Thermal Properties of Matter - Result Question 24
24. A cylindrical metallic rod in therrnal contact with two reservoirs of heat at its two ends conducts an amount of heat $Q$ in time $t$. The metallic rod is melted and the material is formed into a rod of half the radius of the original rod. What is the amount of heat conducted by the new rod, when placed in thermal contact with the two reservoirs in time $t$ ?
[2010]
(a) $\frac{Q}{4}$
(b) $\frac{Q}{16}$
(c) $2 Q$
(d) $\frac{Q}{2}$
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Answer:
Correct Answer: 24. (b)
Solution:
- (b) The rate of heat flow is given by
$ \frac{Q}{t}=K . A \cdot \frac{\Delta T}{\ell} $
Area of Original $rod A=\pi R^{2}$;
Area of new $rod A^{\prime}=\frac{\pi R^{2}}{4}$
$ A^{\prime}=\frac{A}{4} $
Volume of original rod will be equal to the volume of new rod.
So, $A^{\prime} \ell^{\prime}=A \ell$
or,
$\therefore \pi R^{2} \ell=\pi(\frac{R}{2})^{2} \ell^{\prime}$
$\Rightarrow \frac{\ell^{\prime}}{\ell}=\frac{R^{2}}{(\frac{R^{2}}{4})}=4$
$\therefore \quad \frac{Q^{\prime}}{Q}=\frac{A^{\prime}}{A} \frac{\ell}{\ell^{\prime}}=\frac{1}{4} \cdot \frac{1}{4}=\frac{1}{16}$
$\therefore Q^{\prime}=\frac{Q}{16}$
(a) $E=\frac{S}{S_0} \sigma T^{4}=\frac{4 \pi r^{2}}{4 \pi R^{2}} \sigma T^{4}$
$=\sigma \frac{r^{2}}{R^{2}} T^{4}$
