A jet water issues from a nozzel with a velocity of $20 m / s$ and it impinges normally on a flat plate moving away from it at $10 m / s$ . If the cross-sectional area of the jet is $0.02 m^{2}$ and the density of water is taken as $1000 k g / m^{3}$ , then the force developed on the plate will be

10 N

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100 N

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1000 N

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2000 N

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Solution

The correct option is A $10 N$
A jet water issues from a nozzel with a velocity $= 20 m / s$
It impinges normally on a flat plate moving away from it $= 10 m / s$
Cross-sectional area of the jet $= 0.02 m^{2}$
The density of water is taken $= 1000 k g / m^{3}$
The force developed on the plate will be
Hydrostatic force o the bottom surface of the tank will be
$F_{b} = ρ g (l \times b) \times h F_{b} ρ g \times (1 \times 2) \times 2 F_{b} = 4 ρ g (l b h) \to (1)$
Hydrostatic force on vertical surface will be
$F_{v} = ρ g (l \times h) \times \frac{h}{2} F_{v} = ρ g (2 \times 2) \frac{l h^{2}}{2} F_{v} = 2 ρ g (l h^{2}) \to (2)$
Ratio will be
$\frac{F_{b}}{F_{V}} = \frac{4 ρ g (l b h)}{2 ρ g (l h^{2})} \frac{F_{b}}{F_{V}} = 1 F_{b} = 1 \times 10 = 10 N$

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