Question

Water enters a pipe of diameter 3.0, with a velocity of 3.0. The water encounters a constriction, where its velocity is 12. What is the diameter of the constricted portion of the pipe?

• A. $0.33$
• B. $0.75$
• C. $1.0$
• D. $1.5$

Answer 1

The correct option is D $1.5$

Given

$r_1=\frac{3}{2}=\frac{D_1}{2}$

$r_2=\frac{D_2}{2}$

$v_1=3v_2=12$

So we know that

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

$\pi {\left(\frac{3}{2}\right)}^2\times 3=\pi {\left(\frac{D_2}{2}\right)}^2\times 12$

$\frac{27}{4}=\frac{12D_2^2}{4}$

$D_2^2=\frac{9}{4}\times \frac{3}{3}$

$D_2=\frac{3}{2}=1.5$

Answer:-(D)