IIT-JEE 2009 · Paper 2 · Physics Q43

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PhysicsMultiple correct+4 / -1Question 43 of 57

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)

Two identical vertical solenoids side by side, with ring A resting on top of the left one and ring B on top of the right one.

\rho_A>\rho_B

and

m_A=m_B

\rho_A<\rho_B

and

m_A=m_B

\rho_A>\rho_B

and

m_A>m_B

\rho_A<\rho_B

and

m_A<m_B

Answer and worked solution

Official answer: B, D

Impulse on a ring. The same changing flux induces the same emf in both rings; the current, and so the repulsive impulse $J$ , is inversely proportional to the resistivity. The ring rises to $h=\dfrac{J^2}{2m^2g}$ .

For $h_A>h_B$ we need $\rho_Am_A<\rho_Bm_B$ :

(B) $\rho_A<\rho_B$ , $m_A=m_B$ ✓
(D) $\rho_A<\rho_B$ , $m_A<m_B$ ✓
(A) and (C) make $\rho_Am_A>\rho_Bm_B$ .

(B), (D)

Topics:Faraday's and Lenz's laws

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IIT-JEE 2009 · Paper 2, Physics question 43