Question

A ring 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 should be the conductivity of the new $ADB$ part? (Given $\frac{ACB}{ADB}=3$ )

• A. $\frac{7}{3}K$
• B. $2K$
• C. $\frac{5}{2}K$
• D. $3K$

Answer 1

Answer: (A)

Let $ADB=l$ and $ACB=3l$
$\frac{H_{ACB}}{H_{ADB}}=\frac{l_{ADB}}{l_{ACB}}=\frac{1}{3}$
$H_{ACB}+H_{ADB}=H$
$H_{ACB}=\frac{H}{4}$
$H_{ADB}=\frac{3H}{4}=\frac{KA\left(T_2-T_1\right)}{l}$ ............(i)
Now the $ADB$ part is now replaced with another metal, and total heat is $2H$ .
$H_{ACB}=\frac{H}{4}$
$H_{ACB}+H{\prime }_{ADB}=2H$
In later case $H{\prime }_{ADB}=2H-H_{ACB}$
$H{\prime }_{ADB}=2H-\frac{H}{4}$
$H{\prime }_{ADB}=\frac{7H}{4}=\frac{K_{1}A\left(T_2-T_1\right)}{l}$ ............(ii)
Comparing (i) and (ii)
$K_1=\frac{7K}{3}$