1.

A tank, which is open at the top, contains a liquid up to a height H. A small hole is made in the side of the tank at a distance y below the liquid surface. The liquid emerging from the hole lands at a distance x from the tank :

Answer»

`2sqrt(h(H-h))`
`sqrt(h(H-h))`
`2sqrt((h^(2))/(H-h))`
`sqrt((h^(2))/(H-h))`

Solution :Velocity with which the liquid leaves the hole, `v=sqrt(2gh)`
Time TAKEN by any particle of the liquid to travel from the hole to the GROUND, `t=sqrt((2(H-h))/(g))`
Therefore, `L=vt=(sqrt(2gh))(sqrt((2(H-h))/(g)))=2sqrt(h(H-h))`


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