MHT CET · Physics · Mechanical Properties of Fluids
A streamline flow of a liquid of density ' \(\rho\) ' is passing through a horizontal pipe of cross-sectional area \(A_1\) and \(A_2\) at two ends. If the pressure of liquid is ' P ' at a point where flow speed is ' \(v\) ', then pressure at another point where the flow of speed becomes 3 v is
- A \(\mathrm{P}-\frac{3}{4} \rho v^2\)
- B \(P-2 \rho v^2\)
- C \(\mathrm{P}-3 \rho v^2\)
- D \(P-4 \rho v^2\)
Answer & Solution
Correct Answer
(D) \(P-4 \rho v^2\)
Step-by-step Solution
Detailed explanation
Using Bernoulli's equation,
\(\begin{aligned}
& P_1+\frac{1}{2} \rho v_1^2+\rho g h_1=P_2+\frac{1}{2} \rho v_2^2+\rho g h_2 \\
& P_1+\frac{1}{2} \rho v_1^2=P_2+\frac{1}{2} \rho v_2^2 \quad ext{...(given horizontal pipe)}
\end{aligned}\)
Substituting the given values,
\(\begin{aligned} & P+\frac{1}{2} \rho v_1^2=P_2+\frac{1}{2} \rho(3 v)^2 \\ & P_2=P+\left(\frac{1}{2} \rho v^2-\frac{9}{2} \rho v^2\right)=P-4 \rho v^2\end{aligned}\)
\(\begin{aligned}
& P_1+\frac{1}{2} \rho v_1^2+\rho g h_1=P_2+\frac{1}{2} \rho v_2^2+\rho g h_2 \\
& P_1+\frac{1}{2} \rho v_1^2=P_2+\frac{1}{2} \rho v_2^2 \quad ext{...(given horizontal pipe)}
\end{aligned}\)
Substituting the given values,
\(\begin{aligned} & P+\frac{1}{2} \rho v_1^2=P_2+\frac{1}{2} \rho(3 v)^2 \\ & P_2=P+\left(\frac{1}{2} \rho v^2-\frac{9}{2} \rho v^2\right)=P-4 \rho v^2\end{aligned}\)
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