Thermal Physics: Practice Problem & Solution
A cylindrical rod having temperature $T_1$ and $T_2$ at its ends. The rate of flow of heat $Q_1 text{ cal/sec}$. If all the linear dimensions are doubled keeping temperature constant, then rate of flow of heat $Q_2$ will be: (2001)
Solution Explained:
To solve this problem, we apply the core principles of Thermal Physics. Understanding the underlying formula is key to arriving at the correct answer below:
$Q_1 = frac{K A Delta T}{l} = frac{K (pi r^2) Delta T}{l}$
When dimensions are doubled, $r' = 2r$, $l' = 2l$
$Q_2 = frac{K (pi (2r)^2) Delta T}{2l} = frac{4 K (pi r^2) Delta T}{2l} = 2 Q_1$
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