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A ring consisting of two parts ADB and A...

A ring consisting of two parts `ADB` and `ACB` of same conductivity k carries an amount of heat `H` The `ADB` part is now replaced with another metal keeping the temperature `T_91)` and `T_(2)` constant The heat carried increases to `2H` What should be the conductivity of the new `ADB` Given `(ACB)/(ADB)=3`
.

A

`7/3 K`

B

`2 K`

C

`5/2 K`

D

`3 K`

Text Solution

Verified by Experts

The correct Answer is:
A
`H_(1) + H_(2) = (KA(T_(1) - T_(2)) )/(3l) + (KA(T_(1) - T_(2)))/(l)`
`=4/(3l)KA(T_(1)-T_(2))`
In later case
`H_(2) = 2H - H_(1)`
`= (7KA)/(3l) (T_(1) - T_(2)) = (K'A)/(l) (T_(1) - T_(2))`
`rArr K' = 7/3 K`
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