Question

$A$ and $B$ are two points on a uniform metal ring whose centre is $0$. The angle $AOB=\theta A$ and $B$ are maintained at two different constant temperatures. When $\theta =180^{\circ },$ the rate of total heat flow from $A$ to $B$ is $1.2W$. When $\theta =90^{\circ },$ this rate will be

• A. 0.6 W
• B. 0.9 W
• C. 1.6W
• D. 1.8W

Answer 1

Answer: (C)

When $\theta =180^{\circ }$, then thermal resistance of each branch is $\frac{R}{2}$.
So, net thermal resistance is $\frac{R}{4}$. Further,

Thermal current $=\frac{\Delta T}{R_{\text{net }}}$
$\Rightarrow 1.2=\frac{\Delta T}{\left[\frac{R}{4}\right]}$
$\Rightarrow \Delta T=0.3R...(i)$
When $\theta =90^{\circ }$, then thermal resistance of one branch is
$\frac{R}{4}$ and that of other is $\frac{3R}{4}$ and both
$\Rightarrow R_{\text{net }}=\frac{\left[\frac{R}{4}\right]\left[\frac{3R}{4}\right]}{\frac{R}{4}+\frac{3R}{4}}=\frac{3R}{16}$
$\Rightarrow I^{{}^{\prime }}=\frac{\Delta T}{\left[\frac{3R}{16}\right]}$
$\Rightarrow I^{{}^{\prime }}=\frac{(0.3R)16}{3R}$
$\Rightarrow I^{{}^{\prime }}=1.6W$