Question

A horizontal jet of water coming out of a pipe of area 20cm square hits a vertical wall with a velocity of 10 ms and rebounds with the same speed.the force exerted by water on the wall is?

Answer 1

volume of water coming out of the pipe = A * v
     = 20 cm² * 10 m/sec
   = 0.02  m³/sec

mass of water hitting the wall  = m = 0.02 m³/sec * 1000 kg/m³ 
         = 20 kg/sec

As the water hits the wall normal to the wall and gets rebounded at the same speed, the change of momentum of water hitting the wall is:
            = m (v  - (-v))  =  2 m v
           = 2 * 20 kg /sec *  10 m/sec
           =  400 kg-m/sec²

Since, the change of momentum obtained is per unit time,  it is same as the force.  Thus the force acting on the wall by water = force experienced by water jet.

         = 400 Newtons.

PLEASE MARK AS BRAINLIEST IF HELPFUL!!!

Answer 2

F = dp / t
= mdv / t
= m(u - (-u) / t
= 2mu / t
= 2 × ALd × u / t ………[∵ Mass = Volume × Density; Volume = Area × Length]
= 2Adu² ………[∵ L/t = u]
= 2 × 20 × 10⁻⁴ m² × 10³ kg/m³ × (10 m/s)²
………[∵ Density of water = 10³ kg/m³]
= 400 N

Force exerted by water on the wall is 400 N