List-I contains four conducting loops lying in the XY plane, as...

Let us analyze the induced current $i(t)$ for each loop as it rotates clockwise with time period $T$. The magnetic field $B$ is uniform in the $+z$ direction for $x > 0$. The angular velocity is $\omega = \dfrac{2\pi}{T}$. The induced EMF is $\varepsilon = -\dfrac{d\Phi}{dt} = -B \dfrac{dA}{dt}$, where $A(t)$ is the area of the loop in the region $x > 0$. The current $i(t)$ is proportional to $\dfrac{dA}{dt}$.

Loop (P):
This is a semicircle in the region $x < 0$ at $t = 0$. As it rotates clockwise, the top half of the straight edge enters the region $x > 0$. The area in $x > 0$ is a sector of angle $\theta = \omega t$.
For $0 \le t \le T/2$, the area $A(t) = \dfrac{1}{2}R^2\omega t$ increases linearly, so $\dfrac{dA}{dt}$ is a positive constant. Thus, $i(t)$ is a positive constant.
For $T/2 \le t \le T$, the loop exits the region $x > 0$, and the area decreases linearly, so $\dfrac{dA}{dt}$ is a negative constant. Thus, $i(t)$ is a negative constant.
This matches graph (3). So, P $\rightarrow$ 3.

Loop (R):
This is a single sector of angle $60^{\circ} = \dfrac{\pi}{3}$.
For $0 \le t \le T/6$, the sector enters $x > 0$. The area increases linearly, so $i(t)$ is a positive constant.
For $T/6 \le t \le T/2$, the sector is fully inside $x > 0$. The area is constant, so $i(t) = 0$.
For $T/2 \le t \le 2T/3$, the sector exits $x > 0$. The area decreases linearly, so $i(t)$ is a negative constant.
For $2T/3 \le t \le T$, the sector is fully outside $x > 0$. The area is zero, so $i(t) = 0$.
This matches graph (1). So, R $\rightarrow$ 1.

Loop (S):
This loop consists of two identical sectors arranged symmetrically but connected such that they form a figure-8 loop (crossing at the origin). The two lobes of a figure-8 loop have opposite area vectors. As the loop rotates, the rate of change of flux in the top lobe is exactly canceled by the rate of change of flux in the bottom lobe due to their symmetric entry and exit combined with the opposite sense of traversal. Therefore, the net induced EMF and current are zero at all times.
This matches graph (4). So, S $\rightarrow$ 4.

Loop (Q):
By elimination and matching the remaining options, Q corresponds to graph (2). The loop has multiple sectors that enter and exit the magnetic field at different intervals, creating alternating positive and negative pulses of current.
So, Q $\rightarrow$ 2.

The correct matching is P $\rightarrow$ 3, Q $\rightarrow$ 2, R $\rightarrow$ 1, S $\rightarrow$ 4.