IIT JEE 1993 Physics Question Paper with Answer and Solution

(D) The particle starts from rest at $x = 0$ ($v = 0$ at $t = 0$) and comes to rest at $x = 1$ ($v = 0$ at $t = 1$).
Since the particle moves from $x = 0$ to $x = 1$,its average velocity is positive. For the particle to start from rest and end at rest,it must accelerate initially to gain velocity and decelerate later to come to a stop.
If $\alpha$ remained positive for the entire interval $0 \le t \le 1$,the velocity would be strictly increasing,meaning the particle could not return to rest at $t = 1$.
Therefore,$\alpha$ cannot remain positive for all $t$ in the interval $0 \le t \le 1$,which implies $\alpha$ must change sign during the motion.
Thus,both statements $(a)$ and $(c)$ are correct.

(A) For particle $Q$, the velocity is constant at $v$ along the horizontal direction, so the time taken to cover the horizontal distance $AB$ is ${t_Q} = \frac{AB}{v}$.
For particle $P$, the horizontal component of velocity $v_x$ is given by $v_x = v \cos \theta$, where $\theta$ is the angle the velocity vector makes with the horizontal. As the particle slides down the bowl, it gains speed due to gravity. The horizontal component of its velocity $v_x$ at any point is greater than or equal to the initial horizontal component $v$ because the particle accelerates as it moves towards the lowest point $C$ and then decelerates, but its speed remains higher than the initial speed $v$ throughout the path until it reaches $B$. Since the horizontal component of velocity for $P$ is always greater than or equal to $v$ throughout the motion, the average horizontal velocity of $P$ is greater than $v$. Therefore, the time taken by $P$ to cover the same horizontal distance $AB$ is less than the time taken by $Q$. Thus, ${t_P} < {t_Q}$.

(D) The gravitational potential $V$ at any point $P$ due to the solid sphere with cavities is given by the principle of superposition: $V = V_{\text{large sphere}} - V_{\text{cavity A}} - V_{\text{cavity B}}$.
$1$. At the origin $O(0, 0, 0)$,the gravitational field $\vec{E}$ due to the large sphere is zero. The fields due to the two cavities are equal in magnitude and opposite in direction $(\vec{E}_A = -\vec{E}_B)$,so the net field at the origin is zero. Thus,option $(A)$ is correct.
$2$. The gravitational potential at any point $(x, y, z)$ due to a sphere with a cavity is symmetric about the axis passing through the centres of the cavities. For the circle ${y^2} + {z^2} = R^2$ in the plane $x = 0$,the distance from any point on the circle to the centres $A(-2, 0, 0)$ and $B(2, 0, 0)$ is the same. Specifically,for any point $(0, y, z)$ on the circle,the distance to $A$ is $\sqrt{(-2-0)^2 + y^2 + z^2} = \sqrt{4 + R^2}$ and the distance to $B$ is $\sqrt{(2-0)^2 + y^2 + z^2} = \sqrt{4 + R^2}$. Since the distances are equal,the potentials due to the cavities are equal,and the potential due to the large sphere is constant on this circle. Thus,the total potential is constant on the circles ${y^2} + {z^2} = 4$ and ${y^2} + {z^2} = 36$. Options $(B)$ and $(C)$ are also correct.
Therefore,the correct option is $(D)$.

(D) When block $C$ collides with block $A$,they stick together (assuming perfectly inelastic collision) or transfer momentum. Since $C$ and $A$ have equal mass $m$,after the collision,$C$ stops and $A$ moves with velocity $v$.
Now,consider the system of blocks $A$ and $B$ connected by the spring. The initial momentum of the system is $mv$.
Let $V$ be the common velocity of blocks $A$ and $B$ at the moment of maximum compression. By conservation of linear momentum:
$mv = (m + m)V \Rightarrow V = \frac{v}{2}$.
At maximum compression $x$,the kinetic energy of the system is converted into potential energy of the spring and the remaining kinetic energy of the center of mass.
Using conservation of energy:
$\frac{1}{2}mv^2 = \frac{1}{2}(m+m)V^2 + \frac{1}{2}Kx^2$
$\frac{1}{2}mv^2 = \frac{1}{2}(2m)(\frac{v}{2})^2 + \frac{1}{2}Kx^2$
$\frac{1}{2}mv^2 = \frac{1}{4}mv^2 + \frac{1}{2}Kx^2$
$\frac{1}{4}mv^2 = \frac{1}{2}Kx^2 \Rightarrow x^2 = \frac{mv^2}{2K} \Rightarrow x = v\sqrt{\frac{m}{2K}}$.
At maximum compression,the kinetic energy of the $A-B$ system is:
$K.E. = \frac{1}{2}(2m)V^2 = m(\frac{v}{2})^2 = \frac{mv^2}{4}$.
Thus,both statements $(b)$ and $(c)$ are correct.

