Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

The cylindrical tube of a spray pump has radius $R$ , one end of which has $n$ fine holes, each of radius $r$ . If the speed of the liquid in the tube is $v$ , the speed of the ejection of the liquid through the holes is:

$\frac{vR^2}{n^2r^2}$

$\frac{vR^2}{nr^2}$

$\frac{vr^2}{n^2R^2}$

$\frac{v^2R}{nr}$

Step-by-Step Solution

Principle: According to the Equation of Continuity for an incompressible fluid, the volume flow rate (flux) entering the tube must equal the total volume flow rate exiting through the holes ( $A_1 v_1 = A_2 v_2$ ).
Inlet Parameters:

Radius of the tube = $R$
Cross-sectional area of the tube ( $A_1$ ) = $\pi R^2$
Speed of liquid in the tube ( $v_1$ ) = $v$
Flow rate in = $A_1 v_1 = \pi R^2 v$

Outlet Parameters:

Number of holes = $n$
Radius of each hole = $r$
Total cross-sectional area of $n$ holes ( $A_2$ ) = $n \times \pi r^2$
Speed of ejection ( $v_2$ ) = ?
Flow rate out = $A_2 v_2 = n \pi r^2 v_2$

Calculation: Equating flow rates: $\pi R^2 v = n \pi r^2 v_2$ $v_2 = \frac{\pi R^2 v}{n \pi r^2}$ $v_2 = \frac{v R^2}{n r^2}$

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