Question

Water is flowing in a horizontal pipe of non-uniform area of cross- section. The velocity of water at a place, where the radius of pipe is $0.01\text{m}$ is $25\text{m/s}$. What will be the velocity of water where the radius of pipe is $0.02\text{m}$?

• A. None of these
• B. $6.75\text{m/s}$
• C. $6.25\text{m/s}$
• D. $7.25\text{m/s}$

Answer 1

The correct option is C $6.25\text{m/s}$

Given:
$r_1=0.01\text{m}$
$v_1=25\text{m/s}$.
$r_2=0.02\text{m}$

From principle of continuity

$A_1{v}_1=A_2{v}_2$

$\frac{v_1}{v_2}=\frac{A_2}{A_1}=\frac{r_2^2}{r_1^2}$

$\Rightarrow \frac{25}{v_2}=\frac{(0.02{)}^2}{(0.01{)}^2}$

or $v_2=\frac{25}{4}=6.25\text{m/sec}$

Hence, option $\left(C\right)$ is correct.