Water flows through a horizontal pipe of variable cross-section at the rate of 20 ~L ~min^ -1 . What will be the velocity of water at a point where diameter is 4 cm ?

Correct answer: A. 0.2639 ~ms^ -1 Volume of the water flowing per second, V=20 Lmin^ -1 =(20 × 1000)/(60 ×(100)^3) ~m^3 ~s^ -1 =(1)/(3) × 10^ -3 ~m^3 ~s^ -1 Radius of the pipe, r=(4)/(2)=2 ~cm=0.02 ~m Area of cross-section, r=π r^2=(22)/(7) ×(0.02)^2 ~m^2 Let v be the velocity of the flow of water at the given point. Clearly, V=Av or (1)/(3) × 10^ -3 =(22)/(7) ×(0.02)^2 × v or v= 7 × 10^ -3 3 × 22 ×(0.02)^2 =0.2639 ~ms^ -1

JIPMER 2018

Exam: JIPMER
Year: 2018
Subject: Physics
Chapter: Mechanical Properties of Fluids

JIPMER 2018 · Physics · Mechanical Properties of Fluids

Water flows through a horizontal pipe of variable cross-section at...

Water flows through a horizontal pipe of variable cross-section at the rate of

20 \mathrm{~L} \mathrm{~min}^{-1}

. What will be the velocity of water at a point where diameter is 4 cm ?

Options

A $0.2639 \mathrm{~ms}^{-1}$
B $0.5639 \mathrm{~ms}^{-1}$
C $0.4639 \mathrm{~ms}^{-1}$
D $0.3639 \mathrm{~ms}^{-1}$

Correct answer

A. $0.2639 \mathrm{~ms}^{-1}$

Step-by-step solution

Volume of the water flowing per second,

\mathrm{V}=20 \mathrm{Lmin}^{-1}=\frac{20 \times 1000}{60 \times(100)^3} \mathrm{~m}^3 \mathrm{~s}^{-1}

=\frac{1}{3} \times 10^{-3} \mathrm{~m}^3 \mathrm{~s}^{-1}

Radius of the pipe,

\mathrm{r}=\frac{4}{2}=2 \mathrm{~cm}=0.02 \mathrm{~m}

Area of cross-section,

\mathrm{r}=\pi \mathrm{r}^2=\frac{22}{7} \times(0.02)^2 \mathrm{~m}^2

Let

v

be the velocity of the flow of water at the given point. Clearly,

\mathrm{V}=\mathrm{Av}

\quad \frac{1}{3} \times 10^{-3}=\frac{22}{7} \times(0.02)^2 \times v

\quad \mathrm{v}=\frac{7 \times 10^{-3}}{3 \times 22 \times(0.02)^2}=0.2639 \mathrm{~ms}^{-1}

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