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Question
A solid of density 7600 kgm3 is found to weigh 0.950 kgf in air. If 4/5 volume of solid is completely immersed in a solution of density 900 kgm3, find the apparent weight of solid in a liquid.
Solution
Weight of solid in air = 0.950 kgf
∴ Mass of solid in air = m = 0.950 kg
Density of solid =ρ = 7600 kgm3 ∴ Volume of solid = V = `"m"/rho`
V = `0.950/7600 = 0.000125` m3
Density of solution = ρ' = 900 kgm-3
∵ `4/5` volume of solid is completely immersed in the given solution.
∴ Volume of solution displaced = V = `4/5xx"Volume of solid"`
V' = `4/5 xx 0.000125`
V' = 0.0001 m3
Mass of solution displaced = m' = V' × ρ'
m' = 0.0001 × 900
m' = 0.09 kg
Upthrust = Weight of solution displaced
= m'g
= 0.09 × 10 = 0.9 N
`= 0.9/10` kgf
= 0.09 kgf
Apparent weight of solid in liquid
= Actual weight – Upthrust
= 0.950-0.09 = 0.860 kgf
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