**A solid object weighing 5kg in air, weighs 4kg in kerosene. Find (a) its volume and (b) its density**

Actual weight of the body, W = 5kg

Apparent weight in kerosene, W'^{ }=4kg

Let the volume of the body be V and the density be ρ.

Density of kerosene = 820 kg/m ^{3}

Now,

Buoyancy = Actual weight – Apparent weight = 5kg – 4kg = 1kg

Also,

Buoyancy = volume of kerosene displaced × density of kerosene × g

=> g = V '^{ }×820×g

=> V '^{ }= 1.219× 10 ^{-3 }m ^{3}

This is the volume of kerosene displaced.

**Since, the body immerses completely, it is also the volume of the body.**

**So, V = V ' ^{ }= 1.219× 10 ^{-3 }m ^{3}**

Now, W = 5kg

=> Vρg = 5

**=> ρ = 5/(1.219 × 10 ^{-3 }m ^{3 }) = 4100 kg/m ^{3}**

**This is the density of the body.**

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