Consider a parabola `y=Ax^(2)+B, -x_(0) le x le x_(0)`.

If this curve is rotated about y axis, we get a paraboloid surface. The volume below this surface & above x–z plane is given by `V=(piAx_(0)^(4))/2+piBx_(0)^(2)=("volume of cylinder ABCD")/2+"Volume of cylinder CDEF"`
Use the above result to answer following question.
In the above situation (i.e. fully filled cylinder of radius ‘a’ and height ‘h’), if liquid is rotated at twice the angular velocity found in previous problem, then the amount of liquid left (after spillage) in the cylinder will be :

Options

A. `(pia^(2)h)/2`
B. `(pia^(2)h)/4`
C. `(pia^(2)h)/8`
D. `(pia^(2)h)/3`

Explanation

Some base area will be visible. Let radius of visible base be 'r' origin shifts below base.
`h'=(omega^(2)x^(2))/(2g)=(8gh)/(a^(2))xx(a^(2))/(2g)=4h`
Put x=r and y=h'-h=3h
`3h=(8gh)/(a^(2))xx(r^(2))/(2g)rArr r^(2)=3/4 a^(2)`
`V_("left")=(pia^(2)h^(1))/2-[pi(a^(2)-r^(2))(h'-h)+(pir^(2)(h'-h))/2]=(pia^(2)h)/8`