JEE Challenger
More from Mechanical Properties of Fluids
PhyEasyAdv 2014 • Paper 2

Speed of Air Emerging from Spray Gun Nozzle

Comprehension Passage

A spray gun model is illustrated in the diagram, where a piston drives air through an ejection nozzle. A narrow tube with uniform cross-sectional area connects to this nozzle, having its lower end submerged in a small liquid reservoir. When air is expelled through the nozzle by the piston, the liquid is drawn up from the vessel into the nozzle and sprayed out into the environment. In this arrangement, the radius of the piston cylinder is 20 mm20\text{ mm} while that of the nozzle is 1 mm1\text{ mm}. The top of the reservoir remains exposed to the atmosphere.

Passage Diagram 1

If the piston moves inward at a steady speed of 5 mm s−15\text{ mm s}^{-1}, the speed at which air exits the nozzle is given by

Options

A

0.1 ms−10.1\text{ ms}^{-1}

B

1 ms−11\text{ ms}^{-1}

C

2 ms−12\text{ ms}^{-1}

Correct
D

8 ms−18\text{ ms}^{-1}

Topics & Concepts

Mechanical Properties of FluidsFluid DynamicsBernoulli's Principle

Step-by-Step Solution

To determine the speed at which the air exits the nozzle, we can apply the principle of continuity for fluid flow.

Assuming the air behaves as an incompressible fluid under these steady, low-speed conditions, the volume flow rate inside the cylinder must equal the volume flow rate exiting the nozzle: A1v1=A2v2A_1 v_1 = A_2 v_2

where:

  • A1=πR2A_1 = \pi R^2 is the cross-sectional area of the piston cylinder, with radius R=20 mmR = 20\text{ mm},
  • v1=5 mm s−1v_1 = 5\text{ mm s}^{-1} is the speed of the piston,
  • A2=πr2A_2 = \pi r^2 is the cross-sectional area of the nozzle, with radius r=1 mmr = 1\text{ mm},
  • v2v_2 is the speed of the air emerging from the nozzle.

Rearranging for v2v_2: v2=v1(A1A2)=v1(Rr)2v_2 = v_1 \left( \frac{A_1}{A_2} \right) = v_1 \left( \frac{R}{r} \right)^2

Substituting the given values: v2=5 mm s−1×(20 mm1 mm)2v_2 = 5\text{ mm s}^{-1} \times \left( \frac{20\text{ mm}}{1\text{ mm}} \right)^2 v2=5×400 mm s−1=2000 mm s−1v_2 = 5 \times 400\text{ mm s}^{-1} = 2000\text{ mm s}^{-1}

Converting to meters per second: v2=20001000 m s−1=2 m s−1v_2 = \frac{2000}{1000}\text{ m s}^{-1} = 2\text{ m s}^{-1}

Thus, the speed at which air exits the nozzle is 2 m s−12\text{ m s}^{-1}, which corresponds to Option (C).

PreviousNext