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PART - 2 CH - 9 Mechanical Properties of Fluids — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsPART - 2 CH - 9 Mechanical Properties of Fluids2 Marks
Question
Water of pressure $4 \times 10^4 N / m ^2$ is flowing through a pipe of cross-sectional area 0.02 $m ^2$ with a velocity of $2 m / sec$. If the cross-sectional area of pipe in reduced to $0.01 m^2$ then what will be the value of pressure in the pipe?
✓

Answer

Given :
$\begin{aligned}P_1 & =4 \times 10^{+} N / m^2 \\v_1 & =2 m / sec \\A_1 & =0.02 m^2 \\A_2 & =0.01 m^2 \\P_2 & =? \text { and } v_2=?\end{aligned}$
From continuity equation
$\begin{aligned}A_1 v_1 & =A_2 v_2 \\v_2 & =\frac{A_1 v_1}{A_2}=\frac{0.02 \times 2}{0.01}=4 m / sec\end{aligned}$
From Bernoulli's principle
$\begin{array}{c}P_1+\frac{1}{2} \rho v_1^2=P_2+\frac{1}{2} \rho v_2{ }^2 \\4 \times 10^4+\frac{1}{2} \times 10^3 \times(2)^2=P_2+\frac{1}{2} \times 10^3 \times(4)^2 \\P_2=4 \times 10^4+2 \times 10^3-8 \times 10^3\end{array}$
$\begin{array}{l} P _2^2=34 \times 10^3 \\ P _2=3.4 \times 10^4 N / m ^2\end{array}$
Pressure in the small cross-sectional area of pipe is $3.4 \times 10^4 N / m^2$

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