Answer:
Assuming that the two liquids do not mix:
[tex]L \approx 19.1\; \text{cm}[/tex].
[tex]h_{1} \approx 0.712\; \text{cm}[/tex].
Explanation:
To find the height of the [tex]\rho_{\text{B}} = 1.3 \; {\rm g \cdot cm^{-3}}[/tex] "light liquid" column, divide mass by density to find volume, and divide volume by cross-section area to find height:
[tex]\displaystyle L = \frac{V}{A_{\text{R}}} = \frac{m_{\text{B}} / \rho_{\text{B}}}{A_{\text{R}}} \approx \frac{(97.3 / 1.3)}{3.92} \; \text{cm} \approx 19.0934\; \text{cm}[/tex].
To find the change in the height of the [tex]\rho_{\text{A}} = 11 \; {\rm g \cdot cm^{-3}}[/tex] "heavy liquid" column, make use of the two following facts:
- The total volume of each liquid stays the same.
- At equilibrium, the pressure at any horizontal cross-section should be the same for each liquid column.
From each of the two facts, construct an equation about [tex]h_{1}[/tex] and [tex]h_{2}[/tex].
Note that when the "light liquid" is added, the height of the "heavy liquid" column on the right side decreased by [tex]h_{2}[/tex]. At the same time, the height of the column on the left side increased by [tex]h_{1}[/tex]. Since the total volume of the "heavy liquid" stays unchanged, the volume that went down on the right side [tex]h_{2}\, A_\text{R}[/tex] should match the volume that went up on the left side [tex]h_{1}\, A_\text{L}[/tex]:
[tex]h_{2}\, A_\text{R} = h_{1}\, A_\text{L}[/tex].
In a liquid of density [tex]\rho[/tex], the pressure at a depth of [tex]h[/tex] would be [tex]\rho\, g\, h[/tex], where [tex]g[/tex] is the gravitational field strength. At equilibrium, the pressure at the bottom of the "heavy liquid" column of depth [tex](h_{1} + h_{2})[/tex] is supposed to match that at the bottom of the "light liquid" column of depth [tex]L[/tex]. Thus:
[tex]\rho_{\text{A}}\, g \, (h_{1} + h_{2}) = \rho_{\text{B}}\, g\, L[/tex].
Thus:
[tex]\displaystyle h_{1} + h_{2} = \frac{\rho_{\text{B}}\, L}{\rho_{\text{A}}}[/tex].
To find [tex]h_{1}[/tex], rearrange one of the two equations above to express [tex]h_{2}[/tex] in terms of [tex]h_{1}[/tex], and substitute that into the other equation. For example, from the equation [tex]h_{2}\, A_\text{R} = h_{1}\, A_\text{L}[/tex]:
[tex]\displaystyle h_{2} = \left(\frac{A_{\text{L}}}{A_{\text{R}}}\right)\, h_{1}[/tex].
Substitute this expression for [tex]h_{2}[/tex] into the other equation and solve for [tex]h_{1}[/tex]:
[tex]\displaystyle h_{1} + \left(\frac{A_{\text{L}}}{A_{\text{R}}}\right)\, h_{1} = \frac{\rho_{\text{B}}\, L}{\rho_{\text{A}}}[/tex].
[tex]\displaystyle \left(1 + \frac{A_{\text{L}}}{A_{\text{R}}}\right)\, h_{1} = \frac{\rho_{\text{B}}\, L}{\rho_{\text{A}}}[/tex].
[tex]\begin{aligned} h_{1} &= \left(\frac{\rho_{\text{B}}\, L}{\rho_{\text{A}}}\right)\, \left(\frac{A_{\text{R}}}{A_{\text{L}} + A_{\text{R}}}\right) \\ &\approx \frac{(1.3)\, (19.0934)}{(11)}\, \frac{(3.92)}{(8.5) + (3.92)} \\ &\approx 0.712\; \text{cm} \end{aligned}[/tex].
