Liquid A of relative density 0.8 rests on top of liquid B of relative density 1.4 without mixing - Leaving Cert Applied Maths - Question 9 - 2021

For the immersed part of the rod, let ℓ' be the length immersed. From equilibrium:

$B = T + W$

Where B is the buoyant force, T is the tension in the string, and W is the weight of the rod. We can represent:

$B = \frac{1}{2} ℓ' \cdot \rho g$

Using similar triangles in the inclined rod setup, we find:

$\frac{ℓ'}{ℓ} = sin(\alpha)$

Thus, substituting in for B gives:

$B = ℓ' \cdot \frac{ρg}{2} sin(α)$

Now using $x$ to denote the vertical depth implies:

$x \cdot \frac{ℓ'}{s^2} = W$

Equating these, simplifying leads to:

$x^2 = s^2e^2$

Then substituting for $x$ gives:

$x = ℓ' \sqrt{\frac{s}{ρ}}$

Thus proving the desired result.

Using the relations from earlier, we find:

$h = ℓ \cos(α)$

Rearranging gives:

$cos(α) = \frac{h}{ℓ}$

Then using the derived equations, we have:

$\alpha = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right)$

Thus:

$α = 60°$

From earlier derived relations, substituting T gives:

$B = T + W = 2W$

Substituting into the buoyant force equation:

$B = \frac{2W}{ℓ} = 2 ℓ \frac{2W}{4ph^2}$

Solving this combined with area relations leads to:

$s = \frac{ℓ}{x} = \frac{ℓ}{\frac{3}{2}}$

Simplifying gives:

$s = \frac{4g}{9}$