The cylindrical tube of a spray pump has radius `R`, one end of which has `n` fine holes, each of radius `r`. If the speed of the liquid in the tube is `V`, the speed of the ejection of the liquid through the holes is:

Options

A. `(vR^(2))/(n^(2)r^(2))`
B. `(vR^(2))/(nr^(2))`
C. `(vR^(2))/(n^(3)r^(2))`
D. `(v^(2)R)/(nr)`

Explanation

To solve the problem, we will use the principle of conservation of mass, which is expressed through the continuity equation. The continuity equation states that the product of the cross-sectional area and the velocity of fluid flow must remain constant along a streamline.

### Step-by-Step Solution:

1. **Identify the Areas**:
- The cross-sectional area of the cylindrical tube (A1) is given by:
\[ A_1 = \pi R^2 \]
- The total cross-sectional area of the n holes (A2) is:
\[ A_2 = n \cdot \pi r^2 \]

2. **Apply the Continuity Equation**:
- According to the continuity equation:
\[ A_1 V = A_2 V' \]
- Where \( V \) is the speed of the liquid in the tube and \( V' \) is the speed of the liquid ejected through the holes.

3. **Substitute the Areas into the Equation**:
- Substitute the expressions for \( A_1 \) and \( A_2 \) into the continuity equation:
\[ \pi R^2 V = n \cdot \pi r^2 V' \]

4. **Cancel Common Terms**:
- We can cancel \( \pi \) from both sides of the equation:
\[ R^2 V = n r^2 V' \]

5. **Solve for \( V' \)**:
- Rearranging the equation to solve for \( V' \):
\[ V' = \frac{R^2 V}{n r^2} \]

### Final Answer:
The speed of the ejection of the liquid through the holes is:
\[ V' = \frac{R^2 V}{n r^2} \]