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A cylindrical rod having temperature T₁ and T₂ at its end. The rate of flow of heat is Q₁ cal/sec. If all the linear dimensions are doubled keeping temperature constant, then the rate of flow of heat Q₂ will be:

1. 4Q₁
2. 2Q₁
3. Q₁/4
4. Q₁/2

Correct answer: 4Q₁

Solution

The rate of heat flow (Q) is directly proportional to the cross-sectional area (A) and inversely proportional to the length (L) of the rod. Doubling all linear dimensions increases the cross-sectional area by a factor of 4 and the length by a factor of 2. Thus, the rate of heat flow becomes 4/2 = 2 times the original rate, making Q₂ = 4Q₁.

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