Transmission Of Heat Question 82
Question: A cylindrical rod having temperature $ T _{1} $ and $ T _{2} $ at its ends. The rate of flow of heat is $ Q _{1} $ cal/sec. If all the linear dimensions are doubled keeping temperature constant then rate of flow of heat $ Q _{2} $ will be
[CBSE PMT 2001]
Options:
A) $ 4Q _{1} $
B) $ 2Q _{1} $
C) $ \frac{Q _{1}}{4} $
D) $ \frac{Q _{1}}{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
Rate of heat flow $ ( \frac{Q}{t} )=\frac{k\pi r^{2}({\theta _{1}}-{\theta _{2}})}{L}\propto \frac{r^{2}}{L} $ \ $ \frac{Q _{1}}{Q _{2}}={{( \frac{r _{1}}{r _{2}} )}^{2}}( \frac{l _{2}}{l _{1}} )={{( \frac{1}{2} )}^{2}}\times ( \frac{2}{1} )=\frac{1}{2} $
Therefore $ Q _{2}=2Q _{1} $
BETA
