
$T = 20^{\circ}C$

(B) The energy radiated from a black body is given by the Stefan-Boltzmann law: $U = \sigma A T^4 t$,where $\sigma$ is Stefan's constant,$A$ is the surface area,$T$ is the absolute temperature,and $t$ is the time.
In the first case,the energy collected is $U_1 = \sigma A T^4 t$. The temperature rise is $\Delta T_1 = 11.0^{\circ} C - 10.0^{\circ} C = 1.0^{\circ} C$.
In the second case,the new area is $A' = A/2$ and the new temperature is $T' = 2T$. The energy collected is $U_2 = \sigma (A/2) (2T)^4 t = \sigma (A/2) (16T^4) t = 8 \sigma A T^4 t = 8 U_1$.
Since the energy absorbed by the water is proportional to the temperature rise $(\Delta Q = ms \Delta T)$,we have $\frac{\Delta T_2}{\Delta T_1} = \frac{U_2}{U_1} = 8$.
Therefore,$\Delta T_2 = 8 \times \Delta T_1 = 8 \times 1.0^{\circ} C = 8.0^{\circ} C$.
The final temperature of the water will be $10.0^{\circ} C + 8.0^{\circ} C = 18.0^{\circ} C$.

(A) The total energy of the rocket at the instant of launch must be less than or equal to zero for it to remain bound to the solar system.
The orbital speed of the asteroid is given by $v_0 = \sqrt{\frac{GM}{r_0}}$,which implies $v_0^2 = \frac{GM}{r_0}$.
The total energy $E$ of the rocket relative to the sun is the sum of its gravitational potential energy and its kinetic energy:
$E = -\frac{GMm}{r_0} + \frac{1}{2}mv^2$
Given that the speed of the rocket relative to the sun is $v = \alpha v_0$,we substitute this into the energy equation:
$E = -\frac{GMm}{r_0} + \frac{1}{2}m(\alpha v_0)^2$
For the rocket to remain bound,$E \leq 0$:
$-\frac{GMm}{r_0} + \frac{1}{2}m \alpha^2 v_0^2 \leq 0$
Substituting $v_0^2 = \frac{GM}{r_0}$:
$-\frac{GMm}{r_0} + \frac{1}{2}m \alpha^2 \left(\frac{GM}{r_0}\right) \leq 0$
Dividing by $\frac{GMm}{r_0}$:
$-1 + \frac{1}{2} \alpha^2 \leq 0$
$\frac{1}{2} \alpha^2 \leq 1$
$\alpha^2 \leq 2$
$\alpha \leq \sqrt{2}$
Wait,re-evaluating the launch condition: If the rocket is launched *from* the asteroid,its initial velocity relative to the sun is the vector sum of the asteroid's orbital velocity and the launch velocity. However,the problem states $v = \alpha v_0$ is the speed relative to the sun. Thus,the condition for being bound is $E \leq 0$,leading to $\alpha \leq \sqrt{2}$. The highest value is $\sqrt{2}$.

(D) The given wave equation is $y(x, t) = y_0 \sin \left( \frac{2\pi}{L} x \right) \sin \left( \frac{2\pi}{L} x + \frac{\pi}{4} \right)$.
Comparing this with the general form of a standing wave $k = \frac{2\pi}{\lambda} = \frac{2\pi}{L}$,we get $\lambda = L = 1.2 \,m$.
For a pipe closed at both ends,the allowed wavelengths are $\lambda_n = \frac{2L}{n}$. For $n=2$,$\lambda = L = 1.2 \,m$,which is consistent with the wave equation.
At $x=0$,$y(0, t) = y_0 \sin(0) \sin(\pi/4) = 0$,so there is a node at $x=0$.
At $x=L/2$,$y(L/2, t) = y_0 \sin(\pi) \sin(\pi + \pi/4) = 0$,which implies a node at $x=L/2$. However,the statement in option $(c)$ claims an antinode at $x=L/2$,which is incorrect based on the equation,but the question asks for the incorrect statement among the choices provided.
The fundamental frequency is $f = \frac{v}{\lambda_{max}} = \frac{v}{2L} = \frac{300}{2 \times 1.2} = \frac{300}{2.4} = 125 \,Hz$.
Since $125 \,Hz \neq 137.5 \,Hz$,option $(d)$ is clearly incorrect.

