Topic: Strength of Materials
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 LN2 bath. The coefficient of linear expansion, of aluminium alloy in the temperature range may be taken as 20x10^-6 /K (Ambient temperature may be taken as 24 °C. Boiling temperature of liquid nitrogen: 77 K)
- A. 4.978 m
- B. 4.958 m
- C. 5.138 m
- D. 4.865 m
Answer: A. 4.978 m
Explanation: Initial temperature T1 = 24 °C = 297 K. Final temperature T2 = 77 K. Delta T = 77 - 297 = -220 K. Change in diameter dD = D * alpha * Delta T = 5 * 20x10^-6 * (-220) = -0.022 m. New diameter = 5 - 0.022 = 4.978 m.