The cylinderical 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. `(V^(2)R)/(nr)`
B. `(VR^(2))/(n^(2)r^(2))`
C. `(VR^2)/(nr^2)`
D. `(VR^(2))/(n^(3)r^(2))`

Explanation

To find the speed of ejection of the liquid through the holes in the cylindrical tube, we can use the principle of conservation of mass, specifically the equation of continuity. Here are the steps to solve the problem:

### Step-by-Step Solution:

1. **Identify Variables**:
- Let \( R \) be the radius of the cylindrical tube.
- Let \( r \) be the radius of each fine hole.
- Let \( n \) be the number of holes.
- Let \( V \) be the speed of the liquid in the tube.
- Let \( V_2 \) be the speed of the liquid ejected through the holes.

2. **Calculate the Cross-Sectional Areas**:
- The cross-sectional area of the tube, \( A_1 \), is given by:
\[ A_1 = \pi R^2 \]
- The total cross-sectional area of the \( n \) holes, \( A_2 \), is given by:
\[ A_2 = n \cdot \pi r^2 \]

3. **Apply the Equation of Continuity**:
- According to the equation of continuity, the volume flow rate must be constant. Therefore, we have:
\[ A_1 V = A_2 V_2 \]
- Substituting the areas into the equation gives:
\[ \pi R^2 V = n \cdot \pi r^2 V_2 \]

4. **Simplify the Equation**:
- We can cancel \( \pi \) from both sides:
\[ R^2 V = n r^2 V_2 \]

5. **Solve for \( V_2 \)**:
- Rearranging the equation to solve for \( V_2 \):
\[ V_2 = \frac{R^2 V}{n r^2} \]

6. **Final Expression**:
- Thus, the speed of ejection of the liquid through the holes is:
\[ V_2 = \frac{R^2}{n r^2} V \]