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# Two liter glass flask contains some mercury. It is found that at all temperatures the volume of the air inside the flask remains same. The volume of the mercury in side the flask is$\left(\alpha \text{\hspace{0.17em}}for\text{\hspace{0.17em}\hspace{0.17em}}glass=9×{10}^{-6}{/}^{0}C,\text{\hspace{0.17em}\hspace{0.17em}}{\gamma }_{Hg}=1.8×{10}^{-4}{/}^{0}C\right)$$x×{10}^{-6}{m}^{3}$ where x=

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a
3
b
30
c
300
d
3000
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detailed solution

Correct option is C

at all temperatures the volume of the air inside the flask remains same VCγC=VLγL ; here V is volume of container and volume of liquid; γ coefficient of volume expansion of container and liquid  VL=VCγCγL given VC=2liter;γC=3α=3x9x10-6/C  0;γL=1.8x10-4/C  0, substitute VL=2literx3x9x10-61.8x10-4        since 1liter=10-3m3 VL=300x10-6m3=volume of mercury in the flask

Similar Questions

A glass vessel is partially filled with mercury and when both are heated together, the volume of the unfilled part of vessel remains constant at all temperature. Find the initial volume (in cm3) of mercury if the empty part measures 34 cm3 ${\mathrm{\gamma }}_{\mathrm{Hg}}=18×{10}^{-5}{/}^{\circ }\mathrm{C}]$

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