A tank is filled upto a height h with a liquid and is placed on a 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

Options

A. `x_(m)=2h`
B. `x_(m)=1.5h`
C. `y=h`
D. `y=0.75h`

Answer

A, C

Explanation

`x="velocity of efflux"xx"time of flight"`
`x=vxxT" "impliesx=sqrt(2gy)sqrt((2(h+h-y))/(g))impliesx=2sqrt(y(2h-y))`
for x to be max.
`(d)/(dx)y(2h-y)=0implies " "y=h`
`x_("max") = 2h`