1. Mass flow rate of the liquid jet

The liquid comes out of a hose of radius $a$ with speed $v$.

Cross-sectional area of the hose: $A = \pi a^2$

So, mass of liquid striking the mesh per unit time is $\dot m = \rho A v$

2. Find momentum change for each part of the liquid

Take the initial direction of flow as positive.

(i) 50% passes through unaffected

This part has no change in velocity. $\Delta v = 0$ So, no force contribution.

(ii) 25% loses all of its momentum

For this part,

• initial velocity $= v$
• final velocity $= 0$

Change in velocity: $\Delta v = 0 - v = -v$

Rate of momentum change for this part: $F_1 = 0.25\, \dot m\, v$

(iii) 25% comes back with the same speed

For this part,

• initial velocity $= v$
• final velocity $= -v$

Change in velocity: $\Delta v = -v - v = -2v$

Rate of momentum change for this part: $F_2 = 0.25\, \dot m\, (2v) = 0.5\, \dot m\, v$

3. Total force on the liquid

Total rate of change of momentum of the liquid is $F = F_1 + F_2 = 0.25\dot m v + 0.5\dot m v = 0.75\dot m v$

Thus, $F = \frac{3}{4}\dot m v$

Substitute $\dot m = \rho A v$: $F = \frac{3}{4}(\rho A v)v = \frac{3}{4}\rho A v^2$

4. Pressure on the mesh

Pressure is force per area: $P = \frac{F}{A} = \frac{\frac{3}{4}\rho A v^2}{A}$ $P = \frac{3}{4}\rho v^2$

5. Final answer

$\boxed{P = \frac{3}{4}\rho v^2}$

So the correct option is A.