A liquid kept in a container has a horizontal acceleration 'a' as shown in figure.

Using the pressure equation along the path ABC find the angle `theta`.

Answer

A

Explanation

Writing pressure equation along the path ABC we have
`p_(A)+rho gh-rho ax=pc (rho=`density of liquid)
Point A and C are open to atmosphere.
`:. p_(A)=p_(C)=p_(0)=`atmosphere pressure
` implies rhogh-rho ax=0` or `(h)/(x)=(a)/(g)`
`implies tan theta=(a)/(g)` or `theta=tan^.(-1)(a)/(g)`.