Water is filled up to a height h in a beaker of radius R. The density of water is $ρ$ , the surface tension of water is T and the atmospheric pressure is $P_{0}$ . Consider a vertical section ABCD of the water column through a diameter of the beaker. The force on water on one side of the section by water on the other side of this section has magnitude:

$2 P_{0} R h + π R^{2} ρ g h - 2 R T$

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$2 P_{0} R h + R ρ g h^{2} - 2 R T$

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$P_{0} π R^{2} + R ρ g h^{2} - 2 R T$

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$P_{0} π R^{2} + R ρ g h^{2} + 2 R T$

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Solution

The correct option is B
$2 P_{0} R h + R ρ g h^{2} - 2 R T$

Force, $F = \int_{0}^{h} (P_{0} + ρ g x) 2 R d x - 2 T R = 2 P_{0} R h + R ρ g h^{2} - 2 T R$

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Water is filled up to a height h in a beaker of radius R. The density of water is $ρ$ , the surface tension of water is T and the atmospheric pressure is $P_{0}$ . Consider a vertical section ABCD of the water column through a diameter of the beaker. The force on water on one side of the section by water on the other side of this section has magnitude:

Water is kept in a beaker of radius 5.0 cm. Consider a diameter of the beaker on the surface of the water. Find the force by which the surface on one side of the diameter pulls the surface on the other side. Surface tension of water = 0.075 N m^-1.

Q. A large tank is filled with water (density

ρ

), upto height h. Water is coming out from section (4). Fluid pressure at section (3) is:

(Given :

A_{2} = A_{4} = 2 A_{3}

and

A_{2}, A_{3}, A_{4} << A_{1}

; atmospheric pressure =

P_{0}

. Assume water to be non-viscous and incompressible.)

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The correct option is B 2P0Rh+Rρgh2−2RT Force, F=∫h0(P0+ρgx)2Rdx−2TR=2P0Rh+Rρgh2−2TR

The correct option is B
$2 P_{0} R h + R ρ g h^{2} - 2 R T$

Force, $F = \int_{0}^{h} (P_{0} + ρ g x) 2 R d x - 2 T R = 2 P_{0} R h + R ρ g h^{2} - 2 T R$