Question

A jet of a liquid of density $\rho$ with a cross-sectional area A is incident at an angle $\theta$ on a wall with a velocity $v$ and bounces with no loss of energy. The angle $\theta$ is measured with respect to the wall. The force in a direction normal to the surface of the wall is:

• A. $2\rho A{v}^2\cos \theta$
• B. $2\rho A{v}^2\sin \theta$
• C. $2\rho A{v}^2{\cos }^2\theta$
• D. $2\rho A{v}^2{\sin }^2\theta$

Answer 1

Answer: (B)

$2\rho A{v}^2\sin \theta$

From newton's third law, the force produced normal to the surface of the wall is equal to force on jet due to change in its momentum.

As shown in the FBD we have :
Resolving the incoming and outgoing components along the wall and normal to the wall we have:
Rate of change in momentum of incoming jet normal to the wall $=\rho \times A\times v^2\times sin\theta$ (into the wall).

Similarly, the rate of change in momentum of outgoing jet normal to the wall $=\rho \times A\times v^2\times sin\theta$ (away from the wall).
Therefore, total force in normal direction $=2\times \rho \times A\times v^2\times sin\theta$