Question

A syringe of diameter $1\text{cm}$ having a nozzle of diameter $1\text{mm}$, is placed horizontally at a height $5\text{m}$ from the ground. An incompressible, non-viscous liquid is filled in the syringe and the liquid is compressed by moving the piston at a speed of $0.5{\text{ms}}^{-1}$. The horizontal distance travelled by the liquid jet is $(g=10{\text{ms}}^{-2})$

• A. $12.5\text{m}$
• B. $25\text{m}$
• C. $50\text{m}$
• D. $67.5\text{m}$

Answer 1

The correct option is C $50\text{m}$

Let $A_1$ be the cross-sectional area of the piston of the syringe and $A_2$ be the cross-sectional area of the nozzle.
From principle of continuity for non-viscous liquid,
$A_1{v}_1=A_2{v}_2$
$\Rightarrow \pi r_1^2\times 0.5=\pi r_2^2\times v_2...\left(i\right)$
Given, Speed at which liquid is pushed by piston $v_1=0.5{\text{ms}}^{-1}$
Also, $r_1=0.5\times {10}^{-2}\text{m}\to$ radius of syringe, $r_2=0.5\times {10}^{-3}\text{m}\to$ radius of nozzle
$\Rightarrow \pi \times {\left(0.5\right)}^2\times {10}^{-4}\times 0.5=\pi \times {\left(0.5\right)}^2\times {10}^{-6}\times v_2$
$\Rightarrow v_2=\frac{100}{2}=50{\text{ms}}^{-1}$
The liquid coming out of nozzle will follow the path of a horizontal projectile,
$u_x=v_2=50{\text{ms}}^{-1},u_y=0$
Applying kinematic equation in vertical direction,
$h=\frac{1}{2}g{t}^2$
$5=\frac{1}{2}\times 10\times t^2$
$\therefore t=1\text{s}$
This is the time taken by the water jet to reach the ground. Horizontal distance covered will be,
$R=u_x\times t=v_2\times t$
$\therefore R=50\times 1=50\text{m}$