If the container filled with liquid gets accelerated horizontally or vertically, pressure in liquid gets changed. In liquid (`a_(y)`) for calculation of pressure, effective `g` is used. A closed box horizontal base `6 m` by `6 m` and a height `2m` is half filled with liquid. It is given a constant horizontal acceleration `g//2` and vertical downward acceleration `g//2`.
What is the value of vertical acceleration of box for given horizontal acceleration `(g//2)`, so that no part of the bottom of the box is exposed?

Options

A. `g//2` upwards
B. `g//4` downwards
C. `g//4` upwards
D. not possible

Answer

A

Explanation

As the maximum slope of the free surface is `1/3` for the condition of non exposure of the bottom of the box, so
`(a_(x))/(a_(y)+g)=1/3`

As `a_(x)=g/2`
`:. 3a_(x)=a_(y)+g`
`impliesa_(y)=g/2,` upwards