(B) The free body diagrams for block $1$ and $2$ are shown in the figure.
Let $T$ be the tension in the string.
For block $1$,the forces are tension $T$ upwards,and gravity $mg$ and spring force $k_1 x$ downwards. The equation of motion is:
$T + k_1 x - mg = ma_1$ (assuming downward acceleration $a_1$)
For block $2$,the forces are tension $T$ upwards,and gravity $mg$ downwards,and spring force $k_2 x$ upwards. The equation of motion is:
$k_2 x + mg - T = ma_2$ (assuming upward acceleration $a_2$)
Since the string is inextensible,the magnitudes of accelerations must be equal,so $a_1 = a_2 = a$.
Adding the two equations:
$(T + k_1 x - mg) + (k_2 x + mg - T) = ma_1 + ma_2$
$(k_1 + k_2) x = 2ma$
$a = \frac{(k_1 + k_2) x}{2m}$
Thus,the magnitude of acceleration for both blocks is $a = \frac{(k_1 + k_2) x}{2m}$.

(B) Let $v$ be the velocity of the bob at position $\theta$ after being released from the horizontal position.
From the law of conservation of energy,the potential energy lost equals the kinetic energy gained:
$mgh = \frac{1}{2}mv^2$
Since the height $h$ dropped from the horizontal is $l \cos \theta$,we have:
$mg(l \cos \theta) = \frac{1}{2}mv^2$
$\Rightarrow \frac{v^2}{l} = 2g \cos \theta$
The radial (centripetal) acceleration is $a_c = \frac{v^2}{l} = 2g \cos \theta$.
The tangential acceleration is due to the component of gravity perpendicular to the string:
$a_t = g \sin \theta$.
If the total acceleration vector $\vec{a}$ makes an angle $\phi$ with the string (which is the direction of the radial acceleration),then:
$\tan \phi = \frac{a_t}{a_c}$
Substituting the expressions for $a_t$ and $a_c$:
$\tan \phi = \frac{g \sin \theta}{2g \cos \theta} = \frac{\tan \theta}{2}$
Therefore,$\phi = \tan^{-1}\left(\frac{\tan \theta}{2}\right)$.

(B) According to the law of conservation of linear momentum,the total momentum before the collision is equal to the total momentum after the collision.
Let $V$ be the final velocity of the combined system of mass $(m + m) = 2m$.
$m v + m(0) = (2m)V$
$m v = 2m V$
$V = v / 2$
Now,the kinetic energy of the system after the collision is given by:
$K_f = \frac{1}{2} (2m) V^2$
Substituting the value of $V$:
$K_f = \frac{1}{2} (2m) \left(\frac{v}{2}\right)^2$
$K_f = m \times \frac{v^2}{4} = \frac{m v^2}{4}$

(A) Let the speed of the ball be $v$ at the instant just before and just after the collision. The air resistance force $F_r$ is proportional to the speed,so $F_r = kv$,where $k$ is a constant.
When the ball is moving downwards,the gravitational force $mg$ acts downwards and the air resistance $kv$ acts upwards. The net force is $F_{net} = mg - kv$. The acceleration $a_1$ is given by:
$a_1 = \frac{mg - kv}{m} = g - \frac{k}{m}v$
When the ball is moving upwards,the gravitational force $mg$ acts downwards and the air resistance $kv$ also acts downwards (opposing the motion). The net force is $F_{net} = mg + kv$. The acceleration $a_2$ is given by:
$a_2 = \frac{mg + kv}{m} = g + \frac{k}{m}v$
Comparing the two expressions,it is clear that $a_2 > a_1$ or $a_1 < a_2$.

(D) To minimize the drift,the boat must be steered such that its resultant velocity is perpendicular to the river bank.
Let the boat be directed at an angle $\theta$ with the normal to the river flow,as shown in the diagram.
From the velocity triangle,the component of the boat's velocity $v$ along the river flow must cancel the river's velocity $v/2$.
Thus,$v \sin \theta = v/2$.
$\sin \theta = 1/2$,which gives $\theta = 30^{\circ}$.
The angle with respect to the direction of flow is $90^{\circ} + \theta = 90^{\circ} + 30^{\circ} = 120^{\circ}$.

(B) The correct answer is $(B)$.
Ice is a poor conductor of heat,meaning it has a very low thermal conductivity.
Because of this low thermal conductivity,the ice walls of the Igloo act as an insulator,preventing the heat generated inside (by the occupants or a small fire) from escaping to the cold external environment and preventing the cold from entering the interior.
This allows the temperature inside the Igloo to be maintained at a comfortable level,such as $20^{\circ} C$,even when the outside temperature is extremely low.
The thermal conductivity of ice is approximately $1.6 \, W m^{-1} K^{-1}$ (note: the value $16 \, W m^{-1} K^{-1}$ is often cited in simplified contexts,though the physical value is closer to $1.6 \, W m^{-1} K^{-1}$).

