question_answer

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

Correct Answer: C

Solution :

  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 ,