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 $v$ , then the speed of ejection of the liquid through the holes will be:

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

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

$\frac{vR^2}{n^3r^2}$

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

Step-by-Step Solution

Identify the Principle: The problem is based on the Equation of Continuity for an incompressible fluid, which states that the rate of flow (volume flux) remains constant throughout the pipe ( $A_1 v_1 = A_2 v_2$ ) .
Determine Inlet Conditions:

Radius of the main tube = $R$
Area of cross-section ( $A_1$ ) = $\pi R^2$
Speed of liquid ( $v_1$ ) = $v$
Volume flow rate entering = $A_1 v_1 = \pi R^2 v$

Determine Outlet Conditions:

Number of holes = $n$
Radius of each hole = $r$
Total Area of cross-section ( $A_2$ ) = $n \times \pi r^2$
Speed of ejection ( $v_2$ ) = ?
Volume flow rate exiting = $A_2 v_2 = n \pi r^2 v_2$

Apply Conservation of Mass: 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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