MHT CET · Physics · Mechanical Properties of Fluids
Water is flowing in a conical tube as shown in figure. Velocity of water at area ' \(\mathrm{A}_2\) ' is \(60 \mathrm{~cm} / \mathrm{s}\). The value of ' \(\mathrm{A}_1\) ' and ' \(\mathrm{A}_2\) ' is \(10 \mathrm{~cm}^2\) and \(5 \mathrm{~cm}^2\) respectively. The pressure difference at both the cross-section is
- A \(230 \mathrm{~N} / \mathrm{m}^2\)
- B \(200 \mathrm{~N} / \mathrm{m}^2\)
- C \(135 \mathrm{~N} / \mathrm{m}^2\)
- D \(105 \mathrm{~N} / \mathrm{m}^2\)
Answer & Solution
Correct Answer
(C) \(135 \mathrm{~N} / \mathrm{m}^2\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
A_1 \times V_1 & =A_2 \times V_2 \\
10 \times V_1 & =5 \times 60 \\
V_1 & =30 \mathrm{~cm} / \mathrm{s}
\end{aligned}\)
From Bernoulli's equation
\(\begin{aligned}
\left(\mathrm{P}_{\mathrm{A}}-\mathrm{P}_{\mathrm{B}}\right) & =\frac{1}{2} \rho\left(\mathrm{~V}_2^2-\mathrm{V}_1^2\right) \\
& =\frac{1}{2} \times 10^3\left(60^2-30^2\right) \times 10^{-4} \\
& =\frac{1}{2} \times 10^{-1}(3600-900) \\
& =\frac{1}{2} \times 2700 \\
& =1350 \mathrm{dyne} / \mathrm{cm}^2 \\
& =135 \mathrm{~N} / \mathrm{m}^2
\end{aligned}\)
A_1 \times V_1 & =A_2 \times V_2 \\
10 \times V_1 & =5 \times 60 \\
V_1 & =30 \mathrm{~cm} / \mathrm{s}
\end{aligned}\)
From Bernoulli's equation
\(\begin{aligned}
\left(\mathrm{P}_{\mathrm{A}}-\mathrm{P}_{\mathrm{B}}\right) & =\frac{1}{2} \rho\left(\mathrm{~V}_2^2-\mathrm{V}_1^2\right) \\
& =\frac{1}{2} \times 10^3\left(60^2-30^2\right) \times 10^{-4} \\
& =\frac{1}{2} \times 10^{-1}(3600-900) \\
& =\frac{1}{2} \times 2700 \\
& =1350 \mathrm{dyne} / \mathrm{cm}^2 \\
& =135 \mathrm{~N} / \mathrm{m}^2
\end{aligned}\)
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