Properties Of Solids And Liquids Question 539

Question: $ t=t _{1}+t _{2}=\frac{M}{R}(C\Delta T+L) $ 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? ($ \alpha $ for glass is $ 9\times {{10}^{-6}}/{}^\circ C $ and coefficient of real expansion of mercury is$ 180\times {{10}^{-6}}/{}^\circ C $ )

Options:

A) 0.85 cc

B) 0.46 cc

C) 0.153 cc

D) 0.05 cc

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Answer:

Correct Answer: C

Solution:

Due to volume expansion of both mercury and flask, the change in volume of mercury relative to flask is given by $ \Delta V=V _{0}[ \gamma L-\gamma ]\Delta \theta\text{ = V}[ {\gamma _{m}}-3{\alpha _{g}} ]\Delta \theta $

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

$ =0.153cc $



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