IIT JEE 2002 Physics Question Paper with Answer and Solution

(C) The acceleration of the bob of a simple pendulum in motion consists of two components:
$1$. Centripetal acceleration $(a_c)$: Directed towards the point of suspension along the string.
$2$. Tangential acceleration $(a_t)$: Directed along the tangent to the circular path of the bob.
The net acceleration $\vec a$ is the vector sum of these two components: $\vec a = \vec a_c + \vec a_t$.
Since the bob is moving in a circular arc,the resultant acceleration vector $\vec a$ must point towards the interior of the arc,specifically between the string and the tangent,as shown in option $(C)$.

(D) The relationship between force $F(x)$ and potential energy $U(x)$ is given by $F(x) = -\frac{dU}{dx}$.
Integrating this,we get $U(x) = -\int F(x) dx$.
Substituting $F(x) = -kx + ax^3$,we have $U(x) = -\int (-kx + ax^3) dx = \frac{kx^2}{2} - \frac{ax^4}{4} + C$.
Assuming $U(0) = 0$,we get $C = 0$,so $U(x) = \frac{kx^2}{2} - \frac{ax^4}{4} = \frac{x^2}{4}(2k - ax^2)$.
For $x \ge 0$,$U(x) = 0$ at $x = 0$ and $x = \sqrt{\frac{2k}{a}}$.
Between $x = 0$ and $x = \sqrt{\frac{2k}{a}}$,$U(x)$ is positive. For $x > \sqrt{\frac{2k}{a}}$,$U(x)$ becomes negative.
The graph starts at the origin,rises to a maximum,and then decreases,crossing the $x-$axis at $x = \sqrt{\frac{2k}{a}}$. This matches Graph $D$.

(C) According to Kepler's Third Law of planetary motion,the square of the time period $T$ is proportional to the cube of the orbital radius $r$,i.e.,$T^2 \propto r^3$.
For a geostationary satellite,$T_1 = 24\, h$ and $r_1 = 36000\, km$.
For a satellite orbiting near the Earth's surface,$r_2 \approx R_{\text{Earth}} = 6400\, km$.
Using the ratio formula: $\frac{T_2}{T_1} = \left( \frac{r_2}{r_1} \right)^{3/2}$.
Substituting the values: $T_2 = 24 \times \left( \frac{6400}{36000} \right)^{3/2}$.
$T_2 = 24 \times \left( \frac{64}{360} \right)^{3/2} = 24 \times \left( \frac{8}{18.97} \right)^3 \approx 24 \times (0.177)^{3/2} \approx 24 \times 0.075 \approx 1.8\, h$.
Rounding to the nearest integer,the time period is approximately $2\, hours$.

(D) Initially,the block floats with the coin. According to the law of floatation,the total weight of the block and the coin is balanced by the buoyant force,which is equal to the weight of the water displaced by the submerged part of the block. Let $M$ be the mass of the block and $m$ be the mass of the coin. The total weight is $(M+m)g$. The volume of water displaced is $V_{disp} = (M+m)/\rho_w$,where $\rho_w$ is the density of water.
When the coin falls into the water,it sinks (assuming its density is greater than water). The block now only needs to support its own weight $Mg$. The new volume of water displaced by the block is $V'_{disp} = M/\rho_w$. Since $V'_{disp} < V_{disp}$,the submerged depth $l$ of the block decreases.
Regarding $h$,the total volume of water displaced initially was $V_{disp} = (M+m)/\rho_w$. After the coin falls,the block displaces $M/\rho_w$ and the coin displaces its own volume $V_c = m/\rho_c$. The new total volume of water displaced is $V'_{total} = M/\rho_w + m/\rho_c$. Since the density of the coin $\rho_c > \rho_w$,it follows that $m/\rho_c < m/\rho_w$. Therefore,$V'_{total} < V_{disp}$. As the total volume of water displaced decreases,the water level $h$ also decreases. Thus,both $l$ and $h$ decrease.

(A) Initially,an ideal black body absorbs all the radiant energy incident on it,making it appear as the darkest object in the furnace.
According to Kirchhoff's law of radiation,a good absorber is also a good emitter. Therefore,as the black body heats up to the temperature of the furnace,it radiates the maximum amount of energy compared to other bodies at the same temperature.
Once the black body reaches thermal equilibrium with the furnace,it emits the most intense radiation,making it appear the brightest.

