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Question

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)

A
12.5 m
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B
25 m
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C
50 m
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D
67.5 m
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Solution

The correct option is C 50 m
Let A1 be the cross-sectional area of the piston of the syringe and A2 be the cross-sectional area of the nozzle.
From principle of continuity for non-viscous liquid,

A1v1=A2v2

πr21×0.5=πr22×v2 ...(i)

Given, Speed at which liquid is pushed by piston v1=0.5 ms1

Also, r1=0.5×102 m radius of syringe, r2=0.5×103 m radius of nozzle.

π×(0.5)2×104×0.5=π×(0.5)2×106×v2

v2=1002=50 ms1

The liquid coming out of nozzle will follow the path of a horizontal projectile,

ux=v2=50 ms1, uy=0

Applying kinematic equation in vertical direction,

h=12gt2

5=12×10×t2

t=1 s

This is the time taken by the water jet to reach the ground.Horizontal distance covered will be,

R=ux×t=v2×t

R=50×1=50 m

Hence, option (C) is the correct answer.

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