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A glass flask is filled up to a mark with 50 cc of mercury at $18^{0} C$ . If the flask and contents are heated to $38^{0} C$ , how much mercury will be above the mark? ( $α$ for glass is $9 \times 10^{- 6} /^{0} C$ and coefficient of real expansion of mercury is $180 \times 10^{- 6} /^{0} C$ )

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0.46 cc

0.05 cc

0.85 cc

0.153 cc

answer is C.

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Detailed Solution

Due to volume expansion of both mercury and flask, the change in volume of mercury relative to flask is given by

$Δ V = V_{0} [γ_{L} - γ_{g}] Δ θ$

$= V [γ_{m} - 3 α_{g}] Δ θ$

$= 50 [180 \times 10^{- 6} - 3 \times 9 \times 10^{- 6}] (38 - 18) = 0.153 c c$

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