Question

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

• A. $0.6watt$
• B. $0.9watt$
• C. $1.6watt$
• D. $1.8watt$

Answer 1

Answer: (C)

$1.6watt$

Thermal Current $I_{Thermal}=\frac{\Delta T}{R}$

$\therefore I_{Thermal}\ast R=\Delta T=constant$ since temp diff between A and B is maintained constant.

where $\Delta T$ is the temp difference between the points and R is the Thermal resistance between the same points.
When $\theta ={90}^0$ there will be two resistances in parallel. The resistances are $\frac{R}{4}$ and $\frac{3R}{4}$

$R_{AB}=\frac{(\frac{R}{4}\ast \frac{3R}{4})}{\frac{R}{4}+\frac{3R}{4}}=\frac{3}{16}R$

When $\theta ={180}^0$ there will be two resistances in parallel. The resistances are $\frac{R}{2}$ and $\frac{R}{2}$

${R_{AB}}^{\prime }=\frac{(\frac{R}{2}\ast \frac{R}{2})}{\frac{R}{2}+\frac{R}{2}}=\frac{1}{4}R$

Thus, $I_{AB}{R}_{AB}={I_{AB}}^{\prime }\ast {R_{AB}}^{\prime }$

$\therefore I_{AB}\ast \frac{3}{16}R={I_{AB}}^{\prime }\ast \frac{1}{4}R$

Given ${I_{AB}}^{\prime }=1.2$, $I_{AB}=\frac{4}{3}\ast 1.2=1.6W$