A cube is floating on mercury with `(1//5)^(th)` of its volume submerged. If suficient water Is added to cover the cube, what fraction of its volume will remain immersed in mercury ? Density of mercury = `13600 kg//m^(3) `.

Explanation

To solve the problem step by step, we will analyze the situation of the cube floating on mercury and then determine the fraction of its volume that remains submerged when water is added.

### Step 1: Understand the initial condition
A cube is floating on mercury with \( \frac{1}{5} \) of its volume submerged. This means if the volume of the cube is \( V \), then the volume submerged in mercury is:
\[ V_{\text{submerged}} = \frac{1}{5} V \]

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