Question

A liquid flows in the tube, left to right as shown in figure. $d_1$ and $d_2$ are the diameter of tube as shown in figure. Find the ratio of speed $\frac{V_1}{V_2}$.

• A. $\frac{1}{2}$
• B. $\frac{1}{8}$
• C. $\frac{1}{4}$
• D. $\frac{1}{16}$

Answer 1

The correct option is C $\frac{1}{4}$

The area of cross - section of tube at left side, $A_1=\frac{\pi }{4}{d}_1^2$
The area of cross - section of tube at right side.
$A_2=\frac{\pi }{4}{d}_2^2$
By the equation of continuity
$A_1{V}_1=A_2{V}_2$
$\Rightarrow \frac{V_1}{V_2}=\frac{A_2}{A_1}=\frac{\frac{\pi }{4}{d}_2^2}{\frac{\pi }{4}{d}_1^2}=\frac{d_2^2}{d_1^2}={\left(\frac{2}{4}\right)}^2$
$\Rightarrow \frac{V_1}{V_2}=\frac{1}{4}$