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 $ 0.5\text{ }{{m}^{-1}}, $ the horizontal distance travelled by the liquid jet is $ (g=10\text{ }m{{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}}=50\,m{{s}^{-1}} $ From Torricellis theorem, $ h=\frac{1}{2}g{{t}^{2}} $ $ \therefore $ $ 5=\frac{1}{2}\times 10\times {{t}^{2}} $ $ t=1\,s $ 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=50\,m $ j ,