(A) The thermal expansion of a solid with a hole or a gap behaves as if the hole or gap were made of the same material as the solid.
When the temperature of the ring increases by $\Delta T$,every linear dimension of the ring,including the gap width $d$,expands according to the linear expansion formula $\Delta L = L \alpha \Delta T$.
Therefore,the change in the width of the gap is given by $\Delta d = d \alpha \Delta T$.
Since $\Delta T > 0$,the width of the gap increases by $d \alpha \Delta T$.

(B) For the book to remain in equilibrium,the net force acting on it must be zero.
$1$. The vertical forces acting on the book are its weight $mg$ acting downwards and the static frictional force $f$ acting upwards from the wall.
$2$. For the book not to move vertically,the frictional force must balance the weight of the book: $f = mg$.
$3$. The horizontal forces are the applied force $F$ and the normal reaction $N$ from the wall. Since there is no horizontal motion,$N = F$.
$4$. The maximum possible static friction is $f_{max} = \mu N = \mu F$. However,the actual static friction $f$ adjusts itself to balance the weight,provided $f \le f_{max}$.
$5$. Therefore,the frictional force is $mg$ and acts upwards to prevent the book from falling.

(B) Given,surface area of cube $=$ surface area of sphere.
$6a^2 = 4\pi r^2 \Rightarrow \frac{a}{r} = \sqrt{\frac{4\pi}{6}} = \sqrt{\frac{2\pi}{3}}$.
Buoyant force $F_B = V \cdot \rho_f \cdot g$. Since both are completely submerged in the same fluid,the ratio of buoyant forces is equal to the ratio of their volumes:
$\frac{(F_B)_{\text{cube}}}{(F_B)_{\text{sphere}}} = \frac{V_{\text{cube}}}{V_{\text{sphere}}} = \frac{a^3}{\frac{4}{3}\pi r^3} = \frac{3}{4\pi} \left( \frac{a}{r} \right)^3$.
Substituting $\frac{a}{r} = \sqrt{\frac{2\pi}{3}}$:
$\frac{(F_B)_{\text{cube}}}{(F_B)_{\text{sphere}}} = \frac{3}{4\pi} \left( \sqrt{\frac{2\pi}{3}} \right)^3 = \frac{3}{4\pi} \cdot \frac{2\pi}{3} \cdot \sqrt{\frac{2\pi}{3}} = \frac{1}{2} \sqrt{\frac{2\pi}{3}} = \sqrt{\frac{2\pi}{3 \cdot 4}} = \sqrt{\frac{\pi}{6}}$.
Since $\pi \approx 3.14$,$\frac{\pi}{6} < 1$,therefore $\sqrt{\frac{\pi}{6}} < 1$.
This implies $(F_B)_{\text{cube}} < (F_B)_{\text{sphere}}$.
Thus,the buoyant force is greater for the sphere.

(D) The explosion force that splits the firecracker is internal to the system,so the path of the centre of mass of the system remains unchanged.
The range of the centre of mass of the system is given by $R = \frac{u^2 \sin 2\theta}{g}$.
Here,$u = 30 \, ms^{-1}$ and the angle with the horizontal is $\theta = 90^{\circ} - 75^{\circ} = 15^{\circ}$.
Therefore,$R = \frac{30 \times 30 \times \sin(2 \times 15^{\circ})}{10} = 90 \times \sin(30^{\circ}) = 90 \times 0.5 = 45 \, m$.
So,the position ($x$-coordinate) of the centre of mass is at $45 \, m$ distance from the origin.
Using the formula for the centre of mass,$X_{CM} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$,where $m_1 = m_2 = m$ and $X_{CM} = 45 \, m$:
$45 = \frac{m(27) + m(x)}{m + m}$
$45 = \frac{27 + x}{2}$
$90 = 27 + x \Rightarrow x = 63 \, m$.
However,the first piece could also fall at $-27 \, m$ (behind the origin) if the explosion is strong enough to reverse its direction:
$45 = \frac{m(-27) + m(x)}{2m}$
$90 = -27 + x \Rightarrow x = 117 \, m$.
Thus,the other piece will fall at $63 \, m$ or $117 \, m$ from the shooting point.

