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 0.5 ms^-1, the horizontal distance travelled by the liquid jet is (g = 10 ms^-2)

Let A₁ is cross – sectional area of piston of syringe and A₂ the cross – sectional area of nozzle.

From principle of continuity for non – viscous liquid.

 $$\begin{gathered} A_1{v}_1=A_2{v}_2 \\ \Rightarrow \pi r_1^2\times 0.5=\pi r_2^2\times v_2 \end{gathered}$$

Were r₁ – radius of syringe, r₂ – radius of nozzle

$$\begin{gathered} {10}^{-4}\times {\left(\frac{1}{2}\right)}^2\times 0.5={10}^{-6}\times {\left(\frac{1}{2}\right)}^2{v}_2; \\ {v}_2=50m{s}^{-1} \end{gathered}$$

From Torricelli’s theorem,

$$\begin{gathered} h=\frac{1}{2}g{t}^2, \\ 5=\frac{1}{2}\times 10\times t^2, \\ t=1s \end{gathered}$$

This is the times taken for the water jet to reach up to the ground. Horizontal distance $R=v_2\times t=50\times 1=50m$