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Two metallic rings A and B, identical i...

Two metallic rings ` A and B`, identical in shape and size but having different reistivities `rhoA and rhoB`, 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)`, repectively , with ` h_(A) gt h_(B)` . The possible relation(s) between their resistivities and their masses `m_(A) and m_(B)` is(are)

A

`rho_(A) gt rho_(B) and m_(A)=m_(B)`

B

`rho_(A) lt rho_(B) and m_(A)=m_(B)`

C

`rho_(A) gt rho_(B) and m_(A)gt m_(B)`

D

`rho_(A) lt rho_(B) and m_(A)lt m_(B)`

Text Solution

Verified by Experts

The correct Answer is:
B, D
Since the rings are indentical and the solenoids are identical, magnetic field `B=mu_(0) n I` and hence magnetic flux phi=BxxA (here A = area of the ring) is the same for both the rings. Hence the magnetitude of induced enf `(e)` is the same. Since `h_(A) gt h_(B)`, the force exerted on ring `A` greater than that on `B`. Hence current induced in A greater than that in `B`, i.e.
`I_(A) gt I_(B) implies e/R_(A) gt e/R_(B) implies R_(B) gt R_(A)`
Now `R=(rl)/A`. Hence `rho_(A) lt rho_(B)`. Mass `m_(A)` can be less than, equal to or greater than `m_(B)`. Hence the current choices are `(B)` and `(D)`
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