Question

Consider a horizontally oriented syringe containing water located at a height of $1.25m$ above the ground. The diameter of the plunger is $8mm$ and the diameter of the nozzle is $2mm$. The plunger is pushed with a constant speed of $0.25m/s$. Find the horizontal range of water stream on the ground. Take $g=10m/s^2$

Answer 1

From equation of continuity,

$A_1{v}_1=A_2{v}_2$

$\Rightarrow \frac{A_1}{A_2}{v}_1=\left(\frac{\pi r_1^2}{\pi r_2^2}\right)v_1$

or $v_1{\left(\frac{D}{d}\right)}^2{v}_1={\left(\frac{8\times {10}^{-3}}{2\times {10}^{-3}}\right)}^2\times 0.25m/s$

$=4m/s$ (horizontal)

Vertical component of the velocity is zero.

Now, $H=\frac{1}{2}g{t}^2$

$\Rightarrow t=\sqrt{\frac{2H}{g}}$

Range is given by $R=v_{2}t=v_2$

$\sqrt{\frac{2H}{g}}=4\times \sqrt{\frac{2\times 1.25}{10}}=2m$