Answer:
(a) The density of the object is 316/343 × the density of the oil
(b) The fraction of oil displaced after immersing the object is 0.461 of the oil volume
Explanation:
(a) The volume, V of a cone of height, h and base diameter, D = 2×r is given by the following equation;
[tex]V = \dfrac{\pi r^{2} h}{3}[/tex]
The volume of the object is therefore;
[tex]\dfrac{\pi \times 4^{2} \times 14}{3} = 74\tfrac{2}{3}\pi \, cm^3[/tex]
Where 6 cm is above the oil level we have;
[tex]\dfrac{\pi \times \left (6 \times \dfrac{4}{14} \right )^{2} \times 6}{3} = 5\tfrac{43}{49}\pi \, cm^3[/tex] above the oil level
Therefore, volume of the oil displaced = [tex]68\tfrac{116}{147}\pi[/tex] cm³ = 216.11 cm³
The density of the object is thus;
[tex]\dfrac{68\tfrac{116}{147}\pi}{ 74\tfrac{2}{3}\pi} \times Density \ of \ the \ oil = \dfrac{316}{343} \right ) \times Density \ of \ the \ oil[/tex]
The density of the object = 316/343 × the density of the oil.
(b) The volume of the oil = 2 × Volume of the object = [tex]2 \times 74\tfrac{2}{3}\pi \, cm^3 = 149\tfrac{1}{3}\pi \, cm^3[/tex]
The fraction of the volume displaced, x, after immersing the object is given as follows;
[tex]x = \dfrac{68\tfrac{116}{147}\pi}{ 149\tfrac{1}{3}\pi} = \dfrac{158}{343} = 0.461[/tex]
The fraction of oil displaced after immersing the object = 0.461 of the volume of the oil
