A hole of radius `r_(1)` is made centrally in a uniform circular disc of thickness d and radius `r_(2)` . The inner surface (a cylinder of length d and radius `r_(1)`) is maintained at a temperature `theta_(1)` and the outer surface (a cylinder of length d and radius `r_(2)`) is maintained at a temperature `theta_(2)(theta_(1)gttheta_(2))` .The thermal conductivity of the material of the disc is K. Calculate the heat flowing per unit time through the disc.

Answer

A, B, D

Explanation

`(dQ)/(dt)`=Rate of flow heat
Lets us consider a strip at a distance r from the centre of thickness dr.
`Q/(dt)=(Kxx2 pi xx rd xx d theta)/(dr)`
` [d theta=tempareture dofference across the thickness dr]`
`C=(Kxx2 pi xx rd xx d theta)/(dr)[c=(theta)/t]`
`cxx(dr)/r=K2 pi d d theta`
Integrating
`implies C int_(r_1)^(r_2) (dr)/r=K 2 pi d int _(theta_2)^(theta_1) d theta`
`implies C[log r]_(r_1)^(r_2)=K 2 pi d (theta_1=theta_2)`
`C(log r_2-log r_1)=K 2 pi d (theta_1=theta_2)`
`implies c=(K 2 pi d (theta_1=theta_2))/(log(r_2//r_1))`.