If the piston is pushed at a speed of $5 m m s^{- 1}$ , the air comes out of the nozzle with a speed of

$0.1 m s^{- 1}$
$1 m s^{- 1}$
$2 m s^{- 1}$
$8 m s^{- 1}$

8 m s^{- 1}

1 m s^{- 1}

0.1 m s^{- 1}

2 m s^{- 1}

Open in App

Solution

Verified by Toppr

From equation of continuity: $A_{1} V_{1} = A_{2} V_{2}$
Subscript 1 stands for piston and 2 for nozzle respectively.
We have: $d_{1} = 20 m m$ and $d_{2} = 1 m m$
Therefore, $A_{1} = π (20^{2})$ and $A_{1} = π (1^{2})$
Also, $V_{1} = 5 m / s$
Thus, $π (20^{2}) \times 5 = π (1^{2}) \times V_{2}$
$V_{2} = 2 m / s$

Was this answer helpful?

Similar Questions

If the piston is pushed at a speed of

5 m m s^{- 1}

, the air comes out of the nozzle with a speed of

View Solution

A syringe containing water is held horizontally with its nozzle at a height

h = 1.25 m

above the ground as shown in the figure. The diameter of the piston is

5

times as that of the nozzle. The piston is pushed at a constant speed of

20 {cms}^{- 1}

. If

g = 10 {ms}^{- 2}

, then -

View Solution

A rubber pipe with a diameter of

10 c m

is connected to a nozzle

2 c m

in diameter. Water flowing through the pipe at a speed of

0.6 {m s}^{- 1}

comes out like a jet through the nozzle. The backward force of the nozzle is about:

View Solution

A rubber pipe with a diameter of 10 cm is connected to a nozzle 2 cm in diameter. Water flowing through the pipe at a speed of 0.6 ms

^{- 1}

comes out like a jet through the nozzle. The backward force of the nozzle is about:

View Solution

A spray gun is shown in the figure where a piston pushes air out of a nozzle. A thin tube of uniform cross section is connected to the nozzle. The other end of the tube is in a small liquid container As the piston pushes air through the nozzle, the liquid from the container rises into the nozzle and is sprayed out. For the spray gun shown, the radii of the piston and the nozzle are

20 m m

and

1 m m

respectively. The upper end of the container is open to the atmosphere, if the density of air is

p_{a}

and that of the liquid

p_{l}

then for a given piston speed the rate (volume per unit time) at which the liquid is sprayed will be proportional to:

View Solution

If the piston is pushed at a speed of 5mms−1, the air comes out of the nozzle with a speed of0.1ms−11ms−12ms−18ms−1

From equation of continuity: A1V1=A2V2Subscript 1 stands for piston and 2 for nozzle respectively.We have: d1=20mm and d2=1mmTherefore, A1=π(202) and A1=π(12)Also, V1=5m/sThus, π(202)×5=π(12)×V2 V2=2 m/s

If the piston is pushed at a speed of $5 m m s^{- 1}$ , the air comes out of the nozzle with a speed of

$0.1 m s^{- 1}$
$1 m s^{- 1}$
$2 m s^{- 1}$
$8 m s^{- 1}$