Physics Q32 – JEE-MAIN 2026

Home Physics Mechanical Properties of FluidsQuestion #259

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)

Clue 1 of 3: Principle of Conservation

The volume flow rate of an incompressible fluid must be conserved through a pipe or system of pipes.

🧠 Full Solution Path

Step 1: Identify Given Parameters and Convert Units✦ Active

Identify the cross-sectional area of the hose ( $A_{1}$ ), the speed of water in the hose ( $v_{1}$ ), the number of perforations ( $N$ ), and the area of each perforation ( $A_{p}$ ). Convert all units to a consistent system, such as ${cm}^{2}$ and $cm/s$ .

A_{1} = 30 {cm}^{2}

v_{1} = 50 cm/s

N = 10

A_{p} = 15 {mm}^{2} = 15 \times (10^{- 1} cm)^{2} = 0.15 {cm}^{2}

💡 Teacher's Secret Hint

Pay close attention to unit conversions, especially between ${mm}^{2}$ and ${cm}^{2}$ .

Step 2: Calculate Total Outlet Area and Apply Continuity Equation○ Expand

Calculate the total cross-sectional area of all perforations ( $A_{2}$ ) and then apply the equation of continuity, $A_{1} v_{1} = A_{2} v_{2}$ , to find the speed of water flowing out from each perforation ( $v_{2}$ ). Remember that the total flow out must equal the total flow in.

A_{2} = N \times A_{p} = 10 \times 0.15 {cm}^{2} = 1.5 {cm}^{2}

v_{2} = \frac{A_{1} v_{1}}{A_{2}} = \frac{(30 {cm}^{2}) \times (50 cm/s)}{1.5 {cm}^{2}} = \frac{1500}{1.5} cm/s = 1000 cm/s

Step 3: Convert Final Speed to Required Units○ Expand

Convert the calculated speed $v_{2}$ from $cm/s$ to $m/s$ to match the units in the options.

v_{2} = 1000 cm/s = 1000 \times 10^{- 2} m/s = 10 m/s

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Physics Question 32 – JEE-MAIN 2026

Clue 1 of 3: Principle of Conservation

🧠 Full Solution Path