First slide

Heat transfer

Question

The ends of the two rods of conductivities, radii and lengths in the ratio of 1:2 are maintained at the same temperature difference, if the rate of flow of heat through the bigger rod is ${4 cals}^{-1}$ , in shorter it will be ${(in cals}^{-1})$

Moderate

A

1
B

2
C

3
D

8

Solution

$R a t e o f f l o w o f h e a t i s P P= \frac{{KA(θ}_{1} {-θ}_{2})}{l} = \frac{{Kπr}^{2} {(θ}_{1} {-θ}_{2})}{l} h e r e K i s c o e f f i c i e n t o f t h e r m a l c o n d u c t i v i t y, A i s a r e a o f c r o s s s e c t i o n, θ_{1}, θ_{2} temperature at each end of rod l i s l e n g t h o f r o d s u b s t i t u t e g i v e n v a l u e s l e g i b l y \frac{P_{1}}{P_{2}} = \frac{K_{1}}{K_{2}} . {(\frac{r_{1}}{r_{2}})}^{2} × \frac{l_{2}}{l_{1}} = \frac{1}{2} {(\frac{1}{2})}^{2} x \frac{2}{1} g i v e n P_{2} {=4 calorie second}^{- 1} \frac{P}{4} = \frac{1}{4} r a t e o f f l o w o f h e a t t h r o u g h s h o r t r o d i s {P=1 cals}^{-1}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App