1
A cylindrical rod has temperatures T1 and T2 at its ends. The rate of flow of heat is Q1 cal/s. If all the linear dimensions are doubled keeping the temperatures constant, then the rate of flow of heat Q2 will be
- 4Q1
- 2Q1
- Q14
- Q12
Answer
(2) 2Q1
Solution
Rate of heat flow: Q = KA(T1 - T2)L, with A = r2
Doubling the linear dimensions: r 2r A 4A; L 2L
Q2Q1 = 4A/2LA/L = 2
Q2 = 2Q1
Answer: (2)
AIPMT 2001