A tank is filled upto a height `h` with a liquid and is placed on platform of height `h` from the ground. To get maximum range `x_(m)` a small hole is punched at a distance of `y` from the free surface of the liquid. Then

Answer

A, C

Explanation

Velocity of liquid through orifice `v=sqrt(2gy)`
and time taken by liquid to reach the ground
`t=sqrt((2(h+h-y))/(g))=sqrt((2(2h-y))/(g))`
`because` Horizontal distance covered by liquid.
`x=v.t`
`impliesx^(2)=4y(2h-y)`
for x to be maximum, `(d)/(dy)(x^(2))=0`