(C) Since the block is sliding down the inclined plane with a constant speed,its acceleration is zero. Therefore,the change in kinetic energy $(\Delta KE)$ of the block as it moves from height $h$ to the ground is zero.
According to the work-energy theorem,the total work done on the block is equal to the change in its kinetic energy:
$W_{\text{total}} = \Delta KE = 0$
The total work done is the sum of the work done by gravity $(W_{\text{gravity}})$ and the work done by friction $(W_{\text{friction}})$:
$W_{\text{gravity}} + W_{\text{friction}} = 0$
The work done by gravity as the block descends a vertical height $h$ is $W_{\text{gravity}} = mgh$.
Substituting this into the equation:
$mgh + W_{\text{friction}} = 0$
$W_{\text{friction}} = -mgh$
The energy dissipated by friction is the magnitude of the work done by the friction force:
$\text{Energy dissipated} = |W_{\text{friction}}| = mgh$.

(D) Let $m$ be the mass of ice that melts to reach a final equilibrium temperature of $0^{\circ} C$.
According to the principle of calorimetry,Heat lost by water = Heat gained by ice.
Heat lost by $100 \,g$ of water cooling from $80^{\circ} C$ to $0^{\circ} C$ is $Q_{lost} = m_w \cdot s_w \cdot \Delta T = 100 \times 1 \times (80 - 0) = 8000 \,cal$.
Heat required to melt $m$ grams of ice at $0^{\circ} C$ is $Q_{gained} = m \cdot L_f = m \times 80$.
Equating the two: $80m = 8000$.
Solving for $m$: $m = 100 \,g$.
Since the total mass of ice is $150 \,g$,the amount of ice that does not melt is $150 \,g - 100 \,g = 50 \,g$.

(B) The magnetic field at the centre due to a straight wire of length $L$ carrying current $i$ at a perpendicular distance $r$ is given by $B = \frac{{{\mu _0}i}}{{4\pi r}}(\sin \phi_1 + \sin \phi_2)$.
For one side of the regular polygon,the angle subtended by half the side at the centre is $\theta = \frac{\pi}{n}$.
The perpendicular distance from the centre to the side is $r = a \cos \theta = a \cos(\frac{\pi}{n})$.
The angles at the ends of the side are $\phi_1 = \phi_2 = \theta = \frac{\pi}{n}$.
Thus,the magnetic field due to one side is $B_1 = \frac{{{\mu _0}i}}{{4\pi (a \cos \theta)}} (2 \sin \theta) = \frac{{{\mu _0}i}}{{2\pi a}} \tan \theta = \frac{{{\mu _0}i}}{{2\pi a}} \tan(\frac{\pi}{n})$.
Since there are $n$ such sides,the net magnetic field at the centre is $B_{net} = n \times B_1 = \frac{{n{\mu _0}i}}{{2\pi a}} \tan(\frac{\pi}{n})$.

(D) The energy stored in a capacitor is given by $U = \frac{q^2}{2C}$.
During the discharging of an $RC$ circuit,the charge on the capacitor at any time $t$ is given by $q = q_0 e^{-t / RC}$.
Substituting this into the energy formula,we get $U = \frac{(q_0 e^{-t / RC})^2}{2C} = \frac{q_0^2}{2C} e^{-2t / RC} = U_0 e^{-2t / RC}$,where $U_0$ is the initial energy.
We want the time $t$ when $U = \frac{U_0}{2}$.
Setting the equations equal: $\frac{U_0}{2} = U_0 e^{-2t / RC}$.
Dividing by $U_0$: $\frac{1}{2} = e^{-2t / RC}$.
Taking the natural logarithm on both sides: $\ln(1/2) = -2t / RC$.
Since $\ln(1/2) = -\ln 2$,we have $-\ln 2 = -2t / RC$.
Solving for $t$: $t = \frac{RC \ln 2}{2}$.

(A) In a Young's double slit experiment,the resultant intensity $I$ at any point on the screen is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$,where $I_1 = I_2 = I_0$.
Thus,$I = 2 I_0 + 2 I_0 \cos \phi = 4 I_0 \cos^2(\phi / 2)$.
$1$. The minimum intensity occurs when $\cos^2(\phi / 2) = 0$,so $I_{\min} = 0$.
$2$. The maximum intensity occurs when $\cos^2(\phi / 2) = 1$,so $I_{\max} = 4 I_0$.
$3$. The average intensity $I_{\text{avg}}$ over the entire screen is the average of $I_{\max}$ and $I_{\min}$ for a sinusoidal interference pattern,which is $I_{\text{avg}} = \frac{I_{\max} + I_{\min}}{2} = \frac{4 I_0 + 0}{2} = 2 I_0$.
Therefore,the values are $0, 4 I_0, 2 I_0$.

