A spherical piece of ice of radius 10 cm has a piece of iron embedded in it and floats in water, with 95% of its volume immersed - Leaving Cert Applied Maths - Question 9 - 2017

To find the depth of the mercury, we will set up the equilibrium condition between the buoyant force and the weight of the cylinder:

1. The buoyant force from the mercury is given by:

$B = \rho_{mercury} \cdot \left( \frac{\pi (0.125)^2}{4} \cdot h \right) \cdot g$

2. The weight of the cylinder is:

$W = \rho_{iron} \cdot \left( \frac{\pi (0.125)^2}{4} \cdot 0.1 \right) \cdot g$

3. Setting these equal gives:

$13600 \cdot \left( \frac{\pi (0.125)^2}{4} \cdot h \right) \cdot g = 7800 \cdot \left( \frac{\pi (0.125)^2}{4} \cdot 0.1 \right) \cdot g$

4. Simplifying and solving for $h$ results in:

$h \approx 0.057 m$

To find the volume of oil required, we will consider the overall balance including the oil layer on top of the mercury:

1. Establish the buoyant force due to the mercury and the oil:

$B = \rho_{mercury} \cdot \left( \frac{\pi (0.125)^2}{4} \cdot (0.1 - h) \right) \cdot g + \rho_{oil} \cdot \left( \frac{\pi (0.125)^2}{4} \cdot (1.0) \right) \cdot g$

2. The total weight must equal:

$W = 7800 \cdot \left( \frac{\pi (0.125)^2}{4} \cdot 0.1 \right) \cdot g$

3. Rearranging this gives:

$13600(0.1-h) + 1030(h) = 7800(0.1)$

4. Calculating for $h$ results in:

$h = 0.046 m$

5. As a result, the volume of oil needed is:

$V = \frac{\pi (0.15)^2}{4} \cdot (0.1 - 0.046)$

6. This calculation yields:

$V \approx 9.94 \times 10^{-6} m^3$