Aring consisting of two parts ADB and ACB of same conductivity k carries an amount of heat H . ADB p

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10-2.Transmission of Heat

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Aring consisting of two parts $ADB$ and $ACB$ of same conductivity $k$ carries an amount of heat $H$. The $ADB$ part is now replaced with another metal keeping the temperatures $T_1$ and $T_2$ constant. The heat carried increases to $2H$. What $ACB$ should be the conductivity of the new$ADB$ part? Given $\frac{{ACB}}{{ADB}}= 3$

A

$\frac{7}{3} k$

B

$2 k$

C

$\frac{5}{2}k$

D

$3 k$

Solution

$H_{1}+H_{2}=\frac{K A\left(T_{1}-T_{2}\right)}{3 l}+\frac{K A\left(T_{1}-T_{2}\right)}{l}=\frac{4}{3 l} K A\left(T_{1}-T_{2}\right)$

Now, $H_{2}=2 H-H_{1}=\frac{7 K A\left(T_{1}-T_{2}\right)}{3 l}=\frac{K^{\prime} A\left(T_{1}-T_{2}\right)}{l}$

$\therefore \quad K^{\prime}=\frac{7}{3} K$

Standard 11

Physics

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