A water jet of area $50 c m^{2}$ , moving with a velocity of 25 m/s impinges normally on a plate moving with a velocity of 5 m/s in the same direction. If $ρ_{ω} = 1000 k g / m^{3}$ , then the force exerted on the plate will be:

3750 N

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2000 N

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3125 N

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2500 N

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Solution

The correct option is B
2000 N

F = Rate of change of momentum
$F = ρ A \times v_{s t r i k i n g}^{2}$

= $1000 \times 0.005 \times {(25 - 5)}^{2}$

= $2000$

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A water jet of area 50 cm2, moving with a velocity of 25 m/s impinges normally on a plate moving with a velocity of 5 m/s in the same direction. If ρω=1000 kg/m3, then the force exerted on the plate will be:

The correct option is B 2000 N F = Rate of change of momentum F=ρA×v2striking = 1000×0.005×(25−5)2 = 2000

A water jet of area $50 c m^{2}$ , moving with a velocity of 25 m/s impinges normally on a plate moving with a velocity of 5 m/s in the same direction. If $ρ_{ω} = 1000 k g / m^{3}$ , then the force exerted on the plate will be:

The correct option is B
2000 N

F = Rate of change of momentum
$F = ρ A \times v_{s t r i k i n g}^{2}$

= $1000 \times 0.005 \times {(25 - 5)}^{2}$

= $2000$