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

The Correct Option is A

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]$