Q. Water is in streamline flow along a horizontal pipe with nonuniform cross-section. At a point in the pipe where the area of cross-section is $10\ cm^{2}$, the velocity of water is $1\ ms^{- 1}$ and the pressure is 2000 Pa. The pressure at another point where the cross-sectional area is $5\ cm^{2}$ is
A) 4000 Pa
B) 2000 Pa
C) 1000 Pa
D) 500 Pa
Explanation
(d) According to equation of continuity
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$\therefore v_{2} = \frac{A_{1}v_{1}}{A_{2}} = \frac{10cm^{2} \times 1ms^{- 1}}{5cm^{2}} = 2ms^{- 1}$
For a horizontal pipe, according to Bernoulli’s theorem
$P_{1} + \frac{1}{2}\rho v_{1}^{2} = P_{2} + \frac{1}{2}\rho v_{2}^{2}$
$P_{2} = P_{1} + \frac{1}{2}\rho(v_{1}^{2} - v_{2}^{2})$
$= 2000 + \frac{1}{2} \times 10^{3} \times (1^{2} - 2^{2})$
[$\because$ Density of water, $\rho = 10^{3}\ kg/m^{3}$]
$= 2000 - \frac{1}{2} \times 10^{3} \times 3 = 2000 - 1500 = 500Pa$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.