Question

38. A cylindrical rod is used to conduct heat energy.If it conducts Q cal/sec, then what energy will itconduct if its all linear dimensions are halved?(1) Q = 20(2) Q = 2(3) Q = 4Q(4) Q = 2​

Answer 1

Answer:$Q'=\frac{Q}{2}$

Explanation:

Given

heat transfer by cylindrical rod is Q cal/s

linear dimension are halved

i.e. length and diameter is reduced to half

Heat conducted is given by

$Q=\frac{kAdT}{L}$

where k= thermal conductivity

A=cross-sectional area

dT=temperature difference

L=length over which heat is flowing

thus A changes to $\frac{A}{4}$

L to $\frac{L}{2}$

$Q'=\frac{kAdT}{2L}$

$Q'=\frac{Q}{2}$

Answer 2

Answer:

Q'=Q/2

Explanation:

please see the attachment☝️☝️☝️☝️

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