$A$ conducting loop having a capacitor is moving outward from the magnetic field. Which plate of the capacitor will be positive?

$A$ solenoid is oriented end-on so that its opening is perpendicular to the circuit containing the two light bulbs as drawn in figure $C_1.$ For figures $C_2$ and $C_3,$ a shorting wire of negligible resistance is added as shown. Assume that the magnetic field from the solenoid,shown coming out of the plane of the page,decreases uniformly with time at the same rate for each circuit. Rank the circuits for the brightness of the bulb labeled $R_1$ from brightest to dimmest.

$A$ magnet $NS$ is suspended from a spring and while it oscillates,the magnet moves in and out of the coil $C$. The coil is connected to a galvanometer $G$. Then as the magnet oscillates,

$A$ conducting loop of finite resistance lies on the $x-y$ plane. There is a constant magnetic field in the $z$ direction. The area of the loop varies with time $t$, as $A = A_0 (1 + \sin t)$ in appropriate units. The figure that correctly indicates the qualitative behaviour of the power $P$ dissipated in the loop as a function of time is :

The current $i$ in a coil varies with time as shown in the figure. The variation of induced $emf$ with time would be

The network shown in the figure is part of a complete circuit. If at a certain instant,the current $I$ is $5 \ A$ and it is decreasing at a rate of $10^3 \ A \ s^{-1}$,then $V_B - V_A$ equals.....$V$.