Question:
A water spray gun is attached to a hose of cross sectional area $30 \text{ cm}^2$. The gun comprises of 10 perforations each of cross sectional area of $15 \text{ mm}^2$. If the water flows in the hose with the speed of 50 cm/s, calculate the speed at which the water flows out from each perforation. (Neglect any edge effects)
A water spray gun is attached to a hose of cross sectional area $30 \text{ cm}^2$. The gun comprises of 10 perforations each of cross sectional area of $15 \text{ mm}^2$. If the water flows in the hose with the speed of 50 cm/s, calculate the speed at which the water flows out from each perforation. (Neglect any edge effects)
Updated On: Apr 10, 2026
- 100 m/s
- 10 m/s
- 1000 m/s
- $15 \times 10^2$ m/s
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
Since water is an incompressible fluid, the volume flow rate entering the hose must equal the total volume flow rate exiting through all the perforations combined. This is a direct application of the Equation of Continuity.
Step 2: Key Formula or Approach:
Equation of Continuity:
$A_1 v_1 = n \cdot A_2 v_2$
where $A_1, v_1$ are the area and velocity in the main hose, $n$ is the number of holes, and $A_2, v_2$ are the area and velocity of a single hole.
Step 3: Detailed Explanation:
First, convert all units to a common SI standard (meters and seconds).
Main hose area $A_1 = 30 \text{ cm}^2 = 30 \times 10^{-4} \text{ m}^2$.
Water speed in hose $v_1 = 50 \text{ cm/s} = 0.5 \text{ m/s}$.
Number of perforations $n = 10$.
Perforation area $A_2 = 15 \text{ mm}^2 = 15 \times 10^{-6} \text{ m}^2$.
Substitute into the continuity equation to solve for $v_2$:
$v_2 = \frac{A_1 v_1}{n A_2}$
$v_2 = \frac{(30 \times 10^{-4}) \times (0.5)}{10 \times (15 \times 10^{-6})}$
Simplify the terms:
Numerator: $30 \times 0.5 \times 10^{-4} = 15 \times 10^{-4}$.
Denominator: $150 \times 10^{-6} = 15 \times 10^{-5}$.
$v_2 = \frac{15 \times 10^{-4}}{15 \times 10^{-5}}$
$v_2 = 10^{-4 - (-5)} = 10^1 = 10 \text{ m/s}$.
Step 4: Final Answer:
The exit speed is 10 m/s.
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