(A) When the conducting rod moves with velocity $v$ in a uniform magnetic field $B$,a motional $emf$ $\varepsilon = Blv$ is induced across the rod.
This $emf$ drives an induced current $I = \frac{\varepsilon}{R} = \frac{Blv}{R}$ through the circuit.
The current-carrying rod in the magnetic field experiences a magnetic force $F_m = IlB = \frac{B^2l^2v}{R}$ acting in the direction opposite to the velocity $v$.
This force causes the rod to decelerate. As the rod moves,the kinetic energy is dissipated as heat in the resistor due to the Joule heating effect $(P = I^2R)$.
Since there is no other external force or energy storage mechanism (like a capacitor or inductor),the entire initial kinetic energy $\frac{1}{2} m v_0^2$ will eventually be dissipated as heat in the resistor $R$ as the rod comes to rest.

(A) The rate of heat generated in the wire is given by $Q_2 = I^2 R$. Since resistance $R = \frac{\rho L}{\pi r^2}$,we have $Q_2 = \frac{I^2 \rho L}{\pi r^2}$.
The rate of heat loss $Q_1$ is proportional to the surface area $A = 2 \pi r L$. Thus,$Q_1 = k (2 \pi r L) \Delta T$,where $k$ is a constant and $\Delta T$ is the temperature difference.
In a steady state,the rate of heat generated equals the rate of heat loss: $Q_2 = Q_1$.
$\frac{I^2 \rho L}{\pi r^2} = k (2 \pi r L) \Delta T$.
Solving for $\Delta T$: $\Delta T = \frac{I^2 \rho L}{\pi r^2 \cdot 2 \pi r L k} = \frac{I^2 \rho}{2 \pi^2 r^3 k}$.
Since the length $L$ cancels out from both sides,the steady-state temperature difference $\Delta T$ is independent of the length $L$ of the wire.

(A) The electrostatic potential energy $E$ of a uniformly charged spherical shell of radius $R$ and charge $e$ is given by $E = \frac{e^2}{8 \pi \varepsilon_0 R}$.
According to Einstein's mass-energy equivalence,$E = m_e c^2$.
Equating the two expressions for energy,we get $\frac{e^2}{8 \pi \varepsilon_0 R} = m_e c^2$.
Rearranging for $R$,we have $R = \frac{e^2}{8 \pi \varepsilon_0 m_e c^2} = \frac{1}{2} \left( \frac{e^2}{4 \pi \varepsilon_0 m_e c^2} \right)$.
Substituting the given values: $R = \frac{(1.6 \times 10^{-19})^2 \times 9 \times 10^9}{2 \times 9.1 \times 10^{-31} \times (3 \times 10^8)^2}$.
$R = \frac{2.56 \times 10^{-38} \times 9 \times 10^9}{2 \times 9.1 \times 10^{-31} \times 9 \times 10^{16}} = \frac{2.56 \times 10^{-29}}{18.2 \times 10^{-15}} \approx 0.14 \times 10^{-14} \, m = 1.4 \times 10^{-15} \, m$.

(A) The decay process is given by $X \xrightarrow{T_{1/2, X} = 40000 \, yr} Y \xrightarrow{T_{1/2, Y} = 20 \, yr} Z$.
Since the number of $Y$ nuclei remains constant over time,the rate of production of $Y$ must equal the rate of decay of $Y$.
This condition is known as secular equilibrium.
Therefore,$\lambda_X N_X = \lambda_Y N_Y$.
Using the relation $\lambda = \frac{\ln 2}{T_{1/2}}$,we get $\frac{\ln 2}{T_X} N_X = \frac{\ln 2}{T_Y} N_Y$.
This simplifies to $\frac{N_X}{T_X} = \frac{N_Y}{T_Y}$.
Rearranging for $N_Y$,we get $N_Y = N_X \times \frac{T_Y}{T_X}$.
Substituting the given values: $N_Y = (4 \times 10^{20}) \times \frac{20}{40000}$.
$N_Y = (4 \times 10^{20}) \times \frac{1}{2000} = 2 \times 10^{17}$ nuclei.

