Jump Heights of Rings on Solenoids and Resistivity Relation

To find the relation between the resistivities and masses of the two rings, we analyze the electromagnetic force and motion of the rings when the current is switched on.

1. Induced EMF and Resistance:
   Since both rings $A$ and $B$ are identical in shape and size and are placed symmetrically on identical solenoids, the rate of change of magnetic flux through each ring is the same during the switching-on of current:
   $\mathcal{E} = -\frac{d\Phi}{dt}$

   The electrical resistance of a ring is given by:
   $R = \rho \frac{l}{A_c}$
   where $\rho$ is the resistivity, $l$ is the perimeter, and $A_c$ is the cross-sectional area of the ring. Since $l$ and $A_c$ are identical for both rings:
   $R \propto \rho$

2. Induced Current and Repulsive Force:
   The induced current $i$ in the ring is:
   $i = \frac{\mathcal{E}}{R} \propto \frac{1}{\rho}$

   The upward electromagnetic force exerted on the ring due to the diverging magnetic field at the end of the solenoid is given by:
   $F_{\text{mag}} = i B_r (2\pi r) \propto i \propto \frac{1}{\rho}$
   where $B_r$ is the radial component of the magnetic field and $r$ is the radius of the ring.

3. Motion and Jump Height:
   The current rises rapidly over a very short time interval $\Delta t$, imparting an upward impulse $J$ to the ring:
   $J = \int F_{\text{mag}} \, dt \propto \frac{1}{\rho}$

   The initial velocity $v$ acquired by the ring immediately after the impulse is:
   $v = \frac{J}{m} \propto \frac{1}{m\rho}$

   Neglecting air resistance, the maximum height $h$ attained by the ring is:
   $h = \frac{v^2}{2g} \propto \frac{1}{m^2 \rho^2}$

   Therefore, the jump height $h$ is inversely dependent on both the mass $m$ and the resistivity $\rho$ of the ring. A ring will jump higher if it has smaller resistivity and/or smaller mass.

4. Evaluating the Given Conditions:
   We are given that:
   $h_A > h_B$

   • Case 1: If $m_A = m_B$, then for $h_A > h_B$, ring $A$ must experience a greater upward force, which requires:
     $\rho_A < \rho_B$
     This matches option (B).

   • Case 2: If $\rho_A < \rho_B$ and $m_A < m_B$, both the lower resistivity and the smaller mass contribute to giving ring $A$ a larger acceleration and initial velocity, definitively ensuring:
     $h_A > h_B$
     This matches option (D).

   • Options (A) and (C): Both state $\rho_A > \rho_B$ with $m_A \ge m_B$, which would result in $h_A < h_B$, contradicting the given condition.

Hence, the possible relations are (B) and (D).