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 :-

Question

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 :- (A)  K1 A1 = K2 A2  (B)  ( K1 A1/ S1) = (K2 A2/S2)  (C)  K2 A1 = K1 A2  (D)  (K2 A1/S2) = (K1 A2/ S1)

Tardigrade Admin · Accepted Answer

Rate of heat loss in rod 1 = Q1 = ( K1 A1 (T1- T2)/ l1) Rate of heat loss in rod 2 = Q2 = ( K2 A2 (T1- T2)/ l2) By problem, Q1 = Q2 ∴ ( K1 A1 (T1- T2)/ l1) = ( K2 A2 (T1- T2)/ l2) ∴ K1 A1 = K2 A 2 25 mm [ ∵ l1 l2] <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/faa40b4b1f0edd57ebc3103d2b73fed0-.png" />

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 :-

$K_{1} A_{1} = K_{2} A_{2}$

$\frac{K _{1} A _{1}}{S _{1}} = \frac{K _{2} A _{2}}{S _{2}}$

$K_{2} A_{1} = K_{1} A_{2}$

$\frac{K _{2} A _{1}}{S _{2}} = \frac{K _{1} A _{2}}{S _{1}}$

Q. 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 :-

K1​A1​=K2​A2​

S1​K1​A1​​=S2​K2​A2​​

K2​A1​=K1​A2​

S2​K2​A1​​=S1​K1​A2​​

Rate of heat loss in rod 1=Q1​=l1​K1​A1​(T1​−T2​)​ Rate of heat loss in rod 2=Q2​=l2​K2​A2​(T1​−T2​)​ By problem, Q1​=Q2​ ∴l1​K1​A1​(T1​−T2​)​=l2​K2​A2​(T1​−T2​)​ ∴K1​A1​=K2​A2​ 25mm [∵l1​l2​]

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 :-

$K_{1} A_{1} = K_{2} A_{2}$

$\frac{K _{1} A _{1}}{S _{1}} = \frac{K _{2} A _{2}}{S _{2}}$

$K_{2} A_{1} = K_{1} A_{2}$

$\frac{K _{2} A _{1}}{S _{2}} = \frac{K _{1} A _{2}}{S _{1}}$