IIT JEE 1979 Physics Question Paper with Answer and Solution

(C) Given the equation: $t = \sqrt{x} + 3$.
Rearranging for $x$: $\sqrt{x} = t - 3$,so $x = (t - 3)^2$.
The velocity $v$ is the derivative of displacement with respect to time: $v = \frac{dx}{dt} = 2(t - 3)$.
At $t = 0 \ s$,the initial velocity $v_1 = 2(0 - 3) = -6 \ m/s$.
At $t = 6 \ s$,the final velocity $v_2 = 2(6 - 3) = 6 \ m/s$.
According to the Work-Energy Theorem,the work done $W$ is equal to the change in kinetic energy: $W = \Delta K = \frac{1}{2}m(v_2^2 - v_1^2)$.
Substituting the values: $W = \frac{1}{2}m(6^2 - (-6)^2) = \frac{1}{2}m(36 - 36) = 0 \ J$.

(B) The electric field $E$ at any point is directed along the tangent to the electric field line at that point.
If a charged particle is placed in an electric field with an initial velocity of $v = 0$,the force $F = qE$ acts on it in the direction of the electric field.
Since the force is tangent to the field line,the particle will accelerate along the field line.
If the particle has an initial velocity at an angle to the field line,the trajectory will generally be curved and not follow the field line.
Therefore,the particle travels along a line of force only if its initial velocity is zero.

(C) In steady state,capacitors act as open circuits,so no current flows through the capacitor branches. The circuit simplifies to a voltage source of $100\,V$ connected across the capacitor network.
Looking at the circuit,the $3\,\mu F$ and $3\,\mu F$ capacitors are in series,giving an equivalent capacitance of $1.5\,\mu F$. Similarly,the $1\,\mu F$ and $1\,\mu F$ capacitors are in series,giving $0.5\,\mu F$.
These two branches are in parallel with each other,and this combination is in series with the $1\,\mu F$ capacitor.
However,analyzing the potential divider: The total potential difference across the network is $100\,V$. The branch with $3\,\mu F$ and $3\,\mu F$ (equivalent $1.5\,\mu F$) and the branch with $1\,\mu F$ and $1\,\mu F$ (equivalent $0.5\,\mu F$) are connected between $A$ and $C$ through the $1\,\mu F$ capacitor.
By calculating the charge distribution: The potential difference ${V_{AB}}$ is $25\,V$ and ${V_{BC}}$ is $75\,V$.