In Fig. 14-39a, a rectangular block is gradually pushed face-down into a liquid. The block has height $\mathbf{d}$; on the bottom and top the face area is $\mathbf{A}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{67}\mathbf{}{\mathbf{cm}}^{\mathbf{2}}$. Figure 14-39b gives the apparent weight ${\mathbf{\text{W}}}_{\mathbf{app}}$of the block as a function of the depth h of its lower face. The scale on the vertical axis is set by ${\mathbf{W}}_{\mathbf{s}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{20}\mathbf{}\mathbf{N}$. What is the density of the liquid?

• Area of the rectangular block,$\mathrm{A}=5.67{\mathrm{cm}}^2\text{or}5.67\times {10}^{-4}{\mathrm{m}}^2$

• Weight on the scale of the vertical axis,$W_s=0.20\mathrm{N}$

After reading the graph, we can determine the value of the height and buoyant force, and the volume can be calculated with the help of area and height.

Formula:

The buoyant force applied on a body by fluid, $\mathrm{F}=\mathrm{\rho gV}$(i)

Volume of the body,$\mathrm{V}=\mathrm{Area}\times \mathrm{Height}$(ii)

From the figure, we can say that the value of${\mathrm{W}}_{\mathrm{app}}$ at $\mathrm{h}=0$would give us the real weight of the object, and the value of ${\mathrm{W}}_{\mathrm{app}}$at $\mathrm{h}=1.5\mathrm{cm}\mathrm{or}1.5\times {10}^{-2}{\mathrm{m}}^2$would give us the apparent weight of the object. So the difference between these two weights would give us the force of buoyancy. So, we can write

$F_{\mathrm{s}}=(0.25-0.10)\mathrm{N}$

$=0.15\mathrm{N}$

From the figure, we can also say that depth$\mathrm{h}=1.5\mathrm{cm}\mathrm{or}1.5\times {10}^{-2}{\mathrm{m}}^2$

And using equation (ii), we can get

$\mathrm{V}=(5.67)(1.5)$

$=8.51{\mathrm{cm}}^3$

$=8.51\times {10}^{-6}{\mathrm{m}}^3$

From equation (i), the density of the liquid is given as:

$\mathrm{\rho }=\frac{\mathrm{F}}{\mathrm{gV}}$

$=\frac{0.15\mathrm{N}}{\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(8.51\times {10}^6{\mathrm{m}}^3\right)}$

$=1.79\times {10}^3$

$\approx 1800\mathrm{kg}/{\mathrm{m}}^3$

Hence, the density of the liquid is$1800\mathrm{kg}/{\mathrm{m}}^3$