(B) Let $x$ be the maximum extension of the spring.
Since the block starts from rest and momentarily comes to rest at the point of maximum extension,the change in kinetic energy is zero.
According to the work-energy theorem,the total work done by all forces is zero.
The forces acting on the block are gravity ($Mg$ downwards) and the spring force ($-Kx$ upwards).
Work done by gravity = $Mgx$
Work done by spring = $-\frac{1}{2}Kx^2$
Equating the total work to zero:
$Mgx - \frac{1}{2}Kx^2 = 0$
$Mgx = \frac{1}{2}Kx^2$
Since $x \neq 0$,we can divide by $x$:
$Mg = \frac{1}{2}Kx$
$x = \frac{2Mg}{K}$

(A) The frequency of vibration of a sonometer wire is given by $n = \frac{p}{2l} \sqrt{\frac{T}{\mu}}$,where $p$ is the number of loops (antinodes),$l$ is the length,$T$ is the tension,and $\mu$ is the mass per unit length.
In the first case,$p_1 = 5$ and $T_1 = 9g$. Thus,$n = \frac{5}{2l} \sqrt{\frac{9g}{\mu}}$.
In the second case,$p_2 = 3$ and $T_2 = Mg$. Thus,$n = \frac{3}{2l} \sqrt{\frac{Mg}{\mu}}$.
Since the tuning fork is the same,the frequency $n$ remains constant. Equating the two expressions:
$\frac{5}{2l} \sqrt{\frac{9g}{\mu}} = \frac{3}{2l} \sqrt{\frac{Mg}{\mu}}$
Squaring both sides:
$25 \times 9g = 9 \times Mg$
$225g = 9Mg$
$M = \frac{225}{9} = 25 \ kg$.

(B) According to the Doppler effect,when an observer moves towards a stationary source,the observed frequency $n'$ is given by:
$n' = n \left( \frac{v + v_0}{v} \right)$
where $n$ is the source frequency,$v$ is the speed of sound,and $v_0$ is the speed of the observer.
For train $A$:
$5.5 = 5 \left( \frac{v + v_A}{v} \right) \implies 1.1 = 1 + \frac{v_A}{v} \implies \frac{v_A}{v} = 0.1$ ... $(i)$
For train $B$:
$6.0 = 5 \left( \frac{v + v_B}{v} \right) \implies 1.2 = 1 + \frac{v_B}{v} \implies \frac{v_B}{v} = 0.2$ ... $(ii)$
Dividing equation $(ii)$ by equation $(i)$:
$\frac{v_B / v}{v_A / v} = \frac{0.2}{0.1} = 2$
Therefore,the ratio of the velocity of train $B$ to that of train $A$ is $2$.

(B) The angular momentum $(L)$ of the system is conserved because there are no external torques acting on the system about the axis of rotation. Thus, $L = I\omega = \text{constant}$.
As the tortoise moves along a chord of the circular platform, its distance $(r)$ from the center of rotation changes. Specifically, the tortoise starts at the edge, moves towards the center (decreasing $r$), reaches the point on the chord closest to the center, and then moves back towards the edge (increasing $r$).
The moment of inertia $(I)$ of the system is given by $I = I_{\text{platform}} + m r^2$, where $m$ is the mass of the tortoise. As the tortoise moves towards the center, $r$ decreases, so $I$ decreases. As it moves away, $r$ increases, so $I$ increases.
Since $L = I\omega$ is constant, $\omega = L/I$. When $I$ decreases, $\omega$ increases, and when $I$ increases, $\omega$ decreases. Therefore, the angular velocity $\omega(t)$ will first increase and then decrease. The relationship between $r$ and time $t$ is $r^2 = r_{\text{min}}^2 + (vt)^2$, making $I$ a quadratic function of time, which implies that $\omega(t) = L / (I_0 + m(r_{\text{min}}^2 + v^2t^2))$ is a non-linear function of time. This corresponds to a smooth, curved graph.

(B) When the cylinder rolls up the inclined plane,its linear velocity $v$ is directed up the incline and its angular velocity $\omega$ is clockwise. Since it is moving up,its linear velocity decreases,meaning it experiences a linear deceleration. To maintain rolling without slipping,it requires an angular deceleration. The torque provided by the frictional force $f$ acting up the incline creates a counter-clockwise torque,which provides the necessary angular deceleration.
When the cylinder rolls down the inclined plane,its linear velocity $v$ is directed down the incline and its angular velocity $\omega$ is counter-clockwise. As it moves down,its linear velocity increases,meaning it experiences linear acceleration. To maintain rolling without slipping,it requires an angular acceleration. The frictional force $f$ acting up the incline creates a torque that results in the necessary angular acceleration to increase the counter-clockwise angular velocity.

