KCET · Physics · Mechanical Properties of Fluids
Water is in streamline flows along a horizontal pipe with non-uniform cross-section. At a point in the pipe where the area of cross-section is \(10 \mathrm{~cm}^{2}\), the velocity of water is \(1 \mathrm{~ms}^{-1}\) and the pressure is \(2000 \mathrm{~Pa}\). The pressure at another point where the cross-sectional area is \(5 \mathrm{~cm}^{2}\) is
- A \(4000 \mathrm{~Pa}\)
- B \(2000 \mathrm{~Pa}\)
- C \(1000 \mathrm{~Pa}\)
- D \(500 \mathrm{~Pa}\)
Answer & Solution
Correct Answer
(D) \(500 \mathrm{~Pa}\)
Step-by-step Solution
Detailed explanation
According to equation of continuity
\(\mathrm{A}_{1} \mathrm{v}_{1} =\mathrm{A}_{2} \mathrm{v}_{2} \)
\(10 \times 1 =5 \times \mathrm{v}_{2} \)
\(\mathrm{v}_{2} =2 \mathrm{~m} / \mathrm{s} \)
\(\frac{\mathrm{p}_{1}}{\rho}+\frac{\mathrm{v}_{1}^{2}}{2} =\frac{\mathrm{p}_{2}}{\rho}+\frac{\mathrm{v}_{2}^{2}}{2} \)
\(\frac{2000}{1000}+\frac{1^{2}}{2} =\frac{\mathrm{p}_{2}}{1000}+\frac{2^{2}}{2} \)
\(\mathrm{p}_{2} =500 \mathrm{~Pa}\)
\(\mathrm{A}_{1} \mathrm{v}_{1} =\mathrm{A}_{2} \mathrm{v}_{2} \)
\(10 \times 1 =5 \times \mathrm{v}_{2} \)
\(\mathrm{v}_{2} =2 \mathrm{~m} / \mathrm{s} \)
\(\frac{\mathrm{p}_{1}}{\rho}+\frac{\mathrm{v}_{1}^{2}}{2} =\frac{\mathrm{p}_{2}}{\rho}+\frac{\mathrm{v}_{2}^{2}}{2} \)
\(\frac{2000}{1000}+\frac{1^{2}}{2} =\frac{\mathrm{p}_{2}}{1000}+\frac{2^{2}}{2} \)
\(\mathrm{p}_{2} =500 \mathrm{~Pa}\)
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