Question

A metal ball immersed in alcohol weight $W_1$ at $0^{o}C$ and $W_2$ at ${50}^{o}C$. The expansion of cubical expansion of the metal is less than that of the alcohol. Assuming that the density of the metal is large compared to that of alcohol, it can be shown that (This problem is mixed concept of heat and fluid mechanics):

• A. $W_1>W_2$
• B. $W_1=W_2$
• C. $W_1<W_2$
• D. none of these

Answer 1

Answer: (C)

$W_1<W_2$

Since, it is given that the coefficient of cubical expansion of alcohol is more than that of the metal.

This statement means that when we raise the temperature, the volume of alcohol increase at a faster rate than as compared to the increase in volume of metal ball.

OR in other words, the density of alcohol decreases at a faster rate as compared to metal ball. As a result, the metal ball experiences a lesser bouyant force as the temperature is raised. Hence, the weight at increased temperatures would be more.

Hence, answer should be $W_1<W_2$