One end of light inelastic string is tied to a helium filled ballloon and its other end is tied bottom of a water filled container at point O. The container lies on a fixed horizontal acceleration of magnitude a. Assuming no relative motion of balloon and water with respect to container, the string will be inclined with vertical line passing through O by an angle (h is acceleration due to gravity)

Options

A. `theta = "tan"^(-1) (a)/(g)`and string will be on right of vertical line passing through O
B. `theta = "tan"^(-1) (g)/(a)` and string will be on right of vertical line passing through O
C. `theta = "tan"^(-1) (a)/(g)` and string will be on left of vertical line passing through O
D. `theta = "tan"^(-1)(g)/(a)` and string will be on left of vertical line passing through O

Answer

A

Explanation

Let the density of gas and water by `rho_(g)` and `rho_(omega)`. The velocity enclosed by balloon is V.
Then the FBD of balloon is as shown
`:.` In equilibrium `tan theta = (F_(x))/(F_(y)) = (a)/(g)`