Question

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

Answer 1

Let the volume of the body be $vcm^{3}$

Density of the body = $dgcm^{-3}$

Weight of the body in air = $vdg$ dyne

Now the buoyant force on the body when it is completely immersed is the weight of displaced water.

$W_{bouyant} = v*1*g$

Weight of the body in water = $W_{water} =W_{air} -W_{bouyant} \\vg(d-1)$

ATP,

$W_{water}= \frac{1}{3} W_{air}$

=>$vg(d-1)=\frac{1}{3} vdg$

=>$d-1=\frac{1}{3} d\\d-\frac{d}{3} =1\\\frac{2d}{3} =1\\d=\frac{3}{2}\\d=1.5gm/cm^{3}$

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