A fluid container is containing a liquid of density `rho` is accelerating upward with acceleration a along the inclined plane of inclination `alpha` as shown in the figure. Then the angle of inclination `theta` of free surface is

Options

A. `tan^(-1)[(a)/(gcosalpha)]`
B. `tan^(-1)[(a+gsinalpha)/(gcosalpha)]`
C. `tan^(-1)[(a-gsinalpha)/(g(1+cosalpha))]`
D. `tan^(-1)[(a-gsinalpha)/(g(1-cosalpha))]`

Answer

B

Explanation

Apply pseudo force on a particle of mass `m`. Net force along the surface is zero.
`macostheta=mgcos[90-(theta-alpha)]`
`(a)/(g)costheta=gsin(theta-alpha)`
`(a)/(g)=(sinthetacosalpha)/(costheta)-(costhetasinalpha)/(costheta)`
`tantheta=((a)/(g)+sinalpha)/(cosalpha)=(a+gsinalpha)/(gcosalpha)`