You have a bucket containing an unknown liquid. You also have a cube-shaped wooden block that you measure to be 8.0 cm on a side, but you don’t know the mass or density of the block. To find the density of the liquid, you perform an experiment. First you place the wooden block in the liquid and measure the height of the top of the floating block above the liquid surface. Then you stack various numbers of U.S. quarter-dollar coins onto the block and measure the new value of h. The straight line that gives the best fit to the data you have collected is shown in Fig. P12.86.[l1] Find the mass of one quarter (seewww.usmint.gov for quarters dated 2012). Use this information and the slope and intercept of the straight-line fit to your data to calculate (a) the density of the liquid (in kg/m3[l2] ) and (b) the mass of the block (in kg).

• The length of cube shaped wooden block is, L= 8.0 cm.
• The mass of single quarter is, M = 5.670 g.

When a body is submerged in water, it feels an up thrust equivalent to the weight of water that has been displaced, which is crucial to the balance of an object floating in still liquid.

The buoyant force counteracts the weight of a floating item.

Let be mass of block, $m_n$be mass of quarters.

The mass of n quarter can be given as, $m_n=nM$

The mass of a single quarter can be given as,

role="math" localid="1668054968014" $M=m_b+m_n$

The submerged volume can be expressed as,

$v_s=\left(L-h\right)L^2$

Here, h is the .

Balancing the forces and the expression of forces can be written as,

$\begin{array}{r}Mg={\rho }_{\text{iquiid}}g{V}_s\\ \left(m_b+m_n\right)g={\rho }_{\text{liquid}}{L}^2(L-h)g\\ h=\frac{{\rho }_{\text{íquid}}{L}^3-m_b}{{\rho }_{\text{iqquid}}{L}^2}-\left(\frac{M}{{\rho }_{\text{liquid}}{L}^2}\right)n\end{array}$

If we plot graph between h versus n.

Slope is $\left(-\frac{M}{{\rho }_{liquid}{L}^2}\right)$. The y-intercept is $\frac{{\rho }_{liquid}{L}^3-m_b}{{\rho }_{liquid}}$.

From the graph, the slope can be evaluated as,

$=\left(\frac{1.2\mathrm{cm}-3.0\mathrm{cm}}{25}\right)=0.072cm$

Substitute value in density equation we get,

${\rho }_{\text{biquid}}=\frac{-(5.670\mathrm{g})}{\left[(-0.0072\mathrm{cm})(8.0\mathrm{cm}{)}^2\right]}=1.24\mathrm{g}/{\mathrm{cm}}^3\approx 1.2\mathrm{g}/{\mathrm{cm}}^3=1200\mathrm{kg}/{\mathrm{m}}^3$

Simplify y-intercept. The expression can be given as,

$y=\frac{L-m_b}{\left[\left({\rho }_{liquid}\right)\left(L^2\right)\right]}$

From the graph, y-intercept is equal to 3.0 cm.

So, the equate both sides of y-intercept expressions and substitute the values in the above equation.

$\begin{array}{r}3.0\mathrm{cm}=\frac{8.0\mathrm{cm}-m_b}{\left[\left(1.24\mathrm{g}/{\mathrm{cm}}^3\right)(8.0\mathrm{cm}{)}^2\right]}\\ m_b=400\mathrm{g}=0.40\mathrm{kg}\end{array}$

Thus, the density of liquid is $1200\mathrm{kg}I{m}^3$and the m mass of single quarter is 0.40 kg .