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Question

A one litre glass flask contains some mercury. It is found that at different temperatures the volume of air inside the flask remains the same. What is the volume of mercury in this flask if coefficient of linear expansion of glass is 9×106/C while of volume expansion of mercury is 1.8×104/C

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Solution

Let initially volume of mercury and ar in the flask be V1 & V2
so, volume of flask,
V=V1+V2(1) [ where v = 1L]
after heating by T, the volume of flask and mercury become
V and V1 respectively, while V2 remain same.
V=V1+V2(2)
From (1) & (2),
VV2=VV1
Applying volmetric expansion,
VV1=V(1+rT)V1(1+r1T)
VrT=V1r1T
Let initially volume of mercury and ar in the flask be V1& V2
so, volume of flask,
V=V1+V2(1) [ where v = 1L]
after heating by T, the volume of flask and mercury become
V and V respectively, while V2 remain same.
V=V1+V2(2)
From (1) & (2),
VV2=VV1
Applying volmetric expansion,
VV1=V(1+rT)V1(1+r1T)
VrT=V1r1T
V1=Vrr1=1×3×9×1061.8×104 [asr=3α]
V1=0.15L=150mL. (Ans)

1068289_1166315_ans_feaf869a8125412db7c1f3855310af69.png

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