Question

A syringe of diameter 1 cm having a nozzle of diameter 1 mm, is placed horizontally at a height 5 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 $0 \cdot 5\dfrac{m}{s}$. The horizontal distance travelled by the liquid jet is $\left( {g = 10\dfrac{m}{{{s^2}}}} \right)$:
A) 90 m
B) 80 m
C) 60 m
D) 50 m

Answer 1

Hint: Volumetric flow is the rate of flow of the volume of liquid. Range is the horizontal distance that a body travels having some initial velocity the body can be on the ground initially and the body can also be at some height and can travel some distance.

Formula used:
The formula of the second equation of the Newton’s law of motion is given by,
$s = ut + \dfrac{1}{2}a{t^2}$
Where s is the distance u is the initial velocity t is the time taken and a is the acceleration due to gravity.

Complete step by step solution:
It is given in the problem that a syringe of diameter 1 cm having a nozzle of diameter 1 mm is placed horizontally at a height 5 m and a fluid which is in compressible is pushed out of the syringe which travels a horizontal distance having initial speed of $0 \cdot 5\dfrac{m}{s}$, also acceleration due to gravity is $g = 10\dfrac{m}{{{s^2}}}$ and we need to find the range of the jet.
Since the product of area and velocity is always constant therefore,
${A_1}{v_1} = {A_2}{v_2}$
Where ${A_1}$ is the area of cross section of the syringe ${A_2}$ is the area of the nozzle of the syringe ${v_1}$ is the velocity of the fluid in the syringe and ${v_2}$ is the velocity of the jet.
$ \Rightarrow {A_1}{v_1} = {A_2}{v_2}$
Replace the value of the area of the springe and area of the nozzle of the syringe.
$ \Rightarrow {A_1}{v_1} = {A_2}{v_2}$
Also the speed of the piston is equal to $0 \cdot 5\dfrac{m}{s}$,
$ \Rightarrow \left[ {\dfrac{\pi }{4}{{\left( {0 \cdot 01} \right)}^2}} \right] \times \left( {0 \cdot 5} \right) = \left[ {\dfrac{\pi }{4}{{\left( {0 \cdot 001} \right)}^2}} \right]{v_2}$
$ \Rightarrow {\left( {0 \cdot 01} \right)^2} \times \left( {0 \cdot 5} \right) = {\left( {0 \cdot 001} \right)^2}{v_2}$
$ \Rightarrow {v_2} = \dfrac{{{{\left( {0 \cdot 01} \right)}^2} \times \left( {0 \cdot 5} \right)}}{{{{\left( {0 \cdot 001} \right)}^2}}}$
$ \Rightarrow {v_2} = 50\dfrac{m}{s}$
The initial velocity of the jet is ${v_2} = 50\dfrac{m}{s}$.
For vertical motion,
Applying the second relation of the Newton’s law motion.
$s = ut + \dfrac{1}{2}a{t^2}$
Where s is the distance u is the initial velocity t is the time taken and a is the acceleration due to gravity.
As the velocity of the jet in y-direction is zero and g will be positive.
$ \Rightarrow s = ut + \dfrac{1}{2}a{t^2}$
$ \Rightarrow s = \dfrac{1}{2}g{t^2}$
$ \Rightarrow s = \dfrac{1}{2}\left( {10} \right){t^2}$
$ \Rightarrow s = 5{t^2}$
As the height is 5m.
$ \Rightarrow s = 5{t^2}$
$ \Rightarrow 5 = 5{t^2}$
$ \Rightarrow t = 1s$.
Since,
${{distance}} = {{speed \times time}}$
$ \Rightarrow {\text{distance}} = {{speed \times time}}$
As the speed of the jet is $50\dfrac{m}{s}$ and time taken is 1s.
$ \Rightarrow {\text{distance}} = \left( {50{{ \times 1}}} \right)$
$ \Rightarrow {\text{distance}} = 50m$.

The range of the jet is 50m. The correct answer for this problem is option (D).

Note: The product of area of the cross section and the velocity is known to be constant for an incompressible fluid this phenomenon is known as volumetric flow. The time taken in the horizontal and vertical direction to reach the ground will be the same.