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Question

A tube has two areas of cross-section as shown in the figure. The diameter of the bigger section is double the diameter of the smaller section. Find the range of water falling on the horizontal surface, if piston is moving with a constant velocity 0.4 m/s and h=2.5 m (g=10 m/s2)


A
1.13 m
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B
1.5 m
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C
2 m
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D
1.51 m
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Solution

The correct option is A 1.13 m
Let d be the diameter of smaller section.
Area of smaller section a=π4d2 and Area of bigger section A=π4(2d)2=π4(4d2)
By continuity equation,
Av1=av2
v2=Aav1=π4(4d2)π4d2×0.4=1.6 m/s
Here, v2 is the horizontal velocity of water coming out of the tube.
Using S=ut+12at2 for vertical direction
h=12gt2[uy=0]
t=2hg=2×2.510=12 s
Horizontal range of water
R=v2×t
=v2×t
=1.6×12
=1.131 m

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