Solution
The heat released by the metal cylinder as it cools down to $0^\circ\text{C}$ is used to melt the ice below it. The volume of ice melted creates a hole of depth $h$.
Given:
Initial Temp of Cylinder ($T_i$) = $50^\circ\text{C}$
Specific Heat of Cylinder ($s$) = $400 \text{ J}/(\text{kg}^\circ\text{C})$
Density of Cylinder ($\rho_c$) = $9.0 \text{ g/cm}^3$ (Note: Standard density for metal is high, text implies 9.0)
Latent Heat of Ice ($L$) = $4.0 \times 10^5 \text{ J/kg}$
Density of Ice ($\rho_i$) = $0.9 \text{ g/cm}^3$
Let $A$ be the base area of the cylinder and $H$ be its total height.
Let $h$ be the depth it sinks (height of ice melted).
Heat Balance Equation:
Heat lost by Cylinder = Heat gained by Ice (Fusion)
$$ m_c s \Delta T = m_{ice} L $$
Mass = Density $\times$ Volume.
$m_c = \rho_c A H$ and $m_{ice} = \rho_i A h$.
$$ (\rho_c A H) s (T_i – 0) = (\rho_i A h) L $$
Solving for the fraction of height sunk ($h/H$):
$$ \frac{h}{H} = \frac{\rho_c s T_i}{\rho_i L} $$
Substituting values (units of density cancel out, so we can use g/cm³ directly):
$$ \frac{h}{H} = \frac{9.0 \times 400 \times 50}{0.9 \times 4.0 \times 10^5} $$
$$ \frac{h}{H} = \frac{180,000}{360,000} = \frac{1}{2} = 0.5 $$
Percentage = $0.5 \times 100\% = 50\%$
Correct Option: (c) 50%