(C) When an unpolarised light of intensity $I_0$ passes through the first polariser,the intensity of the transmitted light becomes $I_1 = \frac{I_0}{2}$.
Now,this polarised light passes through the second polariser,which is at an angle $\theta = 30^{\circ}$ with respect to the first one. According to Malus' Law,the intensity of the emergent light $I'$ is given by:
$I' = I_1 \cos^2 \theta$
Substituting the values:
$I' = \frac{I_0}{2} \times \cos^2 30^{\circ}$
$I' = \frac{I_0}{2} \times \left( \frac{\sqrt{3}}{2} \right)^2$
$I' = \frac{I_0}{2} \times \frac{3}{4}$
$I' = \frac{3 I_0}{8}$

(A) For a radioactive nucleus with multiple decay modes,the total decay constant $\lambda_B$ is the sum of the individual decay constants: $\lambda_B = \lambda_1 + \lambda_2$.
Since the decay constant $\lambda$ is related to the half-life $\tau$ by $\lambda = \ln(2) / \tau$,we can write the relationship as $1 / \tau_B = 1 / \tau_1 + 1 / \tau_2$.
Given that if decay mode $2$ were absent,the half-life is $\tau_1 = \tau_A / 2$.
Given that if decay mode $1$ were absent,the half-life is $\tau_2 = 3 \tau_A$.
Substituting these into the formula for the actual half-life $\tau_B$:
$1 / \tau_B = 1 / (\tau_A / 2) + 1 / (3 \tau_A)$
$1 / \tau_B = 2 / \tau_A + 1 / (3 \tau_A)$
$1 / \tau_B = (6 + 1) / (3 \tau_A) = 7 / (3 \tau_A)$
Therefore,$\tau_B = (3 / 7) \tau_A$,which implies $\tau_B / \tau_A = 3 / 7$.

(A) The energy of the emergent photo-electron is given by Einstein's photoelectric equation: $E_{max} = h\nu - \Phi_0$.
Given,incident photon energy $h\nu = 3 \,eV$ and work function $\Phi_0 = 2.3 \,eV$.
Therefore,the maximum kinetic energy $E_{max} = 3 - 2.3 = 0.7 \,eV$.
This implies that the electron will turn back when it encounters a retarding potential difference of $0.7 \,V$.
The potential difference between the two plates is $1 \,V$ over a distance of $5 \,mm$.
Assuming a uniform electric field,the potential gradient is $\frac{1 \,V}{5 \,mm} = 0.2 \,V/mm$.
The distance $d$ required to reach a potential difference of $0.7 \,V$ is given by $d = \frac{V_{stop}}{\text{gradient}} = \frac{0.7 \,V}{0.2 \,V/mm} = 3.5 \,mm$.
Thus,the photo-electron turns back after traveling $3.5 \,mm$.

(A) Since spheres $A$ and $B$ are connected,they act as a single conductor at the same potential. Let the charge on the combined system $(A+B)$ be $q$ and the charge on sphere $C$ be $Q_C$.
The potential of the system $(A+B)$ is given by:
$V_B = \frac{1}{4 \pi \varepsilon_0} \left( \frac{q}{b} + \frac{Q_C}{c} \right) = V$
Since sphere $C$ is grounded,its potential is zero:
$V_C = \frac{1}{4 \pi \varepsilon_0} \left( \frac{q}{c} + \frac{Q_C}{c} \right) = 0$
From the second equation,we get $q + Q_C = 0$,which implies $q = -Q_C$.
Substituting $q = -Q_C$ into the first equation:
$V = \frac{1}{4 \pi \varepsilon_0} \left( \frac{-Q_C}{b} + \frac{Q_C}{c} \right)$
$V = \frac{Q_C}{4 \pi \varepsilon_0} \left( \frac{1}{c} - \frac{1}{b} \right) = \frac{Q_C}{4 \pi \varepsilon_0} \left( \frac{b-c}{bc} \right)$
Solving for $Q_C$:
$Q_C = 4 \pi \varepsilon_0 V \left( \frac{bc}{b-c} \right) = -4 \pi \varepsilon_0 V \left( \frac{bc}{c-b} \right)$

(A) Let $R$ be the radius of the circular region through which outside objects are visible.
Let $\theta$ be the critical angle for the water-air interface.
Then,$\sin \theta = \frac{1}{\mu} = \frac{1}{4/3} = \frac{3}{4}$.
Now,$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\sin \theta}{\sqrt{1 - \sin^2 \theta}}$.
Substituting the value of $\sin \theta$:
$\tan \theta = \frac{3/4}{\sqrt{1 - (3/4)^2}} = \frac{3/4}{\sqrt{1 - 9/16}} = \frac{3/4}{\sqrt{7/16}} = \frac{3/4}{\sqrt{7}/4} = \frac{3}{\sqrt{7}}$.
From the geometry of the problem,the diver's eyes are at a height $h$ from the bottom,so the distance from the water surface is $(H-h)$.
Thus,$\tan \theta = \frac{R}{H-h}$.
Equating the two expressions for $\tan \theta$:
$\frac{R}{H-h} = \frac{3}{\sqrt{7}}$.
Therefore,$R = \frac{3(H-h)}{\sqrt{7}}$.

