$A$ very long solenoid of radius $R$ is carrying current $I(t) = kt e^{-\alpha t}$ $(k > 0)$,as a function of time $(t \geq 0)$. Counter-clockwise current is taken to be positive. $A$ circular conducting coil of radius $2R$ is placed in the equatorial plane of the solenoid and is concentric with the solenoid. The induced current in the outer coil as a function of time is correctly depicted by:

$A$ conducting circular loop of resistance $20 \Omega$ and cross-sectional area $20 \times 10^{-2} \,m^{2}$ is placed perpendicular to a spatially uniform magnetic field $B$, which varies with time $t$ as $B = 2 \sin(50 \pi t) \,T$. Find the net charge flowing through the loop in $20 \,ms$ starting from $t = 0$. (in $\,C$)

Shown in the figure is a circular loop of radius $r$ and resistance $R$. $A$ variable magnetic field of induction $B = B_0 e^{-t}$ is established inside the coil. If the key $K$ is closed,the electrical power developed right after closing the switch is equal to

$A$ thin conducting rod $MN$ of mass $20 \text{ g}$,length $25 \text{ cm}$ and resistance $10 \text{ }\Omega$ is held on frictionless,long,perfectly conducting vertical rails as shown in the figure. There is a uniform magnetic field $B_0 = 4 \text{ T}$ directed perpendicular to the plane of the rod-rail arrangement. The rod is released from rest at time $t = 0$ and it moves down along the rails. Assume air drag is negligible. Match each quantity in List-$I$ with an appropriate value from List-$II$,and choose the correct option. [Given: The acceleration due to gravity $g = 10 \text{ m s}^{-2}$ and $e^{-1} = 0.4$]

List-$I$	List-$II$
$(P)$ At $t = 0.2 \text{ s}$,the magnitude of the induced emf in Volt	$(1)$ $0.07$
$(Q)$ At $t = 0.2 \text{ s}$,the magnitude of the magnetic force in Newton	$(2)$ $0.144$
$(R)$ At $t = 0.2 \text{ s}$,the power dissipated as heat in Watt	$(3)$ $1.20$
$(S)$ The magnitude of terminal velocity of the rod in $\text{m s}^{-1}$	$(4)$ $0.12$
	$(5)$ $2.00$

The figure shows three circuits with identical batteries,inductors,and resistors. Rank the circuits according to the current through the battery $(i)$ just after the switch is closed and $(ii)$ a long time later,greatest first.

As per the given figure,if $\frac{ dI }{ dt } = -1 \text{ A/s}$,then the value of $V_{AB}$ at this instant will be $.......... \text{ V}$.