The two ends of a rod of length L and a uniform cross-sectional area A are kept at two temperatures T_1 and T_2(T_1>T_2) . The rate of heat transfer, dQ dt , through the rod in a steady state is given by :

Correct answer: A. dQ dt = kA (T_1-T_2) L For steady state dQ dt = kA(T_1-T_2) L

NEET 2009

NEET 2009 · Physics · Thermal Properties of Matter

The two ends of a rod of length L and a uniform cross-sectional...

The two ends of a rod of length

\mathrm{L}

and a uniform cross-sectional area A are kept at two temperatures

T_1

and

T_2\left(T_1>T_2\right)

. The rate of heat transfer,

\frac{\mathrm{dQ}}{\mathrm{dt}}

, through the rod in a steady state is given by :

A\frac{\mathrm{dQ}}{\mathrm{dt}}=\frac{\mathrm{kA} $\left(\mathrm{T}_1-\mathrm{T}_2\right)}{\mathrm{L}}$
B\frac{\mathrm{dQ}}{\mathrm{dt}}=\frac{\mathrm{kL} $\left(\mathrm{T}_1-\mathrm{T}_2\right)}{\mathrm{A}}$
C\frac{\mathrm{dQ}}{\mathrm{dt}}=\frac{\mathrm{k} $\left(\mathrm{T}_1-\mathrm{T}_2\right)}{\mathrm{LA}}$
D $\frac{\mathrm{dQ}}{\mathrm{dt}}=\mathrm{kLA}\left(\mathrm{T}_1-\mathrm{T}_2\right)$

A. \frac{\mathrm{dQ}}{\mathrm{dt}}=\frac{\mathrm{kA} $\left(\mathrm{T}_1-\mathrm{T}_2\right)}{\mathrm{L}}$

For steady state $\frac{\mathrm{dQ}}{\mathrm{dt}}=\frac{\mathrm{kA}\left(\mathrm{T}_1-\mathrm{T}_2\right)}{\mathrm{L}}