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 $1.2 \,W$. 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 $=\frac{\Delta T}{R_{\text {net }}}$
$\Rightarrow 1.2=\frac{\Delta T}{\left[\frac{R}{4}\right]}$
$\Rightarrow \Delta T=0.3 \,R\,\,\,...(i)$
When $\theta=90^{\circ}$, then thermal resistance of one branch is
$\frac{R}{4}$ and that of other is $\frac{3 R}{4}$ and both
$\Rightarrow \quad R_{\text {net }}=\frac{\left[\frac{R}{4}\right]\left[\frac{3 R}{4}\right]}{\frac{R}{4}+\frac{3 R}{4}}=\frac{3 R}{16}$
$\Rightarrow I^{'}=\frac{\Delta T}{\left[\frac{3 R}{16}\right]}$
$\Rightarrow I^{'}=\frac{(0.3 R) 16}{3 R}$
$\Rightarrow I^{'}=1.6 \,W$