Water is filled up to a height texth in a beaker of radius textR as shown in the figure. The density of water is ρ , the surface tension of water is T and the atmospheric pressure is po . Consider a vertical section textABCD of the water column through a diameter of the beaker. The force on water on one side of this section by water on the other side of this section has magnitude

Question

Water is filled up to a height texth in a beaker of radius textR as shown in the figure. The density of water is ρ , the surface tension of water is T and the atmospheric pressure is po . Consider a vertical section textABCD of the water column through a diameter of the beaker. The force on water on one side of this section by water on the other side of this section has magnitude <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-98ujvjltqaxfpnln.png" /> (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

The force from right-hand side liquid on left-hand side liquid,
(i) Due to surface tension force

= 2RT

(towards the right)
(ii) Due to liquid pressure force

= \int_{0}^{h} (p_{0} + ρ gx) 2 R d x

= 2 R p_{0} h + Rρ g h^{2}

Net force is

∣ ∣ 2 p_{o} R h + Rρ g h^{2} - 2 RT ∣ ∣

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

$∣ 2 p_{0} R h + Rρ g h^{2} - 2 RT ∣$

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

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

The force from right-hand side liquid on left-hand side liquid,
(i) Due to surface tension force
$= 2RT$ (towards the right)
(ii) Due to liquid pressure force
$= \int_{0}^{h} (p_{0} + ρ gx) 2 R d x$
$= 2 R p_{0} h + Rρ g h^{2}$
Net force is $∣ ∣ 2 p_{o} R h + Rρ g h^{2} - 2 RT ∣ ∣$

∣2p0​Rh+πR2ρgh−2RT∣

∣2p0​Rh+Rρgh2−2RT∣

∣p0​πR2+Rρgh2−2RT∣

∣p0​πR2+Rρgh2+2RT∣

The force from right-hand side liquid on left-hand side liquid, (i) Due to surface tension force = 2RT (towards the right) (ii) Due to liquid pressure force =∫0h​(p0​+ρgx)2Rdx =2Rp0​h+Rρgh2 Net force is ∣∣​2po​Rh+Rρgh2−2RT∣∣​

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

$∣ 2 p_{0} R h + Rρ g h^{2} - 2 RT ∣$

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

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

The force from right-hand side liquid on left-hand side liquid,
(i) Due to surface tension force
$= 2RT$ (towards the right)
(ii) Due to liquid pressure force
$= \int_{0}^{h} (p_{0} + ρ gx) 2 R d x$
$= 2 R p_{0} h + Rρ g h^{2}$
Net force is $∣ ∣ 2 p_{o} R h + Rρ g h^{2} - 2 RT ∣ ∣$