(B) Initially,according to figure $(i)$,the potential energy of $Q$ is $U_i = \frac{2kqQ}{a}$ ... $(i)$
According to figure $(ii)$,when charge $Q$ is displaced by a small distance $x$,its potential energy becomes:
$U_f = kqQ \left[ \frac{1}{a+x} + \frac{1}{a-x} \right] = kqQ \left[ \frac{(a-x) + (a+x)}{a^2 - x^2} \right] = \frac{2kqQa}{a^2 - x^2}$ ... $(ii)$
Hence,the change in potential energy is:
$\Delta U = U_f - U_i = 2kqQ \left[ \frac{a}{a^2 - x^2} - \frac{1}{a} \right] = 2kqQ \left[ \frac{a^2 - (a^2 - x^2)}{a(a^2 - x^2)} \right] = \frac{2kqQx^2}{a(a^2 - x^2)}$
Since $x << a$,we can approximate $a^2 - x^2 \approx a^2$. Therefore:
$\Delta U \approx \frac{2kqQx^2}{a(a^2)} = \frac{2kqQ}{a^3} x^2$
Thus,$\Delta U \propto x^2$.

(A) The circuit exhibits symmetry about the horizontal axis passing through the central branch. The vertical resistors of resistance $2R$ are connected between nodes that are at the same potential due to this symmetry. Therefore,no current flows through these vertical resistors,and they can be removed.
After removing the vertical resistors,the circuit simplifies to three parallel branches:
$1$. The top branch has two $2R$ resistors in series,giving a total resistance of $2R + 2R = 4R$.
$2$. The middle branch has two $r$ resistors in series,giving a total resistance of $r + r = 2r$.
$3$. The bottom branch has two $2R$ resistors in series,giving a total resistance of $2R + 2R = 4R$.
Now,we have three resistors $4R$,$2r$,and $4R$ in parallel. The equivalent resistance $R_{eq}$ is given by:
$\frac{1}{R_{eq}} = \frac{1}{4R} + \frac{1}{2r} + \frac{1}{4R} = \frac{2}{4R} + \frac{1}{2r} = \frac{1}{2R} + \frac{1}{2r} = \frac{r + R}{2Rr}$
Therefore,$R_{eq} = \frac{2Rr}{R + r}$.

(D) The resistance of a bulb is given by $R = \frac{V^2}{P}$.
For bulb $B_1$ $(100\, W)$,$R_1 = \frac{V^2}{100}$.
For bulbs $B_2$ and $B_3$ $(60\, W)$,$R_2 = R_3 = \frac{V^2}{60}$.
Bulbs $B_1$ and $B_2$ are in series,and this combination is in parallel with bulb $B_3$ across the $250\, V$ source.
The power dissipated in $B_3$ is $W_3 = \frac{(250)^2}{R_3} = \frac{(250)^2}{V^2/60} = 60 \times \frac{(250)^2}{V^2}$.
The current through the series branch ($B_1$ and $B_2$) is $I = \frac{250}{R_1 + R_2} = \frac{250}{\frac{V^2}{100} + \frac{V^2}{60}} = \frac{250}{V^2(\frac{3+5}{300})} = \frac{250 \times 300}{8V^2} = \frac{75000}{8V^2} = \frac{9375}{V^2}$.
The power in $B_1$ is $W_1 = I^2 R_1 = (\frac{9375}{V^2})^2 \times \frac{V^2}{100} = \frac{9375^2}{100 V^2}$.
The power in $B_2$ is $W_2 = I^2 R_2 = (\frac{9375}{V^2})^2 \times \frac{V^2}{60} = \frac{9375^2}{60 V^2}$.
Comparing $W_1$ and $W_2$,since $R_1 < R_2$,$W_1 < W_2$. Since $B_3$ is connected directly across the source,it operates at its rated voltage,while $B_1$ and $B_2$ operate at lower voltages,thus $W_2 < W_3$. Therefore,$W_1 < W_2 < W_3$.

