A plate moves normally with the speed `v_(1)` towads a horizontal jet of uniform area of cross-section. The jet discharge water at the rate of volume `V` per second at a speed of `v_(2)`. The density of water is `rho`. Assume that water splashes along the surface of the plate ar right angles to the original motion. The magnitude of the force action on the plate due to the jet of water is

Options

A. `rho Vv_(1)`
B. `rho V (v_(1) + v_(2))`
C. `(rho V)/(v_(1) + v_(2)) V_(1)^(2)`
D. `rho [(V)/(v_(2))] (v_(1) + v_(2))^(2)`

Answer

D

Explanation

Force acting on plate, `F = (dp)/(dt) = v ((dm)/(dt))`
Mass of water reaching the plate per sec `= (dm)/(dt) = avp = A (v_(1) + v_(2)) rho = (V)/(V_(2)) (V_(1) + V_(2)) rho`
`(v = v_(1) + v_(2) =` velocity of water coming of jet w.r.t. plate)
(A = Area of cross section of jet `= (V)/(V_(2))`)
`F = (dm)/(dt) V = (V)/(V_(2)) (V_(1) + V_(2)) rho xx (V_(1) + V_(2)) = rho [(V)/(V_(2))] (V_(1) + V_(2))^(2)`