A cubical block of ice floating in water has to support a metal piece weighing 0.50 kg. What can be the minimum edge of the block so that it does not sink in the water? The specific gravity of ice = 0.9.
Answers
Let the minimum edge of the block be a m .
Therefore, The volume of the ice block = a³
Density of ice = density of water at 4° C × Specific Gravity.
∴ Density of the ice = 0.9 × 1000 kg/m³
Density of Ice = 900 kg/m³
Now, The Mass of the ice block = Volume × Density
∴ Mass = a³ × 900 kg
Weight = 900a³ × 10 kg.
Thus, Total weight of the ice block and the metal piece =
900a³ × 10 + 0.50 × 10 = 9000a³ + 5
Weight of the water displaced = Vρg
= a³ × 1000 × 10
= 10000a³
Using Archimedes Principle,
10000a³ = 9000a³ + 5
∴ 1000a³ = 5
∴ a³ = 5/1000
a = 0.167 m.
a = 16.7 cm.
Hope it helps .
Let the minimum edge of the block be a m .
Therefore, The volume of the ice block = a³
Density of ice = density of water at 4° C × Specific Gravity.
∴ Density of the ice = 0.9 × 1000 kg/m³
Density of Ice = 900 kg/m³
Now, The Mass of the ice block = Volume × Density
∴ Mass = a³ × 900 kg
Weight = 900a³ × 10 kg.
Thus, Total weight of the ice block and the metal piece =
900a³ × 10 + 0.50 × 10 = 9000a³ + 5
Weight of the water displaced = Vρg
= a³ × 1000 × 10
= 10000a³
Using Archimedes Principle,
10000a³ = 9000a³ + 5
∴ 1000a³ = 5
∴ a³ = 5/1000
a = 0.167 m.
a = 16.7 cm.
Hope it helps .