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



$\text{H}=\frac{\text{d}\text{Q}}{\text{dt}}=-\text{K}\text{A}\frac{\text{d}\text{θ}}{\text{dr}}$

[Negative sign is used as with increase in r, θ decreases.] Now as for cylindrical shell A = 2πrL,

$\text{H}=-2\text{π}\text{r}\text{L}\text{K}\frac{\text{d}\text{θ}}{\text{dr}}$

$\text{or}{\int }_a^b\frac{\text{dr}}{\text{r}}=-\frac{2\text{π}\text{LK}}{\text{H}}{\int }_{{\text{θ}}_1}^{{\text{θ}}_2}\text{d}\text{θ}$

which on integration and simplification gives:

$\text{H}=\frac{\text{d}\text{Q}}{\text{d}\text{t}}=\frac{2\text{π}\text{KL}\left({\text{θ}}_1-{\text{θ}}_2\right)}{{\text{log}}_{\text{e}}\left(b/a\right)}\text{....(i)}$

$\text{Here}\text{H}=\frac{{\text{I}}^2\text{R}}{\text{4.2}}=\frac{{\left(5\right)}^2\times 4}{\text{4.2}}=\text{24 cal/s,}\text{L}=\text{2 m}=\text{200 cm}$

              r₁ = (0.64/2) mm = 0.032 cm

and    r₂ = r₁ + d = 0.032 + 0.006 = 0.038 cm

$\text{So}\left({\text{θ}}_1-{\text{θ}}_2\right)=\frac{\text{24}\times {\text{log}}_{\text{e}}\left(\text{38}/\text{32}\right)}{\text{2}\times \text{3.14}\times \text{200}\times \text{0.16}\times {\text{10}}^{-2}}$

$=\frac{\text{24}\times \text{2.3026}[{\text{log}}_{\text{10}}\text{38}-{\text{log}}_{\text{10}}\text{32}]}{\text{3.14}\times \text{0.64}}$

$\text{or}\left({\text{θ}}_1-{\text{θ}}_2\right)=\frac{\text{55}\times [\text{1.57}-\text{1.50}]}{2}\simeq 2 °\text{C}$