Brass cylinder floating in mercury

Data: brass cylinder $D=37.5$ mm, $H=76.2$ mm, $S_{brass}=8.5$, mercury $S=13.6$, larger tube $d_1=39.5$ mm. The second tube diameter $d_2$ is not legible in the figure, so its effect is kept as a symbol below.

(a) Pressure on the bottom of the cylinder

The floating cylinder's weight acts over its base area $A_c$, so the pressure on the bottom is the weight divided by the area:

$$p=\frac{W}{A_c}=S_{brass}\,\gamma_w\,H=8.5(9810)(0.0762)=6354\ \text{Pa}$$

(about 6.35 kPa gauge.) This equals the hydrostatic pressure of mercury at the immersion depth $d$.

(b) New equilibrium level

Depth of immersion (weight = buoyancy):

$$S_{brass}H=S_{Hg}\,d\ \Rightarrow\ d=\frac{8.5}{13.6}(76.2)=47.6\ \text{mm}$$

Volume of mercury displaced:

$$V=A_c\,d=\frac{\pi}{4}(37.5)^{2}(47.6)=1104.5\times47.6=52\,600\ \text{mm}^3$$

Let the mercury level rise by $h$ in both tubes. The cylinder bottom is then $(d-h)$ below the original level, and the mercury it pushes out from below the original level fills the annulus and the second tube above it:

$$A_c(d-h)=\left(A_1-A_c\right)h+A_2h\ \Rightarrow\ h=\frac{A_c\,d}{A_1+A_2}=\frac{V}{A_1+A_2}$$

where $A_1=\dfrac{\pi}{4}d_1^{2}=\dfrac{\pi}{4}(39.5)^{2}=1225.4\ \text{mm}^2$ is the full area of the larger tube and $A_2=\dfrac{\pi}{4}d_2^{2}$.

• With the second tube neglected: $h=\dfrac{52\,600}{1225.4}=42.9$ mm.
• For example, with $d_2=12.7$ mm ($A_2=126.7\ \text{mm}^2$): $h=\dfrac{52\,600}{1352.1}=38.9$ mm.

Answer: (a) $p\approx6.35$ kPa; (b) immersion depth 47.6 mm and rise $h=\dfrac{52\,600}{1225.4+0.785\,d_2^{2}}$ mm (42.9 mm if the second tube is neglected; 38.9 mm for $d_2=12.7$ mm).