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.

Answer 1

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