Question

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

• A. 0.85 cc
• B. 0.46 cc
• C. 0.153 cc
• D. 0.05 cc

Answer 1

The correct option is C

0.153 cc

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

$△V=V_0[{\gamma }_L-{\gamma }_g]△\theta =V[{\gamma }_m-3{\alpha }_g]△\theta$

= 50 [ 180 $\times {10}^{-6}-3\times 9\times {10}^{-6}\left]\right(38-18)=00.153cc$