in figure (a), a rectangular block is gradually pushed face-down into a liquid. the block has height $d$; on…

in figure (a), a rectangular block is gradually pushed face-down into a liquid. the block has height $d$; on the bottom and top the face area is $a = 5.95\\text{ cm}^2$. figure (b) gives the apparent weight $w_{\\text{app}}$ of the block as a function of the depth $h$ of its lower face. the scale on the vertical axis is set by $w_s = 0.200\\text{ n}$. what is the density (in $\\text{g/cm}^3$) of the liquid?

in figure (a), a rectangular block is gradually pushed face-down into a liquid. the block has height $d$; on the bottom and top the face area is $a = 5.95\\text{ cm}^2$. figure (b) gives the apparent weight $w_{\\text{app}}$ of the block as a function of the depth $h$ of its lower face. the scale on the vertical axis is set by $w_s = 0.200\\text{ n}$. what is the density (in $\\text{g/cm}^3$) of the liquid?

Answer

Explanation:

Step1: Identify the relationship for apparent weight

The apparent weight $W_{app}$ is the actual weight $W$ minus the buoyant force $F_b$. $$W_{app} = W - F_b = W - \rho_{liquid} \cdot g \cdot V_{sub}$$

Step2: Express submerged volume using depth

For a block with area $A$ and depth $h$, the submerged volume is $V_{sub} = A \cdot h$. $$W_{app} = W - \rho_{liquid} \cdot g \cdot A \cdot h$$

Step3: Determine the slope from the graph

The slope of the $W_{app}$ vs $h$ graph is $-\rho_{liquid} \cdot g \cdot A$. From the graph, $W_{app}$ drops from $W_s = 0.200\text{ N}$ at $h = 0$ to $0$ at $h = 1.5\text{ cm}$. $$\text{Slope} = \frac{\Delta W_{app}}{\Delta h} = \frac{0 - 0.200\text{ N}}{1.5\text{ cm} - 0} = -\frac{0.200}{0.015}\text{ N/m}$$

Step4: Calculate the liquid density

Equate the theoretical slope to the experimental slope and solve for $\rho_{liquid}$. $$\rho_{liquid} \cdot g \cdot A = \frac{0.200}{0.015}$$ $$\rho_{liquid} = \frac{0.200}{9.8 \cdot (5.95 \times 10^{-4}) \cdot 0.015}$$

Step5: Convert units to $\text{g/cm}^3$

Calculate the value in $\text{kg/m}^3$ and divide by $1000$. $$\rho_{liquid} \approx \frac{0.200}{0.000087465} \approx 2286.6\text{ kg/m}^3 = 2.2866\text{ g/cm}^3$$

Answer:

2.29 g/cm³