As shown in the figure below, there is a beaker of radius R . Water is filled in the beaker up to a height h . The density of water is ρ , the surface tension of water is T and the atmospheric pressure is p0 . Consider a vertical section ABCD of the water column through a diameter of the beaker. What is the magnitude of the force on the water on one side of this section by the water on the other side of this section ?:

Question

As shown in the figure below, there is a beaker of radius R . Water is filled in the beaker up to a height h . The density of water is ρ , the surface tension of water is T and the atmospheric pressure is p0 . Consider a vertical section ABCD of the water column through a diameter of the beaker. What is the magnitude of the force on the water on one side of this section by the water on the other side of this section ?: <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-wqjq6oa5qzsrzyrl.jpg" /> (A) |. 2 p0 R h + π R2 ρ g h - 2 R T |. (B) |. 2 p0 R h + R ρ g h2 - 2 R T |. (C) |. p0 π R2 + R ρ g h2 - 2 R T |. (D) |. p0 π R2 + R ρ g h2 + 2 R T |.

Tardigrade Admin · Accepted Answer

Force from right hand side liquid on left hand side liquid,
(i) Due to surface tension force
= 2RT (towards right)
(ii) Due to liquid pressure force

= \int_{x = 0}^{x = h} ((p)_{0} + ρ gh) (2 R \cdot x) dx

= (2p₀Rh + Rρgh²) (towards left)
Net force is |2p₀Rh + Rρgh² - 2RT|

$∣ 2 p_{0} R h + π R^{2} ρ g h - 2 RT ∣$

$∣ 2 p_{0} R h + Rρ g h^{2} - 2 RT ∣$

$∣ p_{0} π R^{2} + Rρ g h^{2} - 2 RT ∣$

$∣ p_{0} π R^{2} + Rρ g h^{2} + 2 RT ∣$

Force from right hand side liquid on left hand side liquid,
(i) Due to surface tension force
= 2RT (towards right)
(ii) Due to liquid pressure force
$= \int_{x = 0}^{x = h} ((p)_{0} + ρ gh) (2 R \cdot x) dx$
= (2p₀Rh + Rρgh²) (towards left)
Net force is |2p₀Rh + Rρgh² - 2RT|

∣2p0​Rh+πR2ρgh−2RT∣

∣2p0​Rh+Rρgh2−2RT∣

∣p0​πR2+Rρgh2−2RT∣

∣p0​πR2+Rρgh2+2RT∣

Force from right hand side liquid on left hand side liquid, (i) Due to surface tension force = 2RT (towards right) (ii) Due to liquid pressure force =∫x=0x=h​((p)0​+ρgh)(2R⋅x)dx = (2p0Rh + Rρgh2) (towards left) Net force is |2p0Rh + Rρgh2 - 2RT|

$∣ 2 p_{0} R h + π R^{2} ρ g h - 2 RT ∣$

$∣ 2 p_{0} R h + Rρ g h^{2} - 2 RT ∣$

$∣ p_{0} π R^{2} + Rρ g h^{2} - 2 RT ∣$

$∣ p_{0} π R^{2} + Rρ g h^{2} + 2 RT ∣$

Force from right hand side liquid on left hand side liquid,
(i) Due to surface tension force
= 2RT (towards right)
(ii) Due to liquid pressure force
$= \int_{x = 0}^{x = h} ((p)_{0} + ρ gh) (2 R \cdot x) dx$
= (2p₀Rh + Rρgh²) (towards left)
Net force is |2p₀Rh + Rρgh² - 2RT|