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 20mm and 1mm respectively. The upper end of the container is open to the atmosphere.

If the piston is pushed at a speed of `5mms^-1`, the air comes out of the nozzle with a speed of

Options

A. `sqrt((p_(a))/p_(1))`
B. `sqrt(p_(a)p_(1))`
C. `sqrt((p_(1))/p_(a))`
D. `p_(1)`

Answer

A

Explanation

Let P be the pressure inside nozzle and v be the speed of air. When air comes out it is open to atmosphere and let us negletct its velocity after ejetion. If `P_(0)` represents atmospheric pressure then we can apply Bernoulli.s equation for the nozzle as follows : `P+1/2 p_(a)v^(2)=P_(0)............(i)`
Now, we can apply the same for vertical tube supplying liquid.
`P_(0)=1/2 p_(1)u^(2)+P .........(ii)`
In above equation u is the speed acquired by liquid. From equation (i) and (ii) we get the following. `1/2 p_(1)u^(2)=1/2p_(a)v^(2)`
`rArr u=v sqrt((p_(a))/(p_(1))`
For a given cross section volume flow rate is proportional to speed.