A metal ball of material density 7800 kg/m3 is suspected to have a large number of cavities. It weighs 9.8 kg when weighed directly on a balance and 1.5 kg less when immersed in water. The fraction by volume of the cavities in the metal ball is approximately :
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The apparent weight loss in water equals the buoyant force, which is the weight of water displaced. The actual volume of the ball (metal + cavities) is Vtotal. The volume of water displaced is also Vtotal.
Weight loss = 1.5 kg × g = (Density of water) × Vtotal × g
So, 1.5 = 1000 × Vtotal
Therefore, Vtotal = 1.5 / 1000 = 0.0015 m³
The volume of solid metal, Vmetal = Mass / Density = 9.8 / 7800 ≈ 0.001256 m³
The volume of the cavities, Vcavities = Vtotal - Vmetal = 0.0015 - 0.001256 = 0.000244 m³
The fraction by volume of cavities = Vcavities / Vtotal = 0.000244 / 0.0015 ≈ 0.1627
Final Answer: or approximately 0.163