A hemispherical bowl of radius R is completely filled with water. The density of water is $ρ$ , the surface tension of water is T and the atmospheric pressure is $p_{0}$ . Consider a vertical section ABC of water through a diameter of top of bowl. The force in magnitude on water on one side of the section by water on other side of the section is not given by

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$| \frac{p_{0} π R^{2}}{2} + \frac{2 ρ R^{3} g}{3} - 2 T R - π T R |$

$| 2 p_{0} R^{2} + ρ g R^{3} - 2 T R |$

$| \frac{p_{0} π R^{2}}{2} + \frac{2 ρ g R^{3}}{3} - 2 T R |$

$| \frac{p_{0} π R^{2}}{2} + \frac{π R^{3} ρ g}{4} - 2 T R |$

answer is A, B, C.

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Detailed Solution

Force due to pressure
Force on element
   $d F = (p_{0} + ρ g R \sin θ) (2 R \cos θ) (R d θ \cos θ) ∴ F p r = \int_{0}^{π / 2} d F = \int_{0}^{π / 2} 2 p_{0} R^{2} \cos^{2} θ d θ + \int_{0}^{π / 2} 2 ρ R^{3} \cos^{2} θ \sin θ d θ = \frac{p_{0} π R^{2}}{2} + \frac{2 ρ g R^{3}}{3}$
Question Image

Or : Fpr = (pressure at geometrical centre) (area)
   $= (p_{0} + ρ g \frac{4 R}{3 π}) \frac{π R^{2}}{2} = \frac{p_{0} π R^{2}}{2} + \frac{2 ρ g R^{3}}{3}$
Force due to surface tension
   $F_{s} = (s u r f a c e t e n s i o n) \times (l e n g t h) = T (2 T) F_{n e t} = | F_{p r} - F_{s} |$

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