Electromagnetic Induction And Alternating Current Ques 95
- Two metallic rings $A$ and $B$, identical in shape and size but having different resistivities $\rho _A$ and $\rho _B$, 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 $h _A>h _B$. The possible relation(s) between their resistivities and their masses $m _A$ and $m _B$ is (are)
(2009)
(a) $\rho _A>\rho _B$ and $m _A=m _B$
(b) $\rho _A<\rho _B$ and $m _A=m _B$
(c) $\rho _A>\rho _B$ and $m _A>m _B$
(d) $\rho _A<\rho _B$ and $m _A<m _B$
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Answer:
Correct Answer: 95.(b, d)
Solution:
- Induced emf $e=-\frac{d \varphi}{d t}$. For identical rings induced emf will be same. But currents will be different. Given $h _A>h _B$.
Hence, $v _A>v _B$ as $h=\frac{v^{2}}{2 g}$.
If $\rho _A>\rho _B$, then, $I _A<I _B$. In this case given condition can be fulfilled if $m _A<m _B$.
If $\rho _A<\rho _B$, then $I _A>I _B$. In this case given condition can be fulfilled if $m _A \leq m _B$.
BETA
