QJEEAdv09Ph5fd0: JEE Advanced 2009 Physics - Class 12 JEEAdv09Ph5fd0 Solution: Two metallic rings A and B, identical in shape and size but…

Induced emf is same in both the rings:

$I = {e \over R} = {{\rho A} \over {\rho l}}$

$I \propto {1 \over \rho } \Rightarrow q \propto {1 \over \rho }$ .... (1)

Impulse is

$J = \int {Bil\,dt = mv = Bl\int {I\,dl = mv} }$

That is, $J = Blq = mv \Rightarrow v\left( {{q \over m}} \right)$ ..... (2)

and ${v^2} \propto h$ ...... (3)

From Eqs. (1), (2) and (3), we get

$mv \propto {1 \over \rho }$

$m\sqrt h \propto {1 \over \rho }$

$m\rho \propto {1 \over {\sqrt h }}$

Since {h_A} > {h_B} and for ${m_A} = {m_B},\rho \sqrt h =$ Constant. Therefore, {\rho _A} < {\rho _B} . Also if {m_A} < {m_B} and {\rho _A} < {\rho _B} ,

{m_A}{\rho _A} < {m_B}{\rho _B}

\Rightarrow {h_A} > {h_B} (already given)