# The weight of a body in water is one third of its weight in air. What is the density of the material of the body? Thanks

Dec 22, 2016

Let the volume of the body be $v c {m}^{3}$

the density of the body be $\mathrm{dg} c {m}^{-} 3$

So weight of the body in air${W}_{\text{air}} = v \mathrm{dg}$ dyne,

$\text{where "g->"acceleration due to gravity}$

Now the bouyant force on the body when it is completely imerssed is the weight of displaced water

${W}_{\text{bouyant}} = v \times 1 \times g$ dyne.

Now the weight of the body in water

${W}_{\text{water"=W_"air"-W_"buoyant}} = v g \left(d - 1\right)$ dyne

By the given condition

${W}_{\text{water"=1/3W_"air}}$

$\implies v g \left(d - 1\right) = \frac{1}{3} v \mathrm{dg}$

$\implies d - 1 = \frac{1}{3} d$

$\implies d - \frac{d}{3} = 1$

$\implies \frac{2 d}{3} = 1$

$\implies d = \frac{3}{2} = 1.5 g c {m}^{\text{-3}}$