Assuming pin fin of infinitely long fin

Fin Effectiveness$(ε_f)$

=$\frac{(heat transfer rate with fin)}{(heat transfer rate without fin)}$=$\frac{Q_(with fin)}{Q_(without fin)}$ =$\frac{Q_0}{(hA_c θ_0 )}$

$Q_0$=$\sqrt{(hPA_c K)}∙θ_0$

substituting above

$ε_f$=$\frac{\sqrt{(hPA_c K)}∙θ_0)}{(hA_c θ_0 )}$=$\sqrt{(\frac{Pk}{(hA_c }}$

For first fin which has lesser diameter effectiveness would be

$ε_f1$=$\sqrt{(\frac{4k}{hD})}$ as $\frac{A}{P}$=$\frac{D}{4}$

For second pin fin which has twice the diameter effectiveness would be

$ε_f2$=$\sqrt{\frac{2k}{hD}}$ as $\frac{A}{P}$=$\frac{D}{4}$

Effectiveness of fin with lesser diameter is higher (ans(i))

Fin Efficiency( η)

=(Actual heat transferred by the fin)/(max.heat transfer by fin if entire fin area were at base temperature)

=$\sqrt{\frac{KA_cs}{hPl^2}}$

Fin efficiency for lesser diameter would be

=$\sqrt{\frac{KD}{4hl^2}}$

Fin efficiency for twice the diameter would be

=$\sqrt{\frac{KD}{2hl^2}}$

Efficiency for fin with twice the diameter would be higher(ans(ii))