(A) The current $I$ flows in the negative $z$-direction,i.e.,along $-\hat{k}$.
The magnetic field $\vec{B}$ at a point $P(x, y)$ in the $xy$-plane is given by the right-hand grip rule.
The magnitude of the magnetic field at a distance $r = \sqrt{x^2 + y^2}$ from the wire is $B = \frac{\mu_0 I}{2\pi r}$.
The direction of the magnetic field is perpendicular to the position vector $\vec{r} = x\hat{i} + y\hat{j}$.
Using the cross product $\vec{B} = \frac{\mu_0 I}{2\pi r^2} (\hat{k} \times \vec{r})$,where $\hat{k}$ is the direction of current flow $(-I\hat{k})$,we get:
$\vec{B} = \frac{\mu_0 (-I)}{2\pi r^2} (\hat{k} \times (x\hat{i} + y\hat{j})) = \frac{-\mu_0 I}{2\pi (x^2 + y^2)} (x\hat{j} - y\hat{i}) = \frac{\mu_0 I (y\hat{i} - x\hat{j})}{2\pi (x^2 + y^2)}$.

(B) The particle moves in a straight line until it reaches $x = a$. Upon entering the magnetic field region $(a \le x \le b)$,the magnetic force acts as a centripetal force,causing the particle to move in a circular path of radius $r = \frac{mv}{qB}$.
For the particle to just reach the region $x > b$,the radius of the circular path must be at least equal to the width of the magnetic field region,which is $(b - a)$.
Therefore,the condition is $r \ge (b - a)$.
Substituting the expression for $r$,we get $\frac{mv}{qB} \ge (b - a)$.
Solving for $v$,we find $v \ge \frac{q(b - a)B}{m}$.
Thus,the minimum velocity required is $v_{\min} = \frac{q(b - a)B}{m}$.

(D) is the correct option.
Magnetic field lines form continuous closed loops.
Outside the magnet,the field lines are directed from the north pole to the south pole.
Inside the magnet,the field lines are directed from the south pole to the north pole,ensuring the continuity of the closed loops.

(D) When switch $S$ is closed,the current $I_P$ in loop $P$ increases from zero to a steady value. This creates an increasing magnetic flux through loop $Q$ in the direction from left to right (away from $P$). According to Lenz's law,the induced current $I_{Q_1}$ in $Q$ must oppose this change by creating a magnetic field in the opposite direction (right to left). For an observer $E$ looking at loop $Q$,a magnetic field pointing from right to left corresponds to an anticlockwise current. Thus,$I_{Q_1}$ is anticlockwise.
When switch $S$ is opened,the current $I_P$ decreases to zero. This causes a decreasing magnetic flux through loop $Q$ in the direction from left to right. According to Lenz's law,the induced current $I_{Q_2}$ in $Q$ must oppose this decrease by creating a magnetic field in the same direction (left to right). For an observer $E$,a magnetic field pointing from left to right corresponds to a clockwise current. Thus,$I_{Q_2}$ is clockwise.

(B) The induced electromotive force $(e)$ is given by Faraday's law: $e = -N \frac{d\phi}{dt} = -NA \frac{dB}{dt}$.
Since $A = \pi r_c^2$ (where $r_c$ is the coil radius),$e \propto N$.
The resistance of the wire is $R = \rho \frac{L}{A_w} = \rho \frac{N(2\pi r_c)}{\pi r^2}$,where $r$ is the wire radius.
Thus,$R \propto \frac{N}{r^2}$.
The power dissipated is $P = \frac{e^2}{R} \propto \frac{N^2}{N/r^2} = N r^2$.
Given $N_2 = 4N_1$ and $r_2 = r_1/2$,the new power $P_2$ is:
$P_2 \propto (4N_1) \times (r_1/2)^2 = 4N_1 \times (r_1^2/4) = N_1 r_1^2 = P_1$.
Therefore,the electrical power dissipated remains the same.

(A) The current $i$ is defined as the rate of flow of charge,given by the formula $i = \frac{Ne}{t}$,where $N$ is the number of electrons,$e$ is the elementary charge $(1.6 \times 10^{-19} \text{ C})$,and $t$ is the time.
To find the number of electrons striking the target per second,we rearrange the formula to solve for $\frac{N}{t}$:
$\frac{N}{t} = \frac{i}{e}$
Given $i = 3.2 \text{ mA} = 3.2 \times 10^{-3} \text{ A}$ and $e = 1.6 \times 10^{-19} \text{ C}$:
$\frac{N}{t} = \frac{3.2 \times 10^{-3}}{1.6 \times 10^{-19}}$
$\frac{N}{t} = 2 \times 10^{16} \text{ electrons/sec}$.

