An aluminium ring of mean diameter of 5m at ambient temperature is dipped into a liquid nitrogen bath. What will be the mean diameter of ring in the $$LN_2$$​ bath. The coefficient of linear expansion of aluminium alloy in the temperature range may be taken as 20 × $$10^{−6}$$​ K . (Ambient temperature may be taken as 24oC. Boiling temperature of liquid nitrogen: 77 K).

Correct option is A

​$$\begin{aligned}&{ \textbf{Given:}} \\&\text{Initial diameter } D_0 = 5\, \text{m} \\&\text{Initial temperature } T_0 = 24^\circ\text{C} = 297\, \text{K} \\&\text{Final temperature } T = 77\, \text{K} \\&\text{Coefficient of linear expansion } \alpha = 20 \times 10^{-6}\, \text{K}^{-1} \\[1.5em]&{ \textbf{Temperature change}} \\&\Delta T = T - T_0 = 77 - 297 = -220\, \text{K} \\[1.5em]&{\textbf{Linear contraction formula}} \\&\text{The change in dimension due to temperature is:} \\&\Delta D = D_0 \cdot \alpha \cdot \Delta T = 5 \cdot 20 \times 10^{-6} \cdot (-220) = -0.022\, \text{m} \\[1.5em]&{\textbf{Final diameter}} \\&D = D_0 + \Delta D = 5 - 0.022 = \boxed{4.978\, \text{m}}\end{aligned}$$​​