A flat plate moves normally towards a discharging jet of water at the rate of `3 m//s`. The jet discharges the water at the rate of `0.1 m^(3)//s` and at the speed of `18 m//s`. The force exerted on the plate due to the jet is

Options

A. `1800N`
B. `2100N`
C. `2450N`
D. `1560N`

Explanation

To find the force exerted on the plate due to the jet of water, we can follow these steps:

### Step 1: Determine the volume flow rate and the velocity of the jet
The volume flow rate of the jet is given as \( Q = 0.1 \, \text{m}^3/\text{s} \) and the velocity of the jet is \( V_j = 18 \, \text{m/s} \).

### Step 2: Calculate the cross-sectional area of the jet
The volume flow rate can be expressed as:
\[ Q = A \cdot V_j \]
where \( A \) is the cross-sectional area of the jet. Rearranging this gives:
\[ A = \frac{Q}{V_j} = \frac{0.1 \, \text{m}^3/\text{s}}{18 \, \text{m/s}} = \frac{0.1}{18} \approx 0.00556 \, \text{m}^2 \]

### Step 3: Determine the velocity of approach
The plate is moving towards the jet at a speed of \( V_p = 3 \, \text{m/s} \). The effective velocity of approach \( V_a \) is the sum of the velocity of the jet and the speed of the plate:
\[ V_a = V_j + V_p = 18 \, \text{m/s} + 3 \, \text{m/s} = 21 \, \text{m/s} \]

### Step 4: Calculate the mass flow rate
The mass flow rate \( \frac{dm}{dt} \) can be calculated using the formula:
\[ \frac{dm}{dt} = \rho \cdot Q \]
where \( \rho \) is the density of water, approximately \( 1000 \, \text{kg/m}^3 \). Thus:
\[ \frac{dm}{dt} = 1000 \, \text{kg/m}^3 \cdot 0.1 \, \text{m}^3/\text{s} = 100 \, \text{kg/s} \]

### Step 5: Calculate the force exerted on the plate
The force \( F \) exerted on the plate by the jet can be calculated using the change in momentum:
\[ F = \frac{dm}{dt} \cdot V_a \]
Substituting the values we have:
\[ F = 100 \, \text{kg/s} \cdot 21 \, \text{m/s} = 2100 \, \text{N} \]

### Final Answer
The force exerted on the plate due to the jet is \( 2100 \, \text{N} \).

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