Explanation

(c) Here, $v_{1} = \sqrt{2g(h + x)};v_{2} = \sqrt{2gx}$
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Let $a$= area of cross-section of each hole
$\rho =$ density of the liquid
The momentum of the liquid flowing out per second through lower hole $=$ mass$\times$ velocity
$= av_{1}\rho \times v_{1} = a\ \rho\ v_{1}^{2}$
The force exerted on the lower hole towards left
$= a\ \rho\ v_{1}^{2}$
Similarly, the force exerted on the upper hole towards right $= a\ \rho\ v_{2}^{2}$
Net force on the tank, $F = a\ \rho(v_{1}^{2} - v_{2}^{2})$
$= a\ \rho\left\lbrack 2g(h + x) - 2gx \right\rbrack = 2a\ \rho gh$
$ie,\ F \propto h$