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A cylindrical metallic rod in thermal contact with two reservoirs of heat at its two ends conducts an amount of heat Q in time t. The metallic rod is melted and the material is formed into a rod of half the radius of the original rod. What is the amount of heat conducted by the new rod, when placed in thermal contact with the two reservoirs in time t?

Asked in CBSE AIPMT 2010 · Rate of conduction through a rod

Answer: (2) Q/(16)

Step-by-step solution

Given: the same material recast at half the radius, same reservoirs, same time t.

Idea: melting the rod conserves its volume, so halving the radius lengthens it; then apply the conduction law to the new shape.

New area: A'=π(r/2)²=(π r²)/4=A/4

Volume fixed: AL=A'L'⇒ L'=A/(A')L=4L

Q=(KA Δ T t)/L, so Q'=(KA' Δ T t)/(L')=(K(A/4)Δ T t)/(4L)

Q'=1/(16)×(KA Δ T t)/L=Q/(16)

Why the other options are wrong

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