Q. A body of length $1m$ having cross sectional area $0.75m^{2}$ has heat flow through it at the rate of $6000\ Joule/s$. Then find the temperature difference if $K = 200\ Jm^{- 1}K^{- 1}$
A) $20℃$
B) $40℃$
C) $80℃$
D) $100℃$
Explanation
(b) $\frac{Q}{t} = \frac{KA\mathrm{\Delta}\theta}{l} \Rightarrow 6000 = \frac{200 \times 0.75 \times \mathrm{\Delta}\theta}{1}$
$\therefore\ \mathrm{\Delta}\theta = \frac{6000 \times 1}{200 \times 0.75} = 40℃$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.