KCET · Physics · Mechanical Properties of Fluids
Water flows through a horizontal pipe of varying cross-section at a rate of \(0.314 \mathrm{~m}^3 \mathrm{~s}^{-1}\). The velocity of water at a point where the radius of the pipe is 10 cm is
- A \(0.1 \mathrm{~ms}^{-1}\)
- B \(1 \mathrm{~ms}^{-1}\)
- C \(10 \mathrm{~ms}^{-1}\)
- D \(100 \mathrm{~ms}^{-1}\)
Answer & Solution
Correct Answer
(C) \(10 \mathrm{~ms}^{-1}\)
Step-by-step Solution
Detailed explanation
Given, flow rate, \(Q=0.314 \mathrm{~m}^3 \mathrm{~s}^{-1}\)
\(r=10 \mathrm{~cm}=0.1 \mathrm{~m}\)
\(=10^{-1} \mathrm{~m}\)
Since, \(\quad Q=A v\)
\(0.314=\pi r^2 v\)
\(0.314=314 \times 10^{-2} v\)
\(\Rightarrow \quad v=10 \mathrm{~m} / \mathrm{s}\)
\(r=10 \mathrm{~cm}=0.1 \mathrm{~m}\)
\(=10^{-1} \mathrm{~m}\)
Since, \(\quad Q=A v\)
\(0.314=\pi r^2 v\)
\(0.314=314 \times 10^{-2} v\)
\(\Rightarrow \quad v=10 \mathrm{~m} / \mathrm{s}\)
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