Consider two rods of the 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 temperature T_1 and T_2 at their ends. If the rate of loss

Correct answer: A. K_1 A_1=K_2 A_2 Rate of conduction of heat is given as Δ Q t =(K A)/(L)(T_2-T_1) According to the question aligned & (Δ Q_1)/(Δ t)=(Δ Q_2)/(Δ t) & (K_1 A_1)/(L)(T_1-T_2)=(K_1 A_1)/(L)(T_1-T_2) & K_1 A_1=K_2 A_2 aligned

NEET 2002

Exam: NEET
Year: 2002
Subject: Physics
Chapter: Thermal Properties of Matter

NEET 2002 · Physics · Thermal Properties of Matter

Consider two rods of the same length and different specific heats...

Consider two rods of the same length and different specific heats

\left(S_1, S_2\right)

, conductivities \left(K_1, K_2\right)

and area of cross sections \left(A_1, A_2\right)

and both having temperature

T_1 and T_2

at their ends. If the rate of loss of heat due to conduction is equal then:

Options

A $K_1 A_1=K_2 A_2$
B $\frac{K_1 A_1}{S_2}=\frac{K_2 A_2}{S_1}$
C $K_2 A_1=K_1 A_2$
D $\frac{K_2 A_1}{S_2}=\frac{K_1 A_2}{S_1}$

Correct answer

A. $K_1 A_1=K_2 A_2$

Step-by-step solution

Rate of conduction of heat is given as

\frac{\Delta \mathrm{Q}}{t}=\frac{K A}{L}\left(T_2-T_1\right) According

to the question

\begin{aligned} & \frac{\Delta Q_1}{\Delta t}=\frac{\Delta Q_2}{\Delta t} \\ & \Rightarrow \frac{K_1 A_1}{L}\left(T_1-T_2\right)=\frac{K_1 A_1}{L}\left(T_1-T_2\right) \\ & \Rightarrow K_1 A_1=K_2 A_2 \end{aligned}

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