Two metallic ring A and B, identical in shape and size but having different resistivities P_A and P_B , are kept on top of two
identical solenoids as shown in the figure. When current I is switched on in both the solenoids in identical manner, the rings A and B jump to heights h_A and h_B , respectively, with $h_{A} \geq h_{B}$ . The possible relation (s) between their resistivities and their masses $m_{A} and m_{B}$ is (are)

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$p_{A} \geq p_{B} and m_{A} = m_{B}$

$p_{A} < p_{B} and m_{A} = m_{B}$

$p_{A} \geq p_{B} and m_{A} \geq m_{B}$

$p_{A} \geq p_{B} a n d m_{A} \geq m_{B}$

answer is B, D.

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Detailed Solution

When current is switched on an induced current is produced in the ring in opposite direction . So repulsive force or impulse will act on ring. This impulse (sayJ) is proportional to induced current in the loop. Hence $J = αI = ΔP = mv$ and $v = \sqrt{2 gh}$
$∴ h α \frac{I^{2}}{m^{2}}$
For identical rings induced emf will be same.
If $p_{A} < p_{B}$ , then $I_{A} > I_{B}$ . In this case given condition can be fulfilled if $m_{A} \leq m_{B} .______$