A solid of density $5000 K g m^{- 3}$ weight $0.5 K g f$ in air. It is completely immersed in water of density $1000 K g m^{- 3}$ . Calculate the apparent weight of the solid in water.

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Solution

Density $ρ$ = 5000 $k g / m^{3}$
Weight = 0.5 kg f in air
mass M = 0.5 kg
Volume V = M / $ρ$
= 0.5 / 5000 = $10^{- 4} m^{3}$
mass of water displaced = $V^{}$ density of water
= $10^{- 4} 1000$ = 0.1 kg

Apparent weight = wt in air - wt of water displaced
= 0.5 kg f - 0.1 kg f
= 0.4 kg f

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A solid of density 5000 Kg m−3 weight 0.5 Kgf in air. It is completely immersed in water of density 1000 Kg m−3. Calculate the apparent weight of the solid in water.

Density ρ = 5000 kg/m3Weight = 0.5 kg f in airmass M = 0.5 kg Volume V = M / ρ = 0.5 / 5000 = 10−4m3mass of water displaced = V∗ density of water = 10−4∗1000 = 0.1 kg Apparent weight = wt in air - wt of water displaced = 0.5 kg f - 0.1 kg f = 0.4 kg f

A solid of density $5000 K g m^{- 3}$ weight $0.5 K g f$ in air. It is completely immersed in water of density $1000 K g m^{- 3}$ . Calculate the apparent weight of the solid in water.

Density $ρ$ = 5000 $k g / m^{3}$
Weight = 0.5 kg f in air
mass M = 0.5 kg
Volume V = M / $ρ$
= 0.5 / 5000 = $10^{- 4} m^{3}$
mass of water displaced = $V^{}$ density of water
= $10^{- 4} 1000$ = 0.1 kg

Apparent weight = wt in air - wt of water displaced
= 0.5 kg f - 0.1 kg f
= 0.4 kg f