(A) To find the equivalent resistance between points $A$ and $B$,we first note that the shorting wires along $AC$ and $BD$ make the potential at $A$ equal to the potential at $C$ $(V_A = V_C)$ and the potential at $B$ equal to the potential at $D$ $(V_B = V_D)$.
By collapsing the nodes $A$ and $C$ into a single node and $B$ and $D$ into another single node,we can redraw the circuit.
In this configuration,the ten resistors are arranged such that they form two parallel branches,each consisting of five resistors in a series-parallel combination. Specifically,the symmetry of the cube allows us to simplify the network into two parallel paths,each having an equivalent resistance of $R$.
Thus,the total equivalent resistance $R_{AB}$ is given by the parallel combination of these two paths:
$R_{AB} = \frac{R \times R}{R + R} = \frac{R}{2} \, \Omega$.

(C) The correct answer is $(c)$.
In an external circuit,free electrons experience an electrostatic force in the direction opposite to the electric field because they carry a negative charge.
The electric field is always directed from higher potential to lower potential.
Therefore,electrons naturally flow from lower potential to higher potential in an external circuit to reach equilibrium.
However,inside a power source (like a battery),non-electrostatic forces (chemical or magnetic) do work to move electrons from higher potential to lower potential,maintaining the potential difference in the circuit.
Thus,electrons flow from lower to higher potential,except through power sources.

(B) According to Snell's Law,when a light ray travels from a medium with refractive index $\mu_1$ to a medium with refractive index $\mu_2$,it bends towards the normal if $\mu_2 > \mu_1$.
At point $Q$,the light ray bends towards the normal,which implies that $\mu_2 > \mu_1$.
At point $R$,the light ray bends away from the normal,which implies that $\mu_3 < \mu_2$.
Since the emergent ray $RS$ is parallel to the incident ray $PQ$,the total deviation produced by the two interfaces must be zero. This occurs when the refractive indices of the first and third media are equal,i.e.,$\mu_1 = \mu_3$.
Combining these observations,we get $\mu_1 = \mu_3 < \mu_2$.

(A) The correct option is $A$.
When a ray of light is incident on a spherical surface such that it passes through the center $C$ of the sphere,the angle of incidence $i$ is $0^\circ$ because the ray is normal to the surface.
According to Snell's law,$n_1 \sin(i) = n_2 \sin(r)$,which implies $\sin(r) = 0$,so the angle of refraction $r$ is also $0^\circ$.
Since the ray passes through the center,it remains undeviated at both the entry and exit points of the spherical drop.
Because there is no deviation or refraction at an angle,there is no dispersion of white light into its constituent colors.
Therefore,the emergent light remains white and emerges without deviating.

(A) For the final image to coincide with the object,the light rays must strike the plane mirror normally (perpendicularly).
This happens if the light rays emerging from the convex lens are parallel to the principal axis.
Light rays from an object placed at the focus of a convex lens become parallel to the principal axis after refraction.
Given the focal length of the convex lens is $f = 15 \,cm$,the object must be placed at a distance of $15 \,cm$ from the lens on the side opposite to the mirror.
When rays from the object at $15 \,cm$ pass through the lens,they become parallel to the principal axis.
These parallel rays strike the plane mirror perpendicularly and reflect back along the same path.
After passing through the lens again,they converge at the same point where the object is located,thus forming the final image at the object's position.

(D) Let the resistance of each bulb be $R$ and the voltage of each battery be $V$.
For circuit $(P)$: Three bulbs are in series with one battery. Total resistance $= 3R$. Power $P_P = \frac{V^2}{3R} \approx 0.33 \frac{V^2}{R}$.
For circuit $(Q)$: Three bulbs are in parallel with one battery. Total resistance $= R/3$. Power $P_Q = \frac{V^2}{R/3} = \frac{3V^2}{R} = 3 \frac{V^2}{R}$.
For circuit $(R)$: One bulb is connected to one battery. Total resistance $= R$. Power $P_R = \frac{V^2}{R} = 1 \frac{V^2}{R}$.
For circuit $(S)$: One bulb is connected to two batteries in series. Total voltage $= 2V$. Total resistance $= R$. Power $P_S = \frac{(2V)^2}{R} = \frac{4V^2}{R} = 4 \frac{V^2}{R}$.
Comparing the values: $0.33 \frac{V^2}{R} < 1 \frac{V^2}{R} < 3 \frac{V^2}{R} < 4 \frac{V^2}{R}$.
Therefore,the order of increasing power consumption is $P < R < Q < S$.

