A 2 mm diameter electric wire is covered by 2 mm thick plastic of conductivity 0.5 W/(m·K). Air is at 320 K with external convection coefficient 25 W/(m²·K). Wire surface remains at 420 K with or without insulation. For steady heat loss, find the nearest insulated/bare heat-loss ratio per metre. Use π=3.14.
Wire radius is $r_1=.001$ m and outer insulation radius $r_2=.003$ m. Bare resistance per metre is $R_b=1/(h2\pi r_1)$. Insulated resistance is $R_i=\ln(r_2/r_1)/(2\pi k)+1/(h2\pi r_2)$. The common temperature difference cancels, giving $Q_i/Q_b=R_b/R_i=1/[hr_1\ln3/k+r_1/r_2]\approx2.5756$. Hence **2.58,A**. Heat loss increases because these radii lie below the critical radius $k/h=.020$ m.
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