An infinitely long pin fin, attached to an isothermal hot surface, transfers heat at a steady rate of 𝑸̇_𝟏 to the ambient air. If the thermal conductivity of the fin material is doubled, while keeping everything else constant, the rate of steadystate heat transfer from the fin becomes 𝑸̇_𝟐. The ratio 𝑸̇_𝟐/𝑸̇_𝟏 is

The correct option is √2.

For an infinitely long pin fin, the steady-state heat transfer rate (${\dot{Q}}_{\mathrm{fin}}$) from the fin to the ambient air is given by the formula:
${\dot{Q}}_{\mathrm{fin}}=\sqrt{hPk{A}_c}(T_b-T_{���})$
where:
• $h$ is the convective heat transfer coefficient,
• $P$ is the perimeter of the fin,
• $k$ is the thermal conductivity of the fin material,
• $A_c$ is the cross-sectional area of the fin,
• $T_b$ is the base temperature of the fin, and
• $T_{\infty }$ is the ambient air temperature.

Since all parameters except the thermal conductivity ($k$) are kept constant, we can see that the heat transfer rate is directly proportional to the square root of the thermal conductivity:
$\dot{Q}\propto \sqrt{k}$

Let the initial thermal conductivity be $k_1$ and the initial heat transfer rate be ${\dot{Q}}_1$.
According to the problem, the thermal conductivity of the fin material is doubled, so the new thermal conductivity is:
$k_2=2k_1$

The new heat transfer rate ${\dot{Q}}_2$ is given by:
${\dot{Q}}_2\propto \sqrt{k_2}$

Taking the ratio of the new heat transfer rate to the initial heat transfer rate, we get:
$\frac{{\dot{Q}}_2}{{\dot{Q}}_1}=\sqrt{\frac{k_2}{k_1}}$

Substitute $k_2=2k_1$ into the equation:
$\frac{{\dot{Q}}_2}{{\dot{Q}}_1}=\sqrt{\frac{2k_1}{k_1}}=\sqrt{2}$

Therefore, the ratio of the steady-state heat transfer rate ${\dot{Q}}_2/{\dot{Q}}_1$ is √2.