Question

The ratio of two radii of 2 cylindrical rods of same material is 2:1 and ratio of their lengths is 23 if the rate of flow of heat in the longer rod is 2 cal sec inverse then that in the shorter rod will be

Answer 1

we know, rate of heat flow , P = KA dT/l

where K is thermal conductivity, A is cross sectional area , dT is temperature change and l is length of rod.

here, ratio of radii of two cylindrical rods = 2 : 1

so, ratio of their cross sectional areas, $A_1:A_2$ = (2)² : 1² = 4 : 1 [ because we know, cross sectional area , A= πr² ]

ratio of their lengths, $l_1:l_2$ = 2 : 3

now, $\frac{P_1}{P_2}=\frac{\frac{kA_1dT}{l_1}}{\frac{kA_2dT}{l_2}}$

$\frac{P_1}{P_2}=\frac{A_1}{A_2}\frac{l_2}{l_1}$

or, $\frac{P_1}{P_2}=\frac{4}{1}\frac{3}{2}=\frac{6}{1}$

given, rate of flow of heat in longer rod is 2cal/s

and we know, longer rod is l2

so, P2 = 2cal/s

then, P1/2cal = 6/1

or, P1 = 12 cal/s

hence, rate of flow of heat in shorter rod is 12 cal/s