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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 msâˆ’1. The horizontal distance travelled by the liquid jet is (g=10 msâˆ’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 mLet 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 ms−1 Also, r1=0.5×10−2 m→ radius of syringe, r2=0.5×10−3 m→ radius of nozzle. ⇒π×(0.5)2×10−4×0.5=π×(0.5)2×10−6×v2 ⇒v2=1002=50 ms−1 The liquid coming out of nozzle will follow the path of a horizontal projectile, ux=v2=50 ms−1, 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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