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JEE Mains · Physics · STD 11 - Transmission of heat
  • JEE Mains 2021
  • Hard
  • MCQ

Two thin metallic spherical shells of radii \({r}_{1}\) and \({r}_{2}\) \(\left({r}_{1}<{r}_{2}\right)\) are placed with their centres coinciding. A material of thermal conductivity \({K}\) is filled in the space between the shells. The inner shell is maintained at temperature \(\theta_{1}\) and the outer shell at temperature \(\theta_{2}\left(\theta_{1}<\theta_{2}\right)\). The rate at which heat flows radially through the material is :-

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Answer & Solution

Correct Answer

(A) \(\frac{4 \pi {Kr}_{1} {r}_{2}\left(\theta_{2}-\theta_{1}\right)}{{r}_{2}-{r}_{1}}\)

Step-by-step Solution

Detailed explanation

Thermal resistance of spherical sheet of thicleness \(dr\) and radius \(r\) is

\({d} {R}=\frac{{dr}}{{K}\left(4 \pi {r}^{2}\right)}\)

\({R}=\int_{{r}_{1}}^{{r}_{2}} \frac{{dr}}{{K}\left(4 \pi {r}^{2}\right)}\)

\({R}=\frac{1}{4 \pi {K}}\left(\frac{1}{{r}_{1}}-\frac{1}{{r}_{2}}\right)=\frac{1}{4 \pi {K}}\left(\frac{{I}_{2}-{r}_{1}}{{r}_{1} {I}_{2}}\right)\)

Thermal current (i) \(=\frac{\theta_{2}-\theta_{1}}{R}\)

\({i} =\frac{4 \pi {Kr}_{1} {r}_{2}\left(\theta_{2}-\theta_{1}\right)}{{r}_{2}-{r}_{1}}\)

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