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`
.

Options

A. `7/3 K`
B. `2 k`
C. `5/2 K`
D. `3 K `

Answer

A

Explanation

Heat carried increases to 2 times. Therefore, new net thermal resistance will remain half.
`therefore` `R^(') = R/2 rArr (R_(1)^(')R_(2)^('))/(R_(1)^(')+R_(2)^('))=(R_(1)R_(2))/(2(R_(1)+R_(2))`
or `((3l//KA)(l//K^(')A))/((3l//KA)+(l//K^(')A))=((3l//KA)(l//KA))/(2{(3l)/(KA)+l/(KA)})`
Solving, this we get `K^(') = 7/3K`