Q. A manometer connected to a close tap reads $3.5 \times 10^{5}\ {Nm}^{- 2}$. When the valve is opened, the reading of manometer falls to$\ 3.0 \times 10^{5}\ {Nm}^{- 2}$ , then velocity of flow of water is
A) $100\ {ms}^{- 1}$
B) $10\ {ms}^{- 1}$
C) $1\ {ms}^{- 1}$
D) $10\sqrt{10}\ \ {ms}^{- 1}$
Explanation
(b) Bernoulli’s theorem for unit mass of liquid
$\frac{P}{\rho} + \frac{1}{2}v^{2}\ constant$
As the liquid starts flowing , it pressure energy decreases
$\frac{1}{2}v^{2} = \frac{P_{1} - P_{2}}{\rho}$
$\Longrightarrow \ \ \ \ \frac{1}{2}v^{2} = \frac{3.5 \times 10^{5} - 3 \times 10^{5}}{10^{3}}$
$\Longrightarrow \ \ v^{2} = \frac{2 \times 0.5 \times 10^{5}}{10^{3}}$
$\Longrightarrow v^{2} = 100$
$\Longrightarrow v = 10\ {ms}^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.