91Ó°ÊÓ

Tension of string when in air$\left(T_{air}\right)=39.2N$

Tension of the string when in water$\left(T_{water}\right)=28.4N$

Tension of string when in unknown liquid$\left(T_{liquid}\right)=21.5N$

Density of water$\left({\mathrm{ÒÏ}}_{\mathrm{w}}\right)=10000\mathrm{kg}/{\mathrm{m}}^3$

The buoyant force can be defined as the upward pull of a fluid on a solid mass in space. In fluids, Archimedes' principle states that a buoyant force equals the object's weight.

Tension on string is acting upwards and weight of string is acting downwards. As rock is suspended as well as it is at rest total net force will be zero.

$$\begin{gathered} \underset{}{∑}F=T_{air}-W=0 \\ W=T_{air} \end{gathered}$$

Tension on string is acting upwards and weight of string is acting downwards as the rock is immersed in water and is also in rest, so total net force will be zero.

$\begin{array}{r}\mathrm{Î\pounds ¹ó}={\mathrm{T}}_{\text{water}}−\mathrm{W}+\mathrm{B}=0\\ \text{As,}\mathrm{W}={\mathrm{T}}_{\text{sir}}\\ {\mathrm{T}}_{\text{water}}−{\mathrm{T}}_{\text{air}}+{\mathrm{ÒÏ}}_{\text{watter}}{\mathrm{V}}_{\text{rook}}\mathrm{g}=0\\ {\mathrm{V}}_{\text{rock}}=\frac{{\mathrm{T}}_{\text{air}}−{\mathrm{T}}_{\text{water}}}{{\mathrm{ÒÏ}}_{\text{water}}\mathrm{g}}\\ {\mathrm{V}}_{\text{rock}}=\frac{39.2−28.4}{\left(1000\right)(9.81)}=0.0011{\mathrm{m}}^3\\ \mathrm{Tension}\mathrm{on}\mathrm{string}\mathrm{is}\mathrm{acting}\mathrm{upwards}\mathrm{and}\mathrm{weight}\mathrm{of}\mathrm{string}\mathrm{is}\mathrm{acting}\mathrm{downwards}.\mathrm{As}\mathrm{rock}\mathrm{is}\mathrm{in}\mathrm{unknown}\mathrm{liquid}\mathrm{and}\mathrm{also}\mathrm{in}\mathrm{rest},\mathrm{so}\mathrm{total}\mathrm{net}\mathrm{force}\mathrm{will}\mathrm{be}\mathrm{zero}.\\ \mathrm{Î\pounds ¹ó}={\mathrm{T}}_{\text{liquid}}−\mathrm{W}+\mathrm{B}=0\\ \text{As,}\mathrm{W}={\mathrm{T}}_{\text{sir}}\\ {\mathrm{T}}_{\text{iqquid}}−{\mathrm{T}}_{\text{air}}+{\mathrm{ÒÏ}}_{\text{iqquid}}{\mathrm{V}}_{\text{rook}}\mathrm{g}=0\\ {\mathrm{ÒÏ}}_{\mathrm{l}}=\frac{{\mathrm{T}}_{\text{air}}−{\mathrm{T}}_{\text{iqquid}}}{{\mathrm{V}}_{\text{rook}}\mathrm{g}}\\ =\frac{39.2−21.5}{(0.0011)(9.81)}=1.64×{10}^3\mathrm{kg}/{\mathrm{m}}^3\end{array}$