91Ó°ÊÓ

The given data can be listed as,

• The mass of the rock is,$m=1.80\mathrm{kg}$.
• The tension in the string attached to the rock is, $T=12.8\mathrm{N}$.

The forces acting on a particular system must always be balanced, as the buoyant force acting on the specific object is in an upward direction and always opposite to gravity.

Apply Archimedes’s principle, and the expression can be given by,

$m={\mathrm{ÒÏ}}_{w}V$

Here,${\mathrm{ÒÏ}}_w$is the density of water, and V is the volume.

The net force can be expressed as,

$T+B=mg$.

Here,$T$is the tension in the string, $V$is the volume of the rock, and ${\mathrm{ÒÏ}}_w$is the density of water.

By given relation, the above expression can be given as,

$$\begin{gathered} T+{\mathrm{ÒÏ}}_{w}Vg=mg \\ V=\frac{\left(mg-T\right)}{{\mathrm{ÒÏ}}_{w}g} \end{gathered}$$

For floating, the equation can be expressed as,

$B=mg$

So, substitute all the values in the above equation.

$$\begin{gathered} \mathrm{ÒÏ}Vg=mg \\ \mathrm{ÒÏ}g=\frac{m}{{\displaystyle \frac{mg-T}{{\mathrm{ÒÏ}}_{w}g}}} \\ \mathrm{ÒÏ}=\frac{m{\mathrm{ÒÏ}}_{w}g}{mg-T} \end{gathered}$$

Substitute all values in the above equation.

$$\begin{gathered} \mathrm{ÒÏ}=\frac{\left(1.80\mathrm{kg}\right)\left(1000\mathrm{kg}/{\mathrm{m}}^3\right)\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)}{\left(1.80\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left({\displaystyle \frac{1\mathrm{N}}{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2}}\right)-12.8\mathrm{N}}\left(\frac{1\mathrm{N}}{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2}\right) \\ =3645\mathrm{kg}/{\mathrm{m}}^2 \end{gathered}$$

Thus, the density of the liquid at which the rock can float is $3645\mathrm{kg}/{\mathrm{m}}^2$.