CBSE Class 12-science Answered
Sir please give me the answer
Asked by minipkda | 18 May, 2018, 06:34: PM
Expert Answer
Let V0 be the initial volume of flask. Let v0 be the initial volume of mercury. Hence air volume (V0-v0).
After expansion, volume of flask = V0(1+γf×ΔT)
After expansion, volume of mercury = v0(1+γm×ΔT)
where γf , γm are volume expansion coefficient of flask and mercury respectively. ΔT is increase in temperature
Volume of air after expansion = V0(1+γf×ΔT) - v0(1+γm×ΔT) = V0 - v0 ......................(1)
we have considered in eqn.(1) that volume of air remains constant even after expansion (given in the question)
after simplifying we get from eqn.(1) V0 = v0×(γf / γm ) ................(2)
we are given that V0 = 1 litre = 1000 CC and ( γf /γm ) = 1/20 . Substituting this values in eqn.(2), we get v0 = 50 CC
Answered by Thiyagarajan K | 21 May, 2018, 04:08: PM
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