Find the rate of heat flow through a cross section of the rod shown in figure `(theta_(2)gttheta_(1))` . Thermal conductivity of the material of the rod is K.

Answer

A, B

Explanation

`Ten Phi =(r_2-r_1)/L=(y-r_1)/x`
`Differentating w,r to x
`r_2-r_1=(Ldy)/(dx)-0`
`implies(dy)/(dx)=(r_2-r_1)/L`
`implies(r_2-r_1)/(L)dx=(dyL)/(r_2-r_1) …(i)`
`(Q/t)=(k pi y^2 d theta )/(dx)`
`Now (Qdx)/t=(k pi y^2 d theta )/(dx)`
`implies(QLdy)/(t(r-2-r_1))=(k pi y^2 Delta theta )`
[From equation (i)]`
`implies d theta=(QLdy)/(t(r_2-r_1)k pi y^2)`
Intrgrating both side
`impliesint_(theta_(1))^(theta_(2)) d theta=(QL)/(t(r_2-r_1)k pi int_(r_(2))^(r_(1)) (dy)//(y^2)`
`implies(theta_2-theta_1)=(QL)/(t(r_2-r_1)k pi) xx[(-1)/(y)]_(r_1)^(r_2)`
`implies(theta_2-theta_1)=(QL)/(t(r_2-r_1)k pi)xx[(1)/(r_1)-(1)/(r_2)]`
`implies(theta_2-theta_1)=(QL)/(t(r_2-r_1)k pi) ((r_2-r_1))/(r_2 r_1)`
`implies (Q/t)=(k pi (r_2 r_1)(theta_1-theta_2)/L`.