Solveeit Logo
Login

Question

Question: A stream of water flowing horizontally with a speed of \(15ms^{- 1}\)gushes out of a tube of cross-s...

A stream of water flowing horizontally with a speed of 15ms−115ms^{- 1}gushes out of a tube of cross-sectional area 10−2m210^{- 2}m^{2}, and hits a vertical wall normally. Assuming that it does not rebound from the wall, the force exerted on the wall by the impact of water is :

A

1.25×10−3N1.25 \times 10^{- 3}N

B

2.25×10−3N2.25 \times 10^{- 3}N

C

3.25×10−3N3.25 \times 10^{- 3}N

D

4.25×10−3N4.25 \times 10^{- 3}N

Answer

2.25×10−3N2.25 \times 10^{- 3}N

Explanation

Solution

Here v=15ms−1v = 15ms^{- 1}

Area of cross section A=10−2m2A = 10^{- 2}m^{2}

Density of water, ρ=103kgm−3\rho = 10^{3}kgm^{- 3}

Mass of water hitting the wall per second

}{= 10^{3}kgm^{- 3} \times 10^{- 2}m^{2} \times 15ms^{- 1} = 150kgs^{- 1}}$$ Force exerted on the wall = Momentum loss of water per second $$= 150kgs^{- 1} \times 15ms^{- 1} = 2250N = 2.25 \times 10^{3}N$$