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Confused! kindly explain, - Properties of Solids and Liquids - JEE Main

The figure shows a system of two concentric spheres of radii r_{1}\: and\: r_{2} and kept at temperatures T_{1}\: and\: T_{2} respectively. The radial rate of flow of heat in a substance between the two concentric spheres is proportional to

  • Option 1)

    \frac{r_{1}r_{2}}{\left ( r_{2}-r_{1} \right )}

  • Option 2)

    \left ( r_{2}-r_{1} \right )

  • Option 3)

    \frac{\left ( r_{2}-r_{1} \right )}{r_{1}r_{2}}

  • Option 4)

    \ln \left ( \frac{r_{2}}{r_{1}} \right )

 
Answers (1)
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As we have learned

Thermal Conductivity -

Q=\frac{KA(\theta_{1}-\theta_{2})t}{l}

K = thermal conductivity

- wherein

 

 

Rate of flow of heat \frac{d\theta }{dt} = \frac{KA\Delta T}{L}

We have a spherical shell of radius r and thickness dx 

A = 4 \pi r^2 , L =dr \\ \frac{d\theta }{dt}= \frac{K4 \pi r^2 dt }{dr }

4 \pi K \int_{T_1}^{T_2}dT= \frac{d\theta }{dt} \int_{r_1}^{r_2}\frac{dr}{r^2} = \frac{d\theta }{dt} \left ( \frac{1}{r_1}-\frac{1}{r_2} \right )

4 \pi K(T_2-T_1)= \frac{d\theta }{dt} \left ( \frac{r_2r_1}{r_2-r_1} \right )

\therefore \frac{d\theta }{dt} \propto \frac{r_2r_1}{r_2-r_1}

 

 

 

 


Option 1)

\frac{r_{1}r_{2}}{\left ( r_{2}-r_{1} \right )}

Option 2)

\left ( r_{2}-r_{1} \right )

Option 3)

\frac{\left ( r_{2}-r_{1} \right )}{r_{1}r_{2}}

Option 4)

\ln \left ( \frac{r_{2}}{r_{1}} \right )

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