A) 2 : p
B) 1 : 2
C) p : 2
D) 3 : 2
Correct Answer: A
Solution :
\[\frac{dQ}{dt}=\frac{KA\Delta \theta }{l}\], For both rods K, A and Dq are same Þ \[\frac{dQ}{dt}\propto \frac{1}{l}\]So \[\frac{{{(dQ/dt)}_{semi\,circular}}}{{{(dQ/dt)}_{straight}}}=\frac{{{l}_{straight}}}{{{l}_{semicircular}}}=\frac{2r}{\pi \,r}=\frac{2}{\pi }\].
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