Question

Consider two rods of same length and different specific heats $(S_1, S_2)$ , conductivities $(K_1, K_2)$ and area of cross-sections $(A_1 , A_2)$ 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. $K_1 A_1 = K_2 A_2$
• B. $\frac{ K_1 A_1}{ S_1} = \frac{K_2 A_2}{S_2}$
• C. $K_2 A_1 = K_1 A_2$
• D. $\frac{K_2 A_1}{S_2} = \frac{K_1 A_2}{ S_1}$

Answer 1

Answer: (A)

$K_1 A_1 = K_2 A_2$

Answer 2

Rate of heat loss in rod $1 = Q_1 = \frac{ K_1 A_1 (T_1- T_2)}{ l_1}$
Rate of heat loss in rod $2 = Q_2 = \frac{ K_2 A_2 (T_1- T_2)}{ l_2}$
By problem, $Q_1 = Q_2$
$\therefore \, \frac{ K_1 A_1 (T_1- T_2)}{ l_1} = \frac{ K_2 A_2 (T_1- T_2)}{ l_2}$
$\therefore \, K_1 A_1 = K_2 A _2$
$25\,mm$ $[ \because \,l_1\,l_2]$