Question

A jet of water coming out of a nozzle with a velocity of 20 m/s strikes a hinged plate at the centre. Nozzle area is $2 \times 10^{-3}$ m$^2$; density of water is 1000 kg/m$^3$. If the angle of the swing of the plate from the vertical is $30^\circ$, then the weight of the plate is:

• A. 800 N
• B. 1000 N
• C. 1200 N
• D. 1600 N

Hint:

In hinged plate problems, the force of the jet ($F$) is usually balanced by the weight component ($W \sin\theta$). Since $\sin(30^\circ) = 1/2$, the weight will always be exactly twice the jet force in this specific configuration.

Answer 1

The Correct Option is D

1. Calculate the Force of the Jet ($F$): The force exerted by a jet striking a vertical plate normally is given by: $$F = \rho \cdot A \cdot V^2$$ Given: $\rho = 1000$ kg/m$^3$, $A = 2 \times 10^{-3}$ m$^2$, $V = 20$ m/s. $$F = 1000 \cdot (2 \times 10^{-3}) \cdot (20)^2$$ $$F = 2 \cdot 400 = 800 \text{ N}$$

2. Equilibrium Conditions at the Hinge: Let $W$ be the weight of the plate and $\theta$ be the angle of swing ($30^\circ$). Let $L$ be the distance from the hinge to the point of impact (center of the plate).

• The moment due to the jet force is $F \cdot (L / \cos\theta)$. However, for a jet striking at the center, the perpendicular distance to the hinge is $L$.

• The moment due to weight acting at the center of gravity is $W \cdot (L \cdot \sin\theta)$.
Equating moments about the hinge: $$F \cdot L = W \cdot L \cdot \sin\theta$$ $$800 = W \cdot \sin(30^\circ)\lt strong\gt 3. Final Calculation:\lt /strong\gt 800 = W \cdot 0.5$$ $$W = \frac{800}{0.5} = 1600 \text{ N}$$ The total weight of the plate is 1600 N.