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 ,You need to login to perform this action.
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