A tank with vertical walls is mounted so that its base is at a height `H` above the horizontal ground. The tank is filled with water to a depth `h`. A hole is punched in the side wall of the tank at a depth `x` below the water surface. To have maximum range of the emerging stream, the value of `x` is

Options

A. `x=2[D(H+D)]^(1//2)`
B. `x=2[D(H-D)]^(1//2)`
C. `x=2(g*D)^(1//2)`
D. None of these

Answer

C

Explanation

The velocity of efflux (v) through an orifice at a depth D below the free surface of the liquid is given by
`v=sqrt(2gD)`
After coming out of the hole, the stream of watermoves along a parabolic path (as in the case of projectile at the highest point in its path), and strikes the ground at C at a distance x from the bottom of the wall
It covers the vertical distance (H-D) in time t
Using s=ut`+1/2at^2`, we get (H-D)=`1/2gt^2`
`:." "t=sqrt(2(H-D)//g)`
`:.` From (1) and (2)
`x=vxxtsqrt((2gDxx2(H-D))/(g))`
`=2sqrt(D(H-D))`