A piece of metal weighs 40 gm in air. When it is immersed in the liquid of specific gravity 2.0 at $28^{\circ} C$ it weights 32 gm, when the temperature of liquid is raised to $32^{\circ} C$ the metal piece weighs 28 gm. If the specific gravity of the liquid at $32^{\circ} C$ is 1.5, then the coefficient of linear expansion of the metal will be

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$\frac{1}{16} /^{\circ} C$

$\frac{1}{12} /^{\circ} C$

$\frac{1}{48} /^{\circ} C$

$\frac{1}{32} /^{\circ} C$

answer is B.

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Detailed Solution

Loss of weight at $28^{0} C = (40 - 32) \times 10^{- 3} g = v_{1} .2 ρ_{l} g ...... (i)$ Loss of weight at $32^{0} C = (40 - 28) \times 10^{- 3} g = v_{2} . (1.5 ρ_{l}) g ...... (i i)$ From (i) and (ii)

$\frac{V_{1}}{V_{2}} = \frac{1}{2} ....... (i i i)$

$V_{2} = V_{1} (1 + 3 α Δ T)$

$α = (1 / 12) /^{0} C$

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