There are two identical small holes on the opposite sides of a tank containing a liquid. The tank is open at the top.The difference in height of the two holes is h shown in the figure. As the liquid comes out of the two holes, the tank will experience a net horizontal force proportional to

h^{\frac{1}{2}}

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h^{\frac{3}{2}}

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h^{2}

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Solution

The correct option is B h
$v_{1} = \sqrt{2 g (h + x)}; v_{2} \sqrt{2 g x}$ ; $ρ =$ density of liquid ;a $=$ cross sectional area of each hole,volume of liquid discharge per second at hole $= a v$ ( $v_{1} = v_{2}$ )
mass of liquid discharge $= a v ρ$
momentum of liquid discharged per second $= a v^{2} ρ$
so,the force exerted at the uper hole to the right $= a ρ {v_{2}}^{2}$
and the force exerted at the lower hole to the left $= a ρ {v_{1}}^{2}$ .
net force on the tank $= a ρ [{v_{1}}^{2} - {v_{2}}^{2}] = a ρ [2 g (h + x) - 2 g x] = 2 a ρ g h$
hence, the tank will experience a net horizontal force proportional to h.

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