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Question
The area of cross section of the wider tube shown in figure is 900 cm2. If the boy standing on the piston weighs 45 kg, find the difference in the levels of water in the two tubes.
Solution
Given:
Area of the wider tube, A = 900 cm2
Weight of the boy, m = 45 kg
Density of water, ρ = 103 kgm−3
Let h be the difference in the levels of water in the tubes and pa be the atmospheric pressure.
As per the figure, we have:
Pa + hρg = Pa +`(mg)/A`
`=>` hρg =`(mg)/A`
`=> h = m/(ρA)`
`=> h = m/(1000×A)`
`=45/(1000xx900xx10^(-4))=1/2`m = 50cm
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