Two metallic rings A and B, identical in shape and size but havin… - Vidyadip

Question

Two metallic rings $\mathrm{A}$ and $\mathrm{B}$, 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 $\mathrm{A}$ and $\mathrm{B}$ 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)

$(A)$ $\rho_A>\rho_B$ 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 $m_A < m_B$

✓

Answer

$F=B_H i l$

$F \Delta t=M V$

$i=\frac{E_{\text {induced }}}{R}$

$h=\frac{v^2}{2 g}$

The horizontanl component of magnetic field due to solenoid will exert force on ring in vertical direction.

$F=B_H i(2 \pi r)$

$F \Delta t=M V$

$i=\frac{(\Delta \phi / \Delta t)}{\left(\rho \frac{(2 \pi r)}{A}\right)}$

$B_H i(2 \pi r) \Delta t=M V$

$V=\frac{B_H \Delta \phi A}{\rho M}=\frac{K}{\rho M}$

$h=\frac{V^2}{2 g}=\frac{K^2}{\rho^2 M^2}$

$h_A > h_B$

$\frac{K^2}{\rho_A^2 M_A^2}>\frac{K^2}{\rho_B^2 M_B^2}$

$\Rightarrow \rho_B M_B>\rho_A M_A$ to Using this we get

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