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

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Answer & Solution

Correct Answer

(B) \(\frac{3}{4}\,\rho {v^2}\)

Detailed explanation

\(F = \frac{1}{4} \times \rho A{v^2} + \frac{1}{4}2\rho a{v^2}\)

\( \Rightarrow \frac{F}{A} = \frac{3}{4}\rho A{v^2} = \rho \)

Same subject

From JEE Mains

List\(-I\)	List\(-II\)
\((a)\) Capacitance, \(C\)	\((i)\) \({M}^{1} {L}^{1} {T}^{-3} {A}^{-1}\)
\((b)\) Permittivity of free space, \(\varepsilon_{0}\)	\((ii)\) \({M}^{-1} {L}^{-3} {T}^{4} {A}^{2}\)
\((c)\) Permeability of free space, \(\mu_{0}\)	\((iii)\) \({M}^{-1} L^{-2} T^{4} A^{2}\)
\((d)\) Electric field, \(E\)	\((iv)\) \({M}^{1} {L}^{1} {T}^{-2} {A}^{-2}\)