Step-by-step Derivation

1. Induced EMF and Current: When the current I is switched on in the solenoids, the magnetic field B changes, leading to a change in magnetic flux Φ through each ring. According to Faraday's law of induction, this induces an electromotive force (EMF) in each ring: $\mathcal{E} = -\frac{d\Phi}{dt}$ Since the solenoids are identical and the current is switched on in an identical manner, the function Φ(t) and hence E(t) are the same for both rings A and B.

   This EMF drives an induced current i in the ring, given by Ohm's law: i = E/R, where R is the resistance of the ring. The resistance of a ring with resistivity ρ, circumference L, and cross-sectional area $A_c$ is $R = ρL/A_c$. Since the rings are identical in shape and size, L and $A_c$ are the same for both. Therefore, the resistance is directly proportional to the resistivity: R ∝ ρ.

   The induced currents in rings A and B are: $i_A = \frac{\mathcal{E}}{R_A} \propto \frac{1}{\rho_A} \quad \text{and} \quad i_B = \frac{\mathcal{E}}{R_B} \propto \frac{1}{\rho_B}$

2. Repulsive Force and Impulse: According to Lenz's law, the induced current flows in a direction that opposes the change in flux. This results in a repulsive magnetic force between the solenoid and the ring. The magnitude of this force F is proportional to the product of the current in the solenoid I(t) and the induced current in the ring i(t). $F(t) \propto i(t) \cdot I(t)$ This repulsive force acts for the short duration Δt while the current I is changing from 0 to its final value. This force imparts an impulse J to the ring. $J = \int F(t) dt \propto \int i(t) I(t) dt$ Substituting i(t) ∝ (1/ρ) dΦ/dt and Φ ∝ I(t), we get i(t) ∝ (1/ρ) dI/dt. $J \propto \int \frac{1}{\rho} \frac{dI}{dt} I(t) dt = \frac{1}{\rho} \int_{0}^{I_{final}} I dI = \frac{I_{final}^2}{2\rho}$ Thus, the impulse delivered to the ring is inversely proportional to its resistivity: J ∝ 1/ρ.

3. Kinetic Energy and Height: The impulse J gives the ring of mass m an initial upward momentum p = J, and an initial kinetic energy: $K.E. = \frac{p^2}{2m} = \frac{J^2}{2m}$ By the principle of conservation of energy, this initial kinetic energy is converted into gravitational potential energy P.E. = mgh as the ring jumps to a height h. $mgh = K.E. = \frac{J^2}{2m}$ $h = \frac{J^2}{2m^2g}$ Since g is constant, the height h is proportional to $J^2/m^2$. $h \propto \frac{J^2}{m^2}$ Substituting J ∝ 1/ρ, we get: $h \propto \frac{(1/\rho)^2}{m^2} = \frac{1}{m^2\rho^2}$

4. Applying the Given Condition: We are given that ring A jumps higher than ring B, i.e., $h_A > h_B$. Using the derived proportionality, we have: $\frac{1}{m_A^2\rho_A^2} > \frac{1}{m_B^2\rho_B^2}$ Taking the reciprocal reverses the inequality sign: $m_A^2\rho_A^2 < m_B^2\rho_B^2$ Since mass m and resistivity ρ are positive quantities, we can take the square root of both sides: $m_A\rho_A < m_B\rho_B$

5. Evaluating the Options: We now check which of the given options satisfy the condition $m_Aρ_A < m_Bρ_B$.

   • A: $ρ_A > ρ_B$ and $m_A = m_B$ If $m_A = m_B$, the condition becomes $ρ_A < ρ_B$. This contradicts the premise $ρ_A > ρ_B$. So, (A) is incorrect.

   • B: $ρ_A < ρ_B$ and $m_A = m_B$ If $m_A = m_B$, the condition becomes $ρ_A < ρ_B$. This matches the premise. So, (B) is a possible relation.

   • C: $ρ_A > ρ_B$ and $m_A > m_B$ Here, $m_A > m_B$ implies $m_A/m_B > 1$, and $ρ_A > ρ_B$ implies $ρ_A/ρ_B > 1$. Then $(m_Aρ_A)/(m_Bρ_B) = (m_A/m_B)(ρ_A/ρ_B) > 1$, which means $m_Aρ_A > m_Bρ_B$. This contradicts our derived condition. So, (C) is incorrect.

   • D: $ρ_A < ρ_B$ and $m_A < m_B$ Here, $m_A < m_B$ implies $m_A/m_B < 1$, and $ρ_A < ρ_B$ implies $ρ_A/ρ_B < 1$. Then $(m_Aρ_A)/(m_Bρ_B) = (m_A/m_B)(ρ_A/ρ_B)$. Since both factors are less than 1, their product is also less than 1. This means $m_Aρ_A < m_Bρ_B$. This is consistent with our derived condition. So, (D) is a possible relation.

Conclusion

The possible relations that satisfy the condition $h_A > h_B$ are given in options (B) and (D).