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Question

Physics Question on thermal properties of matter

A two litre glass flask contains some mercury. It is found that at all temperatures the volume of the air inside the flask remains the same. The volume of the mercury inside the flask is ( α\alpha for glass =9×106/C,γ=9\times {{10}^{-6}}/C,\gamma for mercury = 1.8×104/C=\text{ }1.8\times {{10}^{-4}}/{}^\circ C )

A

1500 cc

B

150 cc

C

3000 cc

D

300 cc

Answer

300 cc

Explanation

Solution

It is given that the volume of air in the flask remains the same. This means that the expansion in volume of the vessel is exactly equal to the volume expansion of mercury, ie, ΔVv=ΔVm\Delta {{V}_{v}}=\Delta {{V}_{m}} or VvγvΔT=VmγmΔT{{V}_{v}}{{\gamma }_{v}}\Delta T={{V}_{m}}{{\gamma }_{m}}\Delta T \therefore Vm=Vvγvγm=Vv×3αγm{{V}_{m}}=\frac{{{V}_{v}}{{\gamma }_{v}}}{{{\gamma }_{m}}}=\frac{{{V}_{v}}\times 3\alpha }{{{\gamma }_{m}}} (γv=3α)(\because \,{{\gamma }_{v}}=3\alpha ) \Rightarrow Vm=2000×3×9×1061.8×104{{V}_{m}}=\frac{2000\times 3\times 9\times {{10}^{-6}}}{1.8\times {{10}^{-4}}} =300cc=300\,cc