A metal ball immersed in Alcohol weights W1 at 0°C and W2 at 50°C. The coefficient of cubical expansion of the metal (γ)m is less than that of alcohol (γ)Al. Assuming that density of metal is large compared to that of alcohol, it can be shown that
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When a metal ball is immersed in alcohol, its apparent weight depends on the buoyant force, which is ρliquidgVball. At higher temperature, both the metal and alcohol expand. Since γm < γAl, the density of alcohol decreases more than that of the metal's volume increases. Thus, ρalcohol decreases significantly, reducing buoyant force. So, apparent weight W increases with temperature.
Therefore, W1 < W2.
Final Answer: W1 < W2