(B) The wire acts as a spring with a force constant $k_1$. From the definition of Young's modulus $Y = \frac{F/A}{\Delta L/L}$,we get $k_1 = \frac{F}{\Delta L} = \frac{YA}{L}$.
The spring constant of the given spring is $k_2 = K$.
Since the wire and the spring are in series,the equivalent spring constant $k_{eq}$ is given by $\frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2}$.
$\frac{1}{k_{eq}} = \frac{L}{YA} + \frac{1}{K} = \frac{KL + YA}{YAK}$.
Therefore,$k_{eq} = \frac{YAK}{YA + KL}$.
The time period of oscillation is $T = 2\pi \sqrt{\frac{m}{k_{eq}}} = 2\pi \sqrt{\frac{m(YA + KL)}{YAK}}$.

(B) According to Ampere's circuital law,the line integral of the magnetic field $\vec{B}$ around any closed loop is equal to $\mu_0$ times the total current $I_{\text{enclosed}}$ passing through the surface bounded by the loop: $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enclosed}}$.
For an infinitely long,thin-walled pipe carrying current $I$ along its length,any closed loop chosen inside the pipe encloses zero current $(I_{\text{enclosed}} = 0)$.
Therefore,$\oint \vec{B} \cdot d\vec{l} = 0$,which implies that the magnetic field $\vec{B}$ at any point inside the pipe is zero.

(C) The initial magnetic moment of the square loop is $\mu_1 = iA = iL^2$,where the direction is perpendicular to the plane of the loop.
When the loop is folded at the middle,it forms two smaller rectangular loops,each of area $A' = L \times (L/2) = L^2/2$.
The magnetic moment of each half is $M = iA' = iL^2/2 = \mu_1/2$.
These two halves lie in perpendicular planes. Let the magnetic moment of the horizontal half be $\overrightarrow{M_h}$ and the vertical half be $\overrightarrow{M_v}$.
Both have magnitude $M = \mu_1/2$. The resultant magnetic moment $\overrightarrow{\mu_2}$ is the vector sum of these two: $\mu_2 = \sqrt{M^2 + M^2} = M\sqrt{2}$.
Substituting $M = \mu_1/2$,we get $\mu_2 = (\mu_1/2) \times \sqrt{2} = \mu_1/\sqrt{2}$.
Therefore,the ratio $\frac{|\overrightarrow{\mu_1}|}{|\overrightarrow{\mu_2}|} = \frac{\mu_1}{\mu_1/\sqrt{2}} = \sqrt{2}$.

(C) The net reaction is: $3(_1H^2) \to _2He^4 + p + n$.
Mass defect $\Delta m = 3 \times M(H^2) - [M(He^4) + M(p) + M(n)]$
$\Delta m = 3(2.014) - [4.001 + 1.007 + 1.008] = 6.042 - 6.016 = 0.026 \, amu$.
Energy released per reaction $Q = 0.026 \times 931.5 \, MeV \approx 24.22 \, MeV$.
$Q = 24.22 \times 10^6 \times 1.6 \times 10^{-19} \, J \approx 3.875 \times 10^{-12} \, J$.
Since $3$ deuterons are used to produce $Q$ energy, energy per deuteron is $E_d = Q/3 = 1.29 \times 10^{-12} \, J$.
Total energy available $E_{total} = N \times E_d = 10^{40} \times 1.29 \times 10^{-12} = 1.29 \times 10^{28} \, J$.
Time taken $t = E_{total} / P = (1.29 \times 10^{28}) / 10^{16} = 1.29 \times 10^{12} \, s$.
Thus, the time is of the order of $10^{12} \, s$.

(D) In the positive half-cycle of the input $V(t)$, diode $D_1$ is forward-biased and $D_2$ is reverse-biased. The current flows through $R_1$ and $R$. The voltage drop across $R$ is determined by the voltage divider rule: $V_{AB, pos} = V(t) \cdot \frac{R}{R + R_1}$.
In the negative half-cycle of the input $V(t)$, diode $D_2$ is forward-biased and $D_1$ is reverse-biased. The current flows through $R_2$ and $R$. The voltage drop across $R$ is: $V_{AB, neg} = |V(t)| \cdot \frac{R}{R + R_2}$.
Since $R_1 = 100 \ \Omega$ and $R_2 = 150 \ \Omega$ are different, the peak voltage values across $R$ during the positive and negative half-cycles will be different.
Therefore, the output is not rectified, and it has different peak values during the positive and negative half-cycles.