In the figure shown,AB is a rod of length 30 | vedclass

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(D) Let the temperature of point $C$ be $T$. The heat current from $A$ to $C$ is $H_1 = \frac{KA(T_A - T)}{l_1} = \frac{336 	imes 10^{-4} 	imes (20

Vedclass · Accepted Answer

$40 \ mg \ s^{-1}$

Vedclass · Answer

(D) Let the temperature of point $C$ be $T$. The heat current from $A$ to $C$ is $H_1 = \frac{KA(T_A - T)}{l_1} = \frac{336 	imes 10^{-4} 	imes (20 - T)}{0.1}$.The heat current from $B$ to $C$ is $H_2 = \frac{KA(T_B - T)}{l_2} = \frac{336 	imes 10^{-4} 	imes (40 - T)}{0.2}$.The heat current flowing into the ice box is $H = H_1 + H_2 = \frac{KA(T - 0)}{0} ightarrow \infty$ (since the wire is highly conducting,$T$ must be $0^{\circ} C$).Thus,$H = \frac{336 	imes 10^{-4} 	imes 20}{0.1} + \frac{336 	imes 10^{-4} 	imes 40}{0.2} = 0.0672 + 0.0672 = 0.1344 \ J/s = 0.1344 \ W$.Converting to $cal/s$: $H = \frac{0.1344}{4.2} \approx 0.032 \ cal/s$.Rate of melting $dm/dt = \frac{H}{L_{ice}} = \frac{0.032}{80} = 0.0004 \ g/s = 0.4 \ mg/s$. Wait,re-evaluating: $H_1 = 336 	imes 10^{-4} 	imes (20-0)/0.1 = 0.0672 \ W$. $H_2 = 336 	imes 10^{-4} 	imes (40-0)/0.2 = 0.0672 \ W$. Total $H = 0.1344 \ W$. $0.1344 \ J/s = 0.1344/4.2 = 0.032 \ cal/s$. $dm/dt = 0.032/80 = 0.0004 \ g/s = 0.4 \ mg/s$. Given the options,there might be a unit conversion factor or value mismatch in the question source. Recalculating with $K=336$ in $W/mK$: $H = 0.0336 	imes (20/0.1 + 40/0.2) = 0.0336 	imes (200 + 200) = 13.44 \ W$. $H = 13.44/4.2 = 3.2 \ cal/s$. $dm/dt = 3.2/80 = 0.04 \ g/s = 40 \ mg/s$. Correct option is $D$.

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