Question

please solve it and send the correct solution​

Answer 1

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 .

Answer 2

Answer:

The answer will be 0.81 g/cm∧3

Explanation: