There are two identical holes on the opposite sides of a tank containing a liquid. The tank is open at the top. The difference in height between the two holes is $2 h$ . As the liquid comes out of the two holes, the tank will experience a net horizontal force proportional to

\sqrt{h}

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h

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

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h^{2}

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Solution

The correct option is B $h$

Here forces are,
$F_{1} = ρ a v_{1}^{2}$ and $F_{2} = ρ a v_{2}^{2}$
But $F_{1} > F_{2}$
$△ F = F_{1} - F_{2} = ρ a (v_{1}^{2} - v_{2}^{2}) = ρ a [2 g h_{1} - 2 g h_{2}]$
$= 2 ρ a g [h_{1} - h_{2}] = 4 ρ a g h$
$\Rightarrow △ F \propto h$

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There are two identical holes on the opposite sides of a tank containing a liquid. The tank is open at the top. The difference in height between the two holes is 2h. As the liquid comes out of the two holes, the tank will experience a net horizontal force proportional to

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There are two identical holes on the opposite sides of a tank containing a liquid. The tank is open at the top. The difference in height between the two holes is $2 h$ . As the liquid comes out of the two holes, the tank will experience a net horizontal force proportional to

The correct option is B $h$

Here forces are,
$F_{1} = ρ a v_{1}^{2}$ and $F_{2} = ρ a v_{2}^{2}$
But $F_{1} > F_{2}$
$△ F = F_{1} - F_{2} = ρ a (v_{1}^{2} - v_{2}^{2}) = ρ a [2 g h_{1} - 2 g h_{2}]$
$= 2 ρ a g [h_{1} - h_{2}] = 4 ρ a g h$
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