Question

A syringe containing water is held horizontally with its nozzle at a height $h=1.25\text{m}$ above the ground as shown in the figure. The diameter of the piston is $5$ times as that of the nozzle. The piston is pushed at a constant speed of $20{\text{cms}}^{-1}$. If $g=10{\text{ms}}^{-2}$, then -

• A. The speed of water emerging from the nozzle is $5{\text{ms}}^{-1}$.
• B. The time taken by water to hit the ground is $0.5\text{s}$.
  
• C. The horizontal range, $R=2.5\text{m}$.
• D. The magnitude of velocity with which the water hits the ground is $5\sqrt{2}{\text{ms}}^{-1}$.

Answer 1

Answer: (A), (B), (C), (D)

The magnitude of velocity with which the water hits the ground is $5\sqrt{2}{\text{ms}}^{-1}$.

Area of piston,
$A=\frac{\pi D^2}{4},D\to$Diameter of piston

Area of nozzle,
$a=\frac{\pi d^2}{4},d\to$Diameter of nozzle

Using equation of continuity,

$AV=au$

$\Rightarrow u=\frac{AV}{a}=\frac{D^2}{d^2}\times V=5^2\times 0.2=5\text{m/s}$

Option $\left(A\right)$ is correct.

Time of flight,

$T=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times 1.25}{10}}=0.5\text{s}$

Option $\left(B\right)$ is correct.

Range,

$R=uT=5\times 0.5=2.5\text{m}$

Option $\left(C\right)$ is correct.

Horizontal velocity,

$v_x=5{\text{ms}}^{-1}$ $[$Remain unchanged$]$

Vertical velocity at $t=0.5\text{s}$,

$v_y=0+a_{y}t$

$\Rightarrow v_y=gT=10\times 0.5=5{\text{ms}}^{-1}$

Further, net speed with which water hits the ground,

$v=\sqrt{5^2+5^2}=5\sqrt{2}\text{m/s}$

Option $\left(D\right)$ is correct.