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10-2.Transmission of Heat
medium
Consider two rods of same length and different specific heats $\left(S_{1}, S_{2}\right)$, conductivities $\left(K_{1}, K_{2}\right)$ and area of cross-sections $\left(A_{1}, A_{2}\right)$ and both having temperatures $T_{1}$ and $T_{2}$ at their ends. If rate of loss of heat due to conduction is equal, then
A
${K_1}{A_2} = {K_2}{A_1}$
B
${K_1}{A_1} = {K_2}{A_2}$
C
${K_1} = {K_2}$
D
${K_1}A_1^2 = {K_2}A_2^2$
(AIPMT-2002)
Solution
(b) ${\left( {\frac{Q}{t}} \right)_1} = \frac{{{K_1}{A_1}({\theta _1} – {\theta _2})}}{l}$ and ${\left( {\frac{Q}{t}} \right)_2} = \frac{{{K_2}{A_2}({\theta _1} – {\theta _2})}}{l}$
given ${\left( {\frac{Q}{t}} \right)_1} = {\left( {\frac{Q}{t}} \right)_2}$
==> ${K_1}{A_1} = {K_2}{A_2}$
Standard 11
Physics
