1. Tardigrade
  2. Question
  3. Physics
  4. Consider two rods of same length and different specific heats (S1, S2) , conductivities (K1, K2) and area of cross-sections (A1 , A2) and both having temperatures T1 and T2 at their ends. If rate of loss of heat due to conduction is equal, then :-

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Q. 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 :-

AIPMTAIPMT 2002

Solution:

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]$
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