Question

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

Assuming no relative motion of balloon and water with respect to the container, the string will be inclined with the vertical line passing through O by an angle:

($g$ is acceleration due to gravity)

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

Answer 1

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

Let the density of gas and water by ${\rho }_g$ and ${\rho }_w$.
The volume enclosed by balloon is $V$.
Then the FBD of balloon is as shown.
Hence, in equilibrium: $\tan \theta =\frac{F_x}{F_y}=\frac{a}{g}$