Q. Water is flowing in a pipe of diameter 4 cm with a velocity $3\ {ms}^{- 1}$. The water then enters in to a pipe of diameter 2 cm. the velocity of water in the other pipe is
A) ${3\ ms}^{- 1}$
B) ${6\ ms}^{- 1}$
C) ${12\ ms}^{- 1}$
D) ${8\ ms}^{- 1}$
Explanation
(a) According to principle of continuity,
$Av = constant$
$or\ \ \ A_{1}v_{1} = A_{2}v_{2}$
$or\ \ \ \pi r_{1}^{2}v_{1} = \pi r_{2}^{2}v_{2}$
$\ Given,\ \ \ r_{1} = \frac{4}{2}\ cm = 0.02\ m,$
$\ \ \ \ \ \ \ \ \ \ \ r_{2} = \frac{2}{2}\ cm = 0.01m,$
${\ \ \ \ \ \ \ \ \ \ v}_{1} = 3{ms}^{- 1}$
$\therefore\ \ \pi(0.02)^{2} \times 3 = \pi(0.01)^{2}v_{2}$
$or\ \ \ \ v_{2} = \left( \frac{0.02}{0.01} \right)^{2} \times 3 = 12\ {ms}^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.