Q57 / 75 · JEE Main 2026 (04 April Shift 2) · Physics
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)
- A$100$ m/s
- Correct answerB$10$ m/s
- C$1000$ m/s
- D$15 \times 10^2$ m/s
Solution
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
Hence, the correct answer is option: 'B'.