JEE Mains · Physics · STD 11 - 9.1 fluid mechanics
A liquid of density \(\rho \) is coming out of a hose pipe of radius \(a\) with horizontal speed \(v\) and hits a mesh. \(50\%\) of the liquid passes through the mesh unaffected. \(25\%\) looses all of its momentum and \(25\%\) comes back with the same speed. The resultant pressure on the mesh will be
- A \(\frac{1}{4}\,\rho {v^2}\)
- B \(\frac{3}{4}\,\rho {v^2}\)
- C \(\frac{1}{2}\,\rho {v^2}\)
- D \(\rho {v^2}\)
Answer & Solution
Correct Answer
(B) \(\frac{3}{4}\,\rho {v^2}\)
Step-by-step Solution
Detailed explanation
\(F = \frac{1}{4} \times \rho A{v^2} + \frac{1}{4}2\rho a{v^2}\) \( \Rightarrow \frac{F}{A} = \frac{3}{4}\rho A{v^2} = \rho \)
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