IIT JEE 1996 Physics Question Paper with Answer and Solution

(B) According to Kepler's third law,the square of the orbital period $(T)$ is proportional to the cube of the semi-major axis $(r)$,i.e.,$T^2 \propto r^3$.
Let $T_1 = 365$ days and $r_1$ be the present distance.
Given $r_2 = \frac{1}{2} r_1$.
Then,$\left( \frac{T_2}{T_1} \right)^2 = \left( \frac{r_2}{r_1} \right)^3 = \left( \frac{1}{2} \right)^3 = \frac{1}{8}$.
Taking the square root on both sides,$\frac{T_2}{T_1} = \sqrt{\frac{1}{8}} = \frac{1}{2\sqrt{2}}$.
Therefore,$T_2 = \frac{T_1}{2\sqrt{2}} = \frac{365}{2 \times 1.414} \approx \frac{365}{2.828} \approx 129$ days.

(A) The speed of sound in a stretched string is given by the formula $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension in the string and $\mu$ is the mass per unit length.
According to Hooke's law,the tension $T$ in a string is directly proportional to the extension $x$,so $T \propto x$.
Substituting this into the velocity formula,we get $v \propto \sqrt{x}$.
Let the initial speed be $v_1 = v$ at extension $x_1 = x$,and the new speed be $v_2$ at extension $x_2 = 1.5x$.
Then,$\frac{v_2}{v_1} = \sqrt{\frac{x_2}{x_1}} = \sqrt{\frac{1.5x}{x}} = \sqrt{1.5}$.
Calculating the value,$\sqrt{1.5} \approx 1.22$.
Therefore,$v_2 = 1.22 \,v$.

(B) The root mean square velocity $(v_{rms})$ of an ideal gas is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}.$
Since $R$ (universal gas constant) and $M$ (molar mass) are constants,we have $v_{rms} \propto \sqrt{T}.$
Given initial temperature $T_1 = 120 \ K$ and initial velocity $v_1 = v.$
Final temperature $T_2 = 480 \ K.$
Using the ratio: $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}.$
Substituting the values: $\frac{v_2}{v} = \sqrt{\frac{480}{120}} = \sqrt{4} = 2.$
Therefore,$v_2 = 2v.$

(D) In the steady state,the rate of heat flow through cube $A$ must be equal to the rate of heat flow through cube $B$.
Let $T$ be the temperature of the interface.
The rate of heat flow $H$ is given by $H = \frac{KA(T_1 - T_2)}{L}$.
Since the cubes have the same size,their cross-sectional area $A$ and length $L$ are equal.
For cube $A$: $H_A = \frac{K_A A (100 - T)}{L}$
For cube $B$: $H_B = \frac{K_B A (T - 0)}{L}$
Equating $H_A = H_B$:
$\frac{K_A A (100 - T)}{L} = \frac{K_B A (T - 0)}{L}$
$K_A (100 - T) = K_B T$
$300(100 - T) = 200T$
$3(100 - T) = 2T$
$300 - 3T = 2T$
$5T = 300$
$T = 60^{\circ}C$
Thus,the temperature of the interface is $60^{\circ}C$.

(C) For a cylindrical capacitor with inner radius $a$ and outer radius $b$,the electric field $E$ at a distance $r$ (where $a < r < b$) from the axis is determined using Gauss's Law.
Consider a Gaussian surface in the form of a cylinder of radius $r$ and length $L$.
The total charge enclosed is $q = \lambda L$,where $\lambda$ is the linear charge density.
According to Gauss's Law,$\oint E \cdot dA = \frac{q_{enclosed}}{\varepsilon_0}$.
$E(2\pi r L) = \frac{\lambda L}{\varepsilon_0}$.
Thus,$E = \frac{\lambda}{2\pi \varepsilon_0 r}$.
This shows that the magnitude of the electric field $E$ varies inversely with the distance $r$ from the axis,i.e.,$E \propto 1/r$.

(D) The electric field lines must be perpendicular to the surface of a conductor at every point. Inside a metallic conductor,the electric field is zero. Therefore,the field lines cannot pass through the sphere; they must terminate on the surface and emerge from the other side,always maintaining a perpendicular orientation to the surface. Path $4$ correctly depicts this behavior,where the lines bend to meet the surface of the sphere normally.

(A) Since the block is metallic,the charge carriers are electrons. For a current $I$ along the positive $x$-axis,the electrons move along the negative $x$-axis,i.e.,$\overrightarrow{v} = -v\hat{i}$.
The magnetic field is along the $y$-axis,i.e.,$\overrightarrow{B} = B\hat{j}$.
The Lorentz force is given by $\overrightarrow{F} = q(\overrightarrow{v} \times \overrightarrow{B})$.
Substituting the values,$\overrightarrow{F} = (-e)(-v\hat{i} \times B\hat{j}) = evB(\hat{i} \times \hat{j}) = evB\hat{k}$.
Since the force on the electrons is in the positive $z$-direction,they accumulate on the face $ABCD$ (as per the coordinate system shown in the figure). The accumulation of negative charges on face $ABCD$ results in the lowering of its potential.

(D) The induced electromotive force $(emf)$ in a conductor moving in a magnetic field is given by the motional $emf$ formula $e = B l v$,where $l$ is the effective length of the conductor perpendicular to the velocity and the magnetic field.
For a semicircular ring of radius $R$ moving with velocity $V$ perpendicular to a uniform magnetic field $B$,the effective length $l$ between the ends $M$ and $Q$ is the diameter of the semicircle,which is $2R$.
Therefore,the induced $emf$ is $e = B(2R)V = 2RBV$.
According to Lenz's law,the induced current will oppose the change in magnetic flux. As the ring moves out of the magnetic field,the flux through the loop decreases. To oppose this,the induced current will create a magnetic field in the same direction as the external field (into the page). By the right-hand rule,this corresponds to a current flowing from $M$ to $Q$ through the ring,making $Q$ the higher potential point.

(C) The binding energy of a nucleus is given by the product of the number of nucleons and the binding energy per nucleon.
For the reactant,two deuterons $(1^2H)$ are involved. Each deuteron has $2$ nucleons.
Total binding energy of two deuterons $= 2 \times (2 \times 1.112 \, MeV) = 4 \times 1.112 \, MeV = 4.448 \, MeV$.
For the product,one $\alpha$-particle $(2^4He)$ is formed. It has $4$ nucleons.
Total binding energy of one $\alpha$-particle $= 4 \times 7.047 \, MeV = 28.188 \, MeV$.
The energy released $Q$ in the fusion reaction is the difference between the total binding energy of the products and the total binding energy of the reactants.
$Q = 28.188 \, MeV - 4.448 \, MeV = 23.74 \, MeV$.
Rounding to the nearest provided option,$Q \approx 23.8 \, MeV$.

(D) In intrinsic semiconductors,both electrons and holes act as charge carriers due to thermal excitation.
In $P-$ type semiconductors,holes are the majority charge carriers.
Therefore,holes are charge carriers in both intrinsic and $P-$ type semiconductors.
Thus,the correct option is $(d)$.

(D) concave lens is a diverging lens that always forms a virtual,erect,and diminished image for any position of the object placed in front of it.
Similarly,a convex mirror is a diverging mirror that always forms a virtual,erect,and diminished image for any position of the object placed in front of it.
Therefore,both the concave lens and the convex mirror satisfy the given condition.
Thus,the correct option is $(d)$.