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Question

Consider a water jar of radius R that has water filled up to height H and is kept on a stand of height h (see figure). Through a hole of radius r (rR) at its bottom, the water leaks out and the stream of water coming down towards the ground has a shape like a funnel as shown in the figure. If the radius of the cross-section of water stream when it hits the ground is x. Then:
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A
x=r(HH+h)14
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B
x=r(HH+h)2
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C
x=r(HH+h)12
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D
x=r(HH+h)
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Solution

The correct option is A x=r(HH+h)14
By energy conservation,

12ρv2112ρv22=ρgHρg(H+h)

Hence, v2v1=H+hH,

Also, since by mass conservation, A1v1=A2v2,

r2v1=x2v2

From the two relations, we get, xr=(HH+h)14.

x=r(HH+h)14.

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