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Question

Physics Question on Faradays laws of induction

Two metallic rings AA and BB, identical in shape and size but having different resistivities ρAandρB\rho_A \, and \, \rho_B, are kept on top of two identical solenoids as shown in the figure. When current II is switched on in both the solenoids in identical manner, the rings AA and BB jump to heights hAandhBh_A \, and \, h_B, respectively, with hA>hBh_A > h_B. The possible relation(s) between their resistivities and their masses mAm_A and mBm_B is (are)

A

ρA>ρB\rho_A > \rho_B and mA=mBm_A = m_B

B

$\rho_A

C

ρA>ρB\rho_A >\, \rho_B and mA>mBm_A >\, m_B

D

ρA<ρB\rho_A < \rho_B and mA<mBm_A < m_B

Answer

ρA<ρB\rho_A < \rho_B and mA<mBm_A < m_B

Explanation

Solution

Induced emf e=dϕdt e = - \frac{d \phi }{dt}. For identical rings induced emf will
be same. But currents will be different. Given hA>hBh_A > h_B.
Hence, vA>vBas(h=v22g).v_A > v_B \, as \bigg( h = \frac{v^2}{2g}\bigg).
If ρA>ρB\rho_A > \rho_B, then, IA<IBI_A < I_B. In this case given condition can be
fulfilled if mA<mBm_A < m_B.
If ρA<ρB\rho_A < \rho_B, then IA>IBI_A > I_B. In this case given condition can be
fulfilled if mAmBm_A \le m_B