TS EAMCET · Physics · Mechanical Properties of Fluids
In a horizontal tube the water pressure changes by \(1500 \mathrm{Nm}^{-2}\) between \(A\) and \(B\) as shown in figure below. The cross-sectional areas at \(A\) and \(B\) of the tube are \(40 \mathrm{~cm}^2\) and \(20 \mathrm{~cm}^2\), respectively. Find the rate of flow of water through the tube.
- A \(1000 \mathrm{~cm}^3 \mathrm{~s}^{-1}\)
- B \(2000 \mathrm{~cm}^3 \mathrm{~s}^{-1}\)
- C \(4000 \mathrm{~cm}^3 \mathrm{~s}^{-1}\)
- D \(6000 \mathrm{~cm}^3 \mathrm{~s}^{-1}\)
Answer & Solution
Correct Answer
(C) \(4000 \mathrm{~cm}^3 \mathrm{~s}^{-1}\)
Step-by-step Solution
Detailed explanation
\(p_A-p_B=\frac{1}{2} \rho\left(v_B^2-v_A^2\right)\)Given that, pressure difference \(p_A-p_B=1500 Nm ^{-2}\)....(i) By using Bernoulli's equation, \(p_A+\frac{1}{2} \rho v_A^2=p_B+\frac{1}{2} \rho v_B^2 \) \(p_A-p_B=\frac{1}{2} \rho\left(v_B^2-v_A^2\right)\)...(ii) Density of…
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