First slide

Fluid dynamics

Question

The cylindrical tube of a spray pump has a radius,H, one end of which has n fine holes, each of radius r. If the speed of flow of the liquid in the tube is y, the speed of ejection of the liquid through the hole is

Easy

A

$\frac{V}{n} (\frac{R}{r})$
B

$\frac{V}{n} {(\frac{R}{r})}^{1 / 2}$
C

$\frac{V}{n} {(\frac{R}{r})}^{3 / 2}$
D

$\frac{V}{n} {(\frac{R}{r})}^{2}$

Solution

Cross-sectional area of the tube $= A = {πR}^{2}$
speed of flow of liquid in the tube = V
cross-sectional area of each hole $= {πr}^{2}$
cross-sectional area of n hole $= {nπr}^{2}$
kt the speed of ejection of liquid through the hole = v
According to continuity equation
$aV = AV ({nπr}^{2}) V = ({πR}^{2}) V V = \frac{V}{n} {(\frac{R}{r})}^{2}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App