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 mm and 1 mm respectively. The upper end of the container is open to the atmosphere. If the piston is pushed at a speed of 5 mms -1, the air comes out of the nozzle with a speed of

Question

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 mm and 1 mm respectively. The upper end of the container is open to the atmosphere. <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/920bb3861dcd45c104bb79226ad683e7-.png" /> If the piston is pushed at a speed of 5 mms -1, the air comes out of the nozzle with a speed of (A) 0.1 ms -1 (B) 1 ms -1 (C) 2 ms -1 (D) 8 ms -1

Tardigrade Admin · Accepted Answer

By A 1 V 1= A 2 V 2 ⇒ π(20)2 × 5=π(1)2 V 2 ⇒ V 2=2 m / s 2

$0.1 m s^{- 1}$

$1 m s^{- 1}$

$2 m s^{- 1}$

$8 m s^{- 1}$

By $A_{1} V_{1} = A_{2} V_{2}$
$\Rightarrow π (20)^{2} \times 5 = π (1)^{2} V_{2}$
$\Rightarrow V_{2} = 2 m / s^{2}$

0.1ms−1

1ms−1

2ms−1

8ms−1

By A1​V1​=A2​V2​ ⇒π(20)2×5=π(1)2V2​ ⇒V2​=2m/s2

$0.1 m s^{- 1}$

$1 m s^{- 1}$

$2 m s^{- 1}$

$8 m s^{- 1}$

By $A_{1} V_{1} = A_{2} V_{2}$
$\Rightarrow π (20)^{2} \times 5 = π (1)^{2} V_{2}$
$\Rightarrow V_{2} = 2 m / s^{2}$