JEE Mains · Physics · STD 11 - 10.2 transmission of heat
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 :-
- A \(\frac{4 \pi {Kr}_{1} {r}_{2}\left(\theta_{2}-\theta_{1}\right)}{{r}_{2}-{r}_{1}}\)
- B \(\frac{\pi{r}_{1} {r}_{2}\left(\theta_{2}-\theta_{1}\right)}{{r}_{2}-{r}_{1}}\)
- C \(\frac{{K}\left(\theta_{2}-\theta_{1}\right)}{{r}_{2}-{r}_{1}}\)
- D \(\frac{{K}\left(\theta_{2}-\theta_{1}\right)\left({r}_{2}-{r}_{1}\right)}{4 \pi {r}_{1} {r}_{2}}\)
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 \({d} {R}=\frac{{dr}}{{K}\left(4 \pi {r}^{2}\right)}\) \({R}=\int_{{r}_{1}}^{{r}_{2}} \frac{{dr}}{{K}\left(4 \pi {r}^{2}\right)}\)…
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