Transmission Of Heat Question 339
Question: A and B are two points on a uniform metal ring whose centre is C. The angle $ ACB=\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 1.2W. When $ \theta =90^{o}, $ this rate will be
Options:
A) 0.6 W
B) 0.9 W
C) 1.6 W
D) 1.8 W
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Let R = total thermal resistance of the ring, $ \Delta T= $ difference in temperature between A and B. For $ \theta =180^{o}, $ two sections of resistance R/2 each are in parallel. Equivalent resistance $ =\frac{R}{4} $ . Rate of total heat flow $ =I _{1}=1.2=\frac{\Delta T}{r/4} $ Or $ 0.3=\frac{\Delta T}{R} $ For $ \theta =90^{o} $ two sections of resistances $ \frac{R}{4} $ and $ \frac{3R}{4} $ are in parallel. Equivalent resistance $ =\frac{(R/4)(3R/4)}{R/4+3R/4}=\frac{3R}{16} $ Rate of total heat flow $ I _{2}=\frac{\Delta T}{3R/16}W=\frac{16}{3}( \frac{\Delta T}{R} )W=\frac{16}{3}\times 0.3W=1.6W $
BETA
