A conducting rod $pq$ is sliding on two parallel conducting rods with constant velocity in a uniform magnetic field of induction $B=2 \mathrm{Wb} {\mathrm{m}}^{-2}$ as shown in the figure. The rods are connected with a circuit. What should be velocity of rod $pq$ (in $\mathrm{m} {\mathrm{s}}^{-1}$), if current in $2 \mathrm{\Omega }$ resistor is $0.1 \mathrm{A}?$

In the circuit shown in the figure, $S_1$ remains closed for a long time and $S_2$ remains open. Now $S_{\mathit{2}}$ is closed and $S_1$ is opened simultaneously. Find out the $\frac{di}{dt}$ $\left(\mathrm{in} \mathrm{Amp} {\sec }^{-1}\right)$ just after that moment

The inductors of two $LR$ circuits are placed next to each other, as shown in the figure. The values of the self-inductance of the inductors, resistances, mutual-inductance and applied voltages are specified in the given circuit. After both the switches are closed simultaneously, then total work done by the batteries against the induced $EMF$ in the inductors by the time the currents reach their steady state values is______________$\mathrm{mJ}$