A cube of ice floats partly in water and partly in K.oil (figure 13-E5). Find the ratio of the volume of ice immersed in water to that in K.oil. The specific gravity of K.oil is 0.8 and that of ice is 0.9.
Answers
Let the volume of ice in water be a m³ and the volume of ice in K.oil be b m³
Therefore, Total Volume = a + b
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
Therefore, total mass of the ice cube = (a + b) × 900 kg
Theretofore, Its weight = (a + b) × 900 × 10
By the law of the Flotation, the total force of buoyancy by the water and the Kerosene oil will be equal to the weight of the ice cube.
∴ The total force of buoyancy = a × 1000 + Y × 800 kg
(a + b) × 900 = a × 1000 + b × 800
900a + 900b = 1000a + 800b
100a = 100b
∴ a = b
Hence, the ratio of there volume is 1 : 1.
Hope it helps .
Answer:
Same as abo e it is correct answer so correct answer is 1 is to 1 ratio