Ncert Exercises & Solutions

5.28 A stream of water flowing horizontally with a speed of 15 m s-1 gushes out of a tube of cross-sectional area 10-2 m2 and hits a vertical wall nearby. What is the force exerted on the wall by the impact of water, assuming it does not rebound?

Speed of the water stream, v = 15 m/s

The cross-sectional area of the tube, A = $10^{- 2} m^{2}$

Volume of water coming out from the pipe per second,

$V = A v = 10^{- 2} \times 15 m^{3} s^{- 1}$ Density of water, ρ = $10^{3} k g / m^{3}$

Mass of water flowing out through the pipe per second = ρ × V = 150 kg/s

The water strikes the wall and does not rebound. Therefore, the force exerted by the water on the wall is given by Newton’s second law of motion as:

F = Rate of change of momentum = $\frac{∆ P}{∆ t}$

$= \frac{mv}{t}$
$= 150 \times 15 = 2250 N$

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