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Question

Consider a horizontally oriented syringe containing water located at a height of 1.25m above the ground. The diameter of the plunger is 8mm and the diameter of the nozzle is 2mm. The plunger is pushed with a constant speed of 0.25m/s. Find the horizontal range of water stream on the ground. Take g=10m/s2
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Solution

From equation of continuity,
A1v1=A2v2
A1A2v1=(πr21πr22)v1
or v1(Dd)2v1=(8×1032×103)2×0.25m/s
=4m/s (horizontal)
Vertical component of the velocity is zero.
Now, H=12gt2
t=2Hg
Range is given by R=v2t=v2
2Hg=4×2×1.2510=2m

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