Question

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

• A. 12.5 m
• B. 25 m
• C. 50 m
• D. 75 m

Answer 1

Answer: (C)

Let ${\text{A}}_1$ is cross-sectional area of piston of syringe and ${\text{A}}_{\text{2}}$ the cross-sectional area of nozzle. From principle of continuity for non-viscous liquid. $A_1{v}_1=A_2{v}_2$ $\therefore$ ${10}^{-4}\times \pi r_1^2\times 0.5={10}^{-6}\times \pi r_2^2\times v_2$ where $r_1=$ radius of syringe $r_2=$ radius of nozzle or ${10}^{-4}\times {\left(\frac{1}{2}\right)}^2\times 0.5=1^{-6}\times {\left(\frac{1}{2}\right)}^2{v}_2$ or $u_2=50m{s}^{-1}$ From Torricellis theorem, $h=\frac{1}{2}g{t}^2$ $\therefore$ $5=\frac{1}{2}\times 10\times t^2$ $t=1s$ This is the time taken for the water zet to reach upto the ground. $\therefore$ Horizontal distance $R=v_2\times t=50\times 1=50m$ j ,