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Orlov [11]
3 years ago
14

A wooden object (conically shaped) has a diameter of 8cm and height of 14cm. It floats in oil with 6cm of its height above oil l

evel. Determine (a)The density of the object (b) If the volume of oil is twice that of the object, what fraction of oil is displaced after immersing the object?
Physics
1 answer:
baherus [9]3 years ago
8 0

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;

V = \dfrac{\pi r^{2} h}{3}

The volume of the object is therefore;

\dfrac{\pi \times 4^{2} \times 14}{3} = 74\tfrac{2}{3}\pi \, cm^3

Where 6 cm is above the oil level we have;

\dfrac{\pi \times \left (6 \times \dfrac{4}{14}   \right )^{2} \times 6}{3} = 5\tfrac{43}{49}\pi \, cm^3 above the oil level

Therefore, volume of the oil displaced = 68\tfrac{116}{147}\pi cm³ = 216.11 cm³

The density of the object is thus;

\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

The density of the object = 316/343 × the density of the oil.

(b) The volume of the oil = 2 × Volume of the object = 2 \times 74\tfrac{2}{3}\pi \, cm^3 = 149\tfrac{1}{3}\pi \, cm^3

The fraction of the volume displaced, x, after immersing the object is given as follows;

x = \dfrac{68\tfrac{116}{147}\pi}{ 149\tfrac{1}{3}\pi} = \dfrac{158}{343} = 0.461

The fraction of oil displaced after immersing the object = 0.461 of the volume of the oil

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