A rectangular tank of height 10m filled with water, is placed near the bottom of a plane inclined at an angle `30^(@)` with horizontal. At height h form bottom a small hole is made (as shown in figure) such that the stream coming out from hole, strikes the inclined plane normally. Calculate h.

Answer

C

Explanation

`v=sqrt(2g(10-h))`
Component of its velocity parallel to the plane is `v cos 30^(@)`
Let the stream strikes the plane after time t. Then
`0=v cos 30^(@)-g sin 30^(@)t`
`:. t=(v cos 30^(@))/(g)`
Further `x=vt=(v^(2)cot 30^(@))/(g)`
`sqrt(3)y`
or, `(v^(2) cot 30^(@))/(g) =sqrt(3)(h-1/2gt^(2))`
`:. (sqrt(3)v^(2))/(g) =sqrt(3)(h-g/2 (v^(2)cot^(2)30^(@))/(G^2))`
or, `(v^(2))/(g)=h-3/2 (v^(2))/(g)`
`:. 5/2 (v^(2))/(g) =h` or `5(10-h)=h`
`:. h=8.33m`