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A tube has two area of cross-section as ...

A tube has two area of cross-section as shown in figure. The diameters of the tube are 8mm and 2mm. Find range of water falling on horizontal surface, if pistion is moving with a constant velocity of `0.25m//s`, `h=1.25m(g=10m//s^2)`

Text Solution

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The correct Answer is:
B
From law of continuity `A_1v_1=A_2v_2`
Given `A_1=pixx(4xx10^-3m)^2, A_2=pixx(1xx10^-3m)^2`

`v_1=0.25m//s`
`:. v_2=(pixx(4xx10^-3)^2xx0.25)/(pixx(1xx10^-3)^2)=4m//s`
Also, `h=1/2g t^2impliest=sqrt((2h)/(g))`
Horizontal range `x=v_2xxt=v_2sqrt((2h)/(g))=4xxsqrt((2xx1.25)/(10))=2m`
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