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JEE Main 20264 AprilEvening Shift
Exam
JEE Main
Year
2026
Date
4 April
Shift
Evening Shift
Subject
Physics
Chapter
Mechanical Properties of Fluids

JEE Main 2026 · 4 April · Evening Shift · Physics · Mechanical Properties of Fluids

A water spray gun is attached to a hose of cross sectional area 30...

A water spray gun is attached to a hose of cross sectional area 3030 cm2^2. The gun comprises of 1010 perforations each of cross sectional area 1515 mm2^2. If the water flows in the hose with the speed of 5050 cm/s, calculate the speed at which the water flows out from each perforation. (Neglect any edge effects)

Options

  1. A100100 m/s
  2. B1010 m/s
  3. C10001000 m/s
  4. D15×10215 \times 10^2 m/s

Correct answer

B. 1010 m/s

Step-by-step solution

From the equation of continuity:

A1v1=nA2v2A_1 v_1 = n A_2 v_2

Here,
A1=30 cm2=30×10−4 m2A_1 = 30 \text{ cm}^2 = 30 \times 10^{-4} \text{ m}^2
v1=50 cm/s=0.5 m/sv_1 = 50 \text{ cm/s} = 0.5 \text{ m/s}
n=10n = 10
A2=15 mm2=15×10−6 m2A_2 = 15 \text{ mm}^2 = 15 \times 10^{-6} \text{ m}^2

Substituting the given values:

(30×10−4)×0.5=10×(15×10−6)×v2(30 \times 10^{-4}) \times 0.5 = 10 \times (15 \times 10^{-6}) \times v_2

15×10−4=150×10−6×v215 \times 10^{-4} = 150 \times 10^{-6} \times v_2

15×10−4=1.5×10−4×v215 \times 10^{-4} = 1.5 \times 10^{-4} \times v_2

v2=15×10−41.5×10−4=10 m/sv_2 = \dfrac{15 \times 10^{-4}}{1.5 \times 10^{-4}} = 10 \text{ m/s}

Answer: 1010 m/s
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