Solveeit Logo

Question

Physics Question on Hydrostatics

An ideal fluid flows through a pipe of circular cross-section made of two sections with diameters 2.5cm2.5\, cm and 3.75cm3.75\, cm. The ratio of the velocities in the two pipes is

A

9:49 : 4

B

3:23 : 2

C

3:2\sqrt{3} : \sqrt{2}

D

2:3\sqrt{2} : \sqrt{3}

Answer

9:49 : 4

Explanation

Solution

As, A1v1=A2v2A_{1}v_{1}=A_{2}v_{2} v1v2=A2A1=(πD22/4)(πD12/4)\Rightarrow \frac{v_{1}}{v_{2}}=\frac{A_{2}}{A_{1}}=\frac{\left(\pi D^{2}_{2}/4\right)}{\left(\pi D^{2}_{1}/4\right)} =(D2D1)2=\left(\frac{D_{2}}{D_{1}}\right)^{2} or v1v2=(3.75cm2.5cm)2\frac{v_{1}}{v_{2}}=\left(\frac{3.75\,cm}{2.5\,cm}\right)^{2} =2.25=9:4=2.25=9:4