Q47 / 57 · JEE Advanced 2009 (Paper 2) · Physics

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) Figure

  • A
    $\rho_A>\rho_B$ and $m_A=m_B$
  • Correct answer
    B
    $\rho_A < \rho_B$ and $m_A=m_B$
  • C
    $\rho_A>\rho_B$ and $m_A>m_B$
  • Correct answer
    D
    $\rho_A < \rho_B$ and $m_A < m_B$

Solution

Induced emf $e=-\frac{d \phi}{d t}$. For identical rings induced emf will be same. But currents will be different. Given $h_A>h_B$. Hence, $v_A>v_B$ as $\left(h=\frac{v^2}{2 g}\right)$. 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 of $m_A \leq m_B$.

Hence, the correct answer is options: 'B', 'D'.

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