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Resistance and Resistors

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Question

Heat flows radially outward through a spherical shell of outside radius $R_{2}$ and inner radius $R_{1}$ . The temperature of inner surface of shell is $Δ_{1}$ and that of outer is $Δ_{2}$ . The radial distance from centre of shell where the temperature is just half way between $Δ_{1}$ and $Δ_{2}$ is :

A

\frac{R_{2} - R_{1}}{2}

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B

\frac{R_{1} R_{2}}{R_{1} + R_{2}}

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C

\frac{2 R_{1} R_{2}}{R_{1} + R_{2}}

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D

R_{1} + \frac{R_{2}}{2}

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Solution

The correct option is A $\frac{R_{2} - R_{1}}{2}$
Given: Heat flows radially outward through a spherical shell of outside radius $R_{2}$ and inner radius $R_{1}$ . The temperature of inner surface of shell is $Δ_{1}$ and that of outer is $Δ_{2}$ .
To find the radial distance from centre of shell where the temperature is just half way between $Δ_{1}$ and $Δ_{2}$
Solution:
By using the current analogy of thermal flow,
change in temperature = potential difference
And, heat flow = current
conductivity = thermal conductivity
And Resistance = thermal resistance
Here
So,
$\frac{Δ_{1} - Δ_{2}}{R_{s h e l l}} = H$ as $\frac{V_{2} - V_{1}}{R} = I$
$⟹ H = \frac{Δ_{1} - Δ_{2}}{\frac{l}{K A}} ⟹ H = \frac{K A (Δ_{1} - Δ_{2})}{R_{2} - R_{1}} . . . . . . . (i)$
as length will be difference in the radius.
Let us consider any cross sectional area, $A$
This same H current will flow through all points across the shell
Let $x$ be the distance from the center where the temperature is half, so
$\frac{((\frac{Δ_{2} + Δ_{1}}{2}) - Δ_{2})}{\frac{R_{2} - x}{K A}} = H ⟹ \frac{K A (\frac{Δ_{2} + Δ_{1} - 2 Δ_{2}}{2})}{R_{2} - x} = \frac{K A (Δ_{1} - Δ_{2})}{R_{2} - R_{1}} ⟹ \frac{(\frac{Δ_{1} - Δ_{2}}{2})}{R_{2} - x} = \frac{(Δ_{1} - Δ_{2})}{R_{2} - R_{1}} ⟹ \frac{(Δ_{1} - Δ_{2})}{2 (R_{2} - x)} = \frac{(Δ_{1} - Δ_{2})}{R_{2} - R_{1}} ⟹ R_{2} - x = \frac{R_{2} - R_{1}}{2} ⟹ x = R_{2} - \frac{R_{2} - R_{1}}{2} ⟹ x = \frac{R_{2} - R_{1}}{2}$
is the radial distance from centre of shell where the temperature is just half way between $Δ_{1}$ and $Δ_{2}$

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