question_answer

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

Correct Answer: C

Solution :

                 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\[=\frac{\Delta T}{{{R}_{net}}}\]                 \[\Rightarrow \]               \[1.2=\frac{\Delta T}{\left[ \frac{R}{4} \right]}\]                 \[\Rightarrow \]               \[\Delta T=0.3R\]                           ...(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 are in parallel                 \[\Rightarrow \]               \[{{R}_{net}}=\frac{\left[ \frac{R}{4} \right]\left[ \frac{3R}{4} \right]}{\frac{R}{4}+\frac{3R}{4}}=\frac{3R}{16}\]                 \[\Rightarrow \]               \[I'=\frac{\Delta T}{\left[ \frac{3R}{16} \right]}\]                 \[\Rightarrow \]               \[I'=\frac{(0.3R)16}{3R}\]                 \[\Rightarrow \]               \[I'=1.6\,W\]