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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.

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प्रश्न

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.

संक्षेप में उत्तर
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उत्तर

 Given:
Area of the wider tube, A = 900 cm2
Weight of the boy, m = 45 kg
Density of water, ρ = 10kgm−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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अध्याय 13: Fluid Mechanics - Exercise [पृष्ठ २७३]

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एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
अध्याय 13 Fluid Mechanics
Exercise | Q 3 | पृष्ठ २७३

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