Consider two rods of same length and different specific heats ( c… - Vidyadip

MCQ

Consider two rods of same length and different specific heats $\left( c _1, c _2\right)$, thermal conductivities $\left( K _1, K_2\right)$ and area of cross-sections $\left(A_1, A_2\right)$ and both having temperatures $\left(T_1, T_2\right)$ at their ends. If their rate of loss of heat due to conduction is equal, then:

A
$\frac{K_2 A_1}{c_2}=\frac{K_1 A_2}{c_1}$
B
$K _1 A_1= K _2 A_2$
C
$K _2 A_1= K _1 A_2$
D
$\frac{K_1 A_1}{c_1}=\frac{K_2 A_2}{c_2}$

✓

Answer

(b) $K_1 A_1=K_2 A_2$
Explanation: Given: $\frac{Q_1}{t}=\frac{Q_2}{t}$
or $\frac{K_1 A_1\left(T_1-T_2\right)}{l}=\frac{K_2 A_2\left(T_1-T_2\right)}{l}$
or $K _1 A_1= K _2 A_2$

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