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A solid cylinder of radius $r_1=2.5 \mathrm{~cm}$, length $l_1=5.0 \mathrm{~cm}$ and temperature $40^{\circ} \mathrm{C}$ is suspended in an environment of temperature $60^{\circ} \mathrm{C}$. The thermal radiation transfer rate for cylinder is $1.0 \mathrm{~W}$. If the cylinder is stretched until its radius becomes $r_2=0.50 \mathrm{~cm}$, the thermal radiation transfer rate is changed to
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$3.35 \mathrm{~W}$
Keeping nature of surface and temperature of body and surroundings same, rate of heat transfer by radiation depends on area of body,
$\Rightarrow$ Rate of heat transfer $\propto$ Surface area of body
$\Rightarrow \quad \frac{Q_2}{Q_1}=\frac{S_2}{S_1} \Rightarrow Q_2=Q_1 \times \frac{S_2}{S_1}$ $\ldots(i)$
Now, volume remains same in stretching, so
$\pi r_1^2 l_1=\pi r_2^2 l_2$
$\Rightarrow \quad l_2=\frac{r_1^2}{r_2^2} \cdot l_1=\frac{(2.5)^2}{(0.5)^2} \times 5=125 \mathrm{~cm}$
So, area before stretching, $S_1=2 \pi r_1\left(l_1+r_1\right)$ and area after stretching, $S_2=2 \pi r_2\left(l_2+r_2\right)$ Hence by Eq. (i), we get
$Q_2=Q_1 \times \frac{S_2}{S_1}=Q_1 \times \frac{2 \pi r_2\left(l_2+r_2\right)}{2 \pi r_1\left(l_1+r_1\right)}$
$=Q_1 \times \frac{r_2\left(l_2+r_2\right)}{r_1\left(l_1+r_1\right)}=\frac{1 \times 0.5 \times 10^{-2}(125+0.5) \times 10^{-2}}{2.5 \times 10^{-2}(5+2.5) \times 10^{-2}}$
$=3.346=3.35 \mathrm{~W}$
$\Rightarrow$ Rate of heat transfer $\propto$ Surface area of body
$\Rightarrow \quad \frac{Q_2}{Q_1}=\frac{S_2}{S_1} \Rightarrow Q_2=Q_1 \times \frac{S_2}{S_1}$ $\ldots(i)$
Now, volume remains same in stretching, so
$\pi r_1^2 l_1=\pi r_2^2 l_2$
$\Rightarrow \quad l_2=\frac{r_1^2}{r_2^2} \cdot l_1=\frac{(2.5)^2}{(0.5)^2} \times 5=125 \mathrm{~cm}$
So, area before stretching, $S_1=2 \pi r_1\left(l_1+r_1\right)$ and area after stretching, $S_2=2 \pi r_2\left(l_2+r_2\right)$ Hence by Eq. (i), we get
$Q_2=Q_1 \times \frac{S_2}{S_1}=Q_1 \times \frac{2 \pi r_2\left(l_2+r_2\right)}{2 \pi r_1\left(l_1+r_1\right)}$
$=Q_1 \times \frac{r_2\left(l_2+r_2\right)}{r_1\left(l_1+r_1\right)}=\frac{1 \times 0.5 \times 10^{-2}(125+0.5) \times 10^{-2}}{2.5 \times 10^{-2}(5+2.5) \times 10^{-2}}$
$=3.346=3.35 \mathrm{~W}$
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