(B) The decay process is represented as: ${ }_{92}^{238} U \longrightarrow{ }_{84}^{214} Po + 6({ }_{2}^{4} He) + n({ }_{-1}^{0} e)$.
Applying the law of conservation of mass number:
$238 = 214 + 6(4) + n(0)$
$238 = 214 + 24 = 238$ (This is satisfied).
Applying the law of conservation of atomic number (charge):
$92 = 84 + 6(2) + n(-1)$
$92 = 84 + 12 - n$
$92 = 96 - n$
$n = 96 - 92 = 4$.
Therefore,the number of electrons emitted is $4$.

(C) The correct answer is $(c)$.
According to the Rutherford model of the atom:
$1$. The entire positive charge and nearly all the mass of the atom are concentrated in a very small region at the center called the nucleus.
$2$. The size of the nucleus $(10^{-15} \ m)$ is extremely small compared to the size of the atom $(10^{-10} \ m)$.
$3$. Electrons revolve around the nucleus in circular orbits.
Since the size of the nucleus is much smaller than the size of the atom,the statement that the size of the nucleus is comparable to the atom is incorrect.

(A) The water molecule $(H_2O)$ is a polar molecule,meaning it has a permanent electric dipole moment.
When a charged rod (whether positively or negatively charged) is brought near a neutral stream of water,it induces a redistribution of charge within the water molecules due to electrostatic induction.
Because the water molecules are polar,the end of the molecule with the opposite charge to the rod is attracted,while the end with the same charge is repelled.
Since the attracted end is closer to the rod than the repelled end,the net force is always attractive.
Therefore,the water stream will bend towards the rod regardless of whether the rod is positively or negatively charged.

(A) In case $I$ (parallel combination):
Total circuit resistance,$R_{\text{eq}, 1} = R_0 + \frac{R}{n} = \frac{n R_0 + R}{n}$.
Circuit current,$i_1 = \frac{E}{R_{\text{eq}, 1}} = \frac{n E}{n R_0 + R}$.
Power dissipated in the $n$ resistors,$P_1 = i_1^2 \cdot \left(\frac{R}{n}\right) = \left(\frac{n E}{n R_0 + R}\right)^2 \cdot \frac{R}{n} = \frac{n E^2 R}{(n R_0 + R)^2}$.
In case $II$ (series combination):
Total circuit resistance,$R_{\text{eq}, 2} = R_0 + n R$.
Circuit current,$i_2 = \frac{E}{R_{\text{eq}, 2}} = \frac{E}{R_0 + n R}$.
Power dissipated in the $n$ resistors,$P_2 = i_2^2 \cdot (n R) = \left(\frac{E}{R_0 + n R}\right)^2 \cdot n R = \frac{n E^2 R}{(R_0 + n R)^2}$.
Given $P_1 = P_2$:
$\frac{n E^2 R}{(n R_0 + R)^2} = \frac{n E^2 R}{(R_0 + n R)^2}$.
This implies $(n R_0 + R)^2 = (R_0 + n R)^2$.
Taking the square root,$n R_0 + R = R_0 + n R$ (since $R_0, R, n > 0$).
$(n - 1) R_0 = (n - 1) R$.
Since $n > 1$,we get $R_0 = R$,so $\frac{R_0}{R} = 1$.

(B) The magnetic field produced by the long wire carrying current $i$ is directed into the plane of the paper (inward) at the location of the loop.
According to Lenz's law,the induced current in the loop will oppose the change in magnetic flux linked with it.
For a clockwise current to be induced in the loop,the magnetic flux through the loop must decrease (as per the right-hand rule,a clockwise current produces an inward magnetic field,which would oppose a decrease in the existing inward flux).
The magnetic flux linked with the loop decreases when the loop is moved away from the current-carrying wire.
Looking at the diagram,moving the loop in the direction $E$ (East) increases the distance from the wire,thereby decreasing the magnetic flux through the loop. Thus,a clockwise current is induced when the loop is pulled towards $E$.