Problem Example 1
An object weighs 36 g in air and has a volume of 8.0 cm3. What will be its apparent weight when immersed in water?
Solution:
When immersed in water, the object is buoyed up by the mass of the water it displaces, which of course is the mass of 8 cm3 of water. Taking the density of water as unity, the upward (buoyancy) force is just 8 g.
The apparent weight will be (36 g) – (8 g) = 28 g.
Solution:
When immersed in water, the object is buoyed up by the mass of the water it displaces, which of course is the mass of 8 cm3 of water. Taking the density of water as unity, the upward (buoyancy) force is just 8 g.
The apparent weight will be (36 g) – (8 g) = 28 g.
Problem Example 2
A balloon having a volume of 5.000 L is placed on a sensitive balance which registers a weight of 2.833 g. What is the "true weight" of the balloon if the density of the air is 1.294 g L–1?
Solution:
The mass of air displaced by the balloon exerts a buoyancy force of
(5.000 L) / (1.294 g L –1) = 3.860 g. Thus the true weight of the balloon is this much greater than the apparant weight:
(2.833 + 3.860) g = 6.69 g.
Solution:
The mass of air displaced by the balloon exerts a buoyancy force of
(5.000 L) / (1.294 g L –1) = 3.860 g. Thus the true weight of the balloon is this much greater than the apparant weight:
(2.833 + 3.860) g = 6.69 g.
Problem Example 3
A piece of metal weighs 9.25 g in air, 8.20 g in water, and 8.36 g when immersed in gasoline.
a) What is the density of the metal?
b) What is the density of the gasoline?
Solution:
When immersed in water, the metal object displaces (9.25 – 8.20) g = 1.05 g of water whose volume is (1.05 g) / (1.00 g cm–3) = 1.05 cm3. The density of the metal is thus (9.25 g) / (1.05 cm3) = 8.81 g cm–3.
The metal object displaces (9.25 - 8.36) g = 0.89 g of gasoline, whose density must therefore be (0.89 g) / (1.05 cm3) = 0.85 g cm–3.
a) What is the density of the metal?
b) What is the density of the gasoline?
Solution:
When immersed in water, the metal object displaces (9.25 – 8.20) g = 1.05 g of water whose volume is (1.05 g) / (1.00 g cm–3) = 1.05 cm3. The density of the metal is thus (9.25 g) / (1.05 cm3) = 8.81 g cm–3.
The metal object displaces (9.25 - 8.36) g = 0.89 g of gasoline, whose density must therefore be (0.89 g) / (1.05 cm3) = 0.85 g cm–3.
