A metal ball immersed in alcohol weighs $W_{1}$ at $0^{0} C$ and $W_{2}$ at $59^{0} C$ . The coefficient of cubical expansion of the metal is less than that of alcohol. Assuming that the density of metal is large compared to that of alcohol, it can be shown that :

W_{1} > W_{2}

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W_{1} = W_{2}

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W_{1} < W_{2}

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W_{2} = (W_{1} / 2)

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Solution

The correct option is C $W_{1} < W_{2}$
Buoyant force acting on a body immersed in liquid is the weight of liquid it displaces $= V_{0} d_{0} g$
With increase in temperature, this buoyant force becomes $V_{0} (1 + γ_{m e t a l} T) \frac{d_{0}}{1 + γ_{l i q u i d} T} g$
If $γ_{m e t a l} < γ_{l i q u i d}$ ,
The new buoyant force is lesser than the initial buoyant force. Hence the weight that is measured would be increased.

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