A 2 m long wire of resistance `4 Omega` and diameter 0.64 mm is coated with plastic insulation of thickness 0.66 mm. A current of 5A flows through the wire. Find the temperature difference across the insulation in the steady state. Thermal conductivity of plastic is `0.16 xx (10^-2) cal//s cm.^@C`.

Explanation

Consider a concenric cylindrical shell of radius r and thickness dr as shown in figure. The radial rate of flow of heat through this shell in steady state will be
`H=(dQ)/(dt)=-KA(d theta)/(dr)` Negative sign isused as with
Increase in r `theta` decreases
Now as for cylindrical shell `A=2 pi rL`
`H=-2piLK(d theta)/(dr)`
or `int_(a)^(b)(dr)/r=-(-2piK)/Hint_(theta_(1))^(theta_(2))d theta`
Which on integration and simplification gives
`H=(dQ)/(dt)=-(-2piLK(theta_(1)-theta_(2)))/("In"(b//a))`..........(1)

Here `H=(I^(2)R)/4.2=((5)^(2)xx4)/4.2=24("cal")/("sec"),L=2m=200cm`
`r_(1)=(0.64//2)mm=0.032cm` and
`r_(2)=r_(1)+d=0.032+0.006=0.038cm`
`(theta_(1)-theta_(2))=(24xx"In"(38//32))/(2xx3.14xx200xx0.16xx10^(-2))`
`implies(theta_(1)-theta_(2))=(55xx[1.57-1.50])/2=2^(@)C`