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Question: A liquid of density \(\rho\) is coming out of the hose pipe of radius a with horizontal speed v and ...

A liquid of density ρ\rho is coming out of the hose pipe of radius a with horizontal speed v and hits a mesh. 50%50\% of the liquid passes through the mesh unaffected. 25% loses all of its momentum and 25% comes back with the same speed. The resultant pressure on the mesh will be:
A. pv2p{v^2}
B. 34pv2\dfrac{3}{4}p{v^2}
C. 12pv2\dfrac{1}{2}p{v^2}
D. 14pv2\dfrac{1}{4}p{v^2}

Explanation

Solution

In this question, we need to determine the resultant pressure on the mesh such that the 50%50\% of the liquid passes through the mesh unaffected out of which 25% loses all of its momentum and 25% comes back with the same speed. For this we will use the relation between the mass per unit volume of the liquid, density of the liquid, area occupied by the liquid and the area occupied by the liquid.

Formula used:
Mass per unit volume of the liquid=ρAV = \rho AV, where ρ\rho the density of the liquid is, AA is the area occupied by the liquid, VV is the velocity of the liquid.
Pressure is the ratio of the net force per unit area. Mathematically, P=FnetAP = \dfrac{{{F_{net}}}}{A}

Complete step by step answer:
Density of the liquid =ρ= \rho
Radius of the hose pipe =a = a
We know mass per unit volume of the liquid is given by the formula =ρAV = \rho AV
It is said that liquid coming out of the hose pipes falls vertically on a mesh through which only 50%50\% of the liquid passes and 25% loses all of its momentum and 25% comes back with the same speed as shown in the figure below

Now since only 25% of liquid hits the mesh where the final velocity comes to zero, so the total force by the liquid hitting the mesh will be

{F_1} = 25\% \left( {\rho AV} \right)\left( {V - 0} \right) \\\ = \dfrac{1}{4}\rho A{V^2} - - (ii) \\\ \end{gathered} $$ When 25% of liquid hits the mesh then 25% comes back with the same speed so the final velocity will be v in negative direction, hence we can write

{F_2} = 25% \left( {\rho AV} \right)\left( {V - \left( { - V} \right)} \right) \\
= 25% \left( {\rho AV} \right)2V \\
= \dfrac{2}{4}\rho A{V^2} \\
= \dfrac{1}{2}\rho A{V^2} - - (iii) \\

Now the resultant pressure of the force $${F_1}$$and $${F_2}$$ on the area A will be $$P = \dfrac{{{F_1} + {F_2}}}{A} - - (iv)$$, hence by substituting the values of force $${F_1}$$and $${F_2}$$from equation (ii) and (iii) will be equal to

P = \dfrac{{{F_1} + {F_2}}}{A} \\
= \dfrac{{\dfrac{1}{4}\rho A{V^2} + \dfrac{1}{2}\rho A{V^2}}}{A} \\
= \dfrac{{\rho A{V^2}\left( {\dfrac{1}{4} + \dfrac{1}{2}} \right)}}{A} \\
= \dfrac{3}{2}\rho {V^2} \\

Hence the resultant pressure on the mesh will be $$ = \dfrac{3}{2}\rho {V^2}$$. **So, the correct answer is “Option B”.** **Note:** It should be noted here that the force must be added arithmetically and not algebraically, which implies that the direction of the force in the pressure formula plays a very important role. Pressure is the ratio of the net force per unit area.