TS EAMCET · Physics · Mechanical Properties of Fluids
Water from a tap emerges vertically downwards with initial velocity \(4 \mathrm{~ms}^{-1}\). The cross-sectional area of the tap is \(A\). The flow is steady and pressure is constant throughout the stream of water. The distance \(h\) vertically below the tap, where the cross-sectional area of the stream becomes \(\left(\frac{2}{3}\right) A\), is \(\left(g=10 \mathrm{~m} / \mathrm{s}^2\right)\)
- A \(0.5 \mathrm{~m}\)
- B \(1 \mathrm{~m}\)
- C \(1.5 \mathrm{~m}\)
- D \(2.2 \mathrm{~m}\)
Answer & Solution
Correct Answer
(B) \(1 \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
The equation of continuity \(A_1 v_1=A_2 v_2\) \(A \times 4=\frac{2}{3} A \times v_2\) \(v_2=6 \mathrm{~ms}^{-1}\) From Bernoulli's theorem \(p+\rho g h_1+\frac{1}{2} \rho v_1^2=p+\rho g h_2+\frac{1}{2} \rho v_2^2\) \(g\left(h_1-h_2\right)=\frac{1}{2}\left(v_2^2-v_1^2\right)\)…
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