JEE Mains · Physics · STD 11 - Fluid mechanics
- JEE Mains 2026
- Medium
- MCQ
A water spray gun is attached to a hose of cross sectional area \(30\) cm\(^2\). The gun comprises of \(10\) perforations each of cross sectional area \(15\) 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)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(10\) m/s
Step-by-step Solution
Detailed explanation
From the equation of continuity:
\(A_1 v_1 = n A_2 v_2\)
Here,
\(A_1 = 30 \text{ cm}^2 = 30 \times 10^{-4} \text{ m}^2\)
\(v_1 = 50 \text{ cm/s} = 0.5 \text{ m/s}\)
\(n = 10\)
\(A_2 = 15 \text{ mm}^2 = 15 \times 10^{-6} \text{ m}^2\)
Substituting the given values:
\((30 \times 10^{-4}) \times 0.5 = 10 \times (15 \times 10^{-6}) \times v_2\)
\(15 \times 10^{-4} = 150 \times 10^{-6} \times v_2\)
\(15 \times 10^{-4} = 1.5 \times 10^{-4} \times v_2\)
\(v_2 = \dfrac{15 \times 10^{-4}}{1.5 \times 10^{-4}} = 10 \text{ m/s}\)
Answer: \(10\) m/s
\(A_1 v_1 = n A_2 v_2\)
Here,
\(A_1 = 30 \text{ cm}^2 = 30 \times 10^{-4} \text{ m}^2\)
\(v_1 = 50 \text{ cm/s} = 0.5 \text{ m/s}\)
\(n = 10\)
\(A_2 = 15 \text{ mm}^2 = 15 \times 10^{-6} \text{ m}^2\)
Substituting the given values:
\((30 \times 10^{-4}) \times 0.5 = 10 \times (15 \times 10^{-6}) \times v_2\)
\(15 \times 10^{-4} = 150 \times 10^{-6} \times v_2\)
\(15 \times 10^{-4} = 1.5 \times 10^{-4} \times v_2\)
\(v_2 = \dfrac{15 \times 10^{-4}}{1.5 \times 10^{-4}} = 10 \text{ m/s}\)
Answer: \(10\) m/s
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