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When coming out of a horizontal tube at a speed V strikes normally a vertical wall close to the mouth of the tube and falls down vertically after impact. When the speed of the water is increased to $2 V .$

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The thrust exerted by the water on the wall will become four times

The energy lost per second by water striking the wall will be increased eight times

The thrust force exerted by the water on the wall will be doubled

The energy lost per second by water striking the wall will also become four times

answer is B, D.

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Detailed Solution

Thrust is given by $F = ρ A v^{2}$ , so on doubling the velocity the force will become four times.

The energy lost by the water is given by $E = \frac{1}{2} ρ A v^{3}$ , so on doubling the velocity the force will become eight times.

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