(C) In a gamma-decay process,the nucleus transitions from an excited state to a lower energy state by emitting a gamma-ray photon.
During this process,there is no change in the mass number $A$ or the atomic number $Z$ of the nucleus.
The process is represented as: $^A{X_Z}^* \to {\,^A}{X_Z} + \gamma$.
Therefore,option $(c)$ correctly represents this process.

(A) The law of radioactive decay states that the activity $A$ at time $t$ is given by $A = A_0 (\frac{1}{2})^n$,where $n$ is the number of half-lives.
Given that the activity decays to $\frac{1}{16}$ of its initial value,we have $\frac{A}{A_0} = \frac{1}{16}$.
Since $\frac{1}{16} = (\frac{1}{2})^4$,we find that $n = 4$.
The total time taken $t$ is given by $t = n \times T_{1/2}$,where $T_{1/2} = 100 \mu s$.
Therefore,$t = 4 \times 100 \mu s = 400 \mu s$.

(B) The second excited state corresponds to the principal quantum number $n = 3$.
According to Bohr's quantization condition,the angular momentum $l$ is given by $l = n \left( \frac{h}{2\pi} \right)$.
Since both are in the same state $n = 3$,their angular momenta are equal: $l_H = l_{Li} = 3 \left( \frac{h}{2\pi} \right)$.
The energy of an electron in a hydrogen-like atom is given by $E = -13.6 \frac{Z^2}{n^2} \text{ eV}$.
For Hydrogen $(Z_H = 1)$ and $Li^{++}$ $(Z_{Li} = 3)$,the energies are $E_H = -13.6 \frac{1^2}{3^2}$ and $E_{Li} = -13.6 \frac{3^2}{3^2}$.
Taking the magnitude,$|E_H| = \frac{13.6}{9} \text{ eV}$ and $|E_{Li}| = 13.6 \text{ eV}$.
Clearly,$|E_H| < |E_{Li}|$. Thus,option $(b)$ is correct.

(B) The line of sight of the observer remains constant,making an angle of $45^{\circ}$ with the normal in the air.
From the geometry of the setup,when the liquid is filled to a height of $2h$,the light ray from the bottom of the rod travels to the liquid surface at a distance $h$ from the vertical axis of the beaker.
The angle of incidence $\theta$ in the liquid is given by $\tan \theta = \frac{h}{2h} = 1/2$.
Thus,$\sin \theta = \frac{1}{\sqrt{1^2 + 2^2}} = \frac{1}{\sqrt{5}}$.
Applying Snell's Law at the liquid-air interface: $\mu \sin \theta = 1 \cdot \sin 45^{\circ}$.
$\mu \left( \frac{1}{\sqrt{5}} \right) = \frac{1}{\sqrt{2}}$.
Therefore,$\mu = \frac{\sqrt{5}}{\sqrt{2}} = \sqrt{\frac{5}{2}}$.

(C) The dispersion produced by a spherical surface depends on its radius of curvature.
For a lens,the total dispersion is the sum of the dispersions produced by its two surfaces.
$A$ lens will not exhibit dispersion if the dispersion produced by one surface is exactly cancelled by the other.
This occurs when the two surfaces have equal radii of curvature,with one being convex and the other concave,such that the lens acts as a parallel-sided plate for the light passing through it.
In the given options,the meniscus lens shown in option $C$ has both surfaces with the same radius of curvature $R$,where one is convex and the other is concave. Thus,it does not exhibit dispersion.

(A) When a glass plate of thickness $t$ and refractive index $\mu$ is introduced in the path of one of the beams,the path difference introduced is $\Delta x = (\mu - 1)t$.
For the intensity to remain unchanged at the position of the original central maximum,the new path difference must be an integer multiple of the wavelength,i.e.,$\Delta x = n\lambda$,where $n = 1, 2, 3, \dots$.
To find the minimum thickness $t_{\min}$,we set $n = 1$:
$(\mu - 1)t_{\min} = 1 \cdot \lambda$
Given $\mu = 1.5$,we have:
$(1.5 - 1)t_{\min} = \lambda$
$0.5 t_{\min} = \lambda$
$t_{\min} = \frac{\lambda}{0.5} = 2\lambda$.