Question

A $2 \, m$ long wire of resistance $4 \, ohm$ and diameter $0.64 \, mm$ is coated with plastic insulation of thickness $0.06 \, mm$ . When a current of $5 \, $ ampere flows through the wire, find the temperature difference across the insulation in steady-state if $K \, = \, 0.16 \, \times \, 10^{- 2} \, cal \, cm^{- 1} \, {}^\circ C^{- 1} \, s^{- 1}$ .

• A. $1 \, {}^\circ C$
• B. $2 \, {}^\circ C$
• C. $3^\circ C$
• D. $4^\circ C$

Answer 1

Answer: (B)

Consider a coaxial cylindrical shell of radius r and thickness dr as shown in diagram. The radial rate of flow of heat through this shell in steady state will be

$H =\frac{ dQ }{ dt }=- KA \frac{ d \theta}{ dr }$
[Negative sign is used as with increase in $r , \theta$ decreases.] Now as for cylindrical shell $A=2 \pi r L$,
$H =-2 \pi rLK \frac{ d \theta}{ dr }$
or $ \int\limits_{a}^{b} \frac{d r }{ r }=-\frac{2 \pi LK }{ H } \int\limits_{\theta_{1}}^{\theta_{2}} d \theta$
which on integration and simplification gives:
$H =\frac{ dQ }{ dt }=\frac{2 \pi KL \theta_{1}-\theta_{2}}{\log _{ e } b / a} \ldots ...$(i)$ $
Here $H =\frac{1^{2} R }{4.2}=\frac{5^{2} \times 4}{4.2}=24 \,cal / s , L =2\, m =200 \,cm$
$r_{1}=(0.64 / 2) mm =0.032\, cm$
and $r_{2}=r_{1}+d=0.032+0.006=0.038\, cm$
So $\theta_{1}-\theta_{2}=\frac{24 \times \log _{ e } 38 / 32}{2 \times 3.14 \times 200 \times 0.16 \times 10^{-2}}$
$=\frac{24 \times 2.3026\left[\log _{10^{38}}-\log _{10^{32}}\right]}{3.14 \times 0.64}$
or $ \theta_{1}-\theta_{2}=\frac{55 \times[1.57-1.50]}{2} \simeq 2{ }^{\circ} C$