a piece of iron d=7.8×10^-3 and v=100 cm^3 is immersed in water calculate the weight of iron piece in air
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Answer:
= 7.8N - 1N = 6.8N
Step-by-step explanation:
density of iron (Dr) = 7.8×10³ kg/m³
volume of iron piece(V) = 100cm³ = 10^(-4) m³
density of water (Dw) = 10³ kg/m³
(a) weight in air = Dr×V×g = 7.8×10³ × 10^(-4) ×10 = 7.8 N
(b) upthrust is the buoyant force exerted by water
upthrust = Dw×V×g = 10³ × 10^(-4) ×10 = 1N
(c) apparent weight in water = weight in air - upthrust
= 7.8N - 1N = 6.8N
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