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A and B are two points on a uniform metal ring whose centre is C. The angle ACB` = theta`. A and B are maintained at two different constant temperatures. When `theta = 180^(@)`,the rate of total heat flow from A to B is 1.2 W.When `theta = 90^(@)`, this rate will be

A

`0.6 W`

B

`0.9 W`

C

`1.6W`

D

`1.8W`

Text Solution

Verified by Experts

The correct Answer is:
C

`R=1/((KA)/(L//2)+(KA)/(L//2))=(KA)/(4L),R'= (1)/((KA)/(3L//4) + (KA)/(L//4)) = (3)/(4)R`
`H_(1) = (DeltaT)/(R) = 1.2W`
`H_(11) = (DeltaT) / (3R//4) = (4)/(3) (DeltaT)/(R) = (4)/(3) xx 1.2 = 1.6W`
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