Question

One end of light inelastic string is tied to a helium filled balloon and its other end is tied to bottom of a water filled container at point $O$. The container lies on a fixed horizontal surface and is pulled horizontally towards right with constant 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. ($g$ is acceleration due to gravity)

• A. $\theta ={\tan }^{-1}\left(a/2g\right)$ and string will be on left of vertical line passing through $O$
• B. $\theta ={\tan }^{-1}\left(a/2g\right)$ and string will be on right of vertical line passing through $O$
• C. $\theta ={\tan }^{-1}\left(a/g\right)$ and string will be on left of vertical line passing through $O$
• D. $\theta ={\tan }^{-1}\left(a/g\right)$ and string will be on right of vertical line passing through $O$.

Answer 1

The correct option is D $\theta ={\tan }^{-1}\left(a/g\right)$ and string will be on right of vertical line passing through $O$.

Draw a free body diagram of given problem.

Calculate angle through which string will be inclined
Let the density of gas and water will be $p_g$ and $p_w$ respectively and volume enclosed by balloon will be $V$.
In equilibrium state

$\tan \theta =\frac{F_x}{F_y}=\frac{a}{g}$