Question

A solid weighs 1.5kgf in air and 0.9kgf in a liquid of density 1.2x10 3 kg m -3 .calculate relative density of solid.

Answer 1

It is simple to find it by using the formula for the relative density of solid.
 
$r.d._{solid}=\frac{W_{air}}{W_{air}-W_{liquid}}*\frac{d_{liquid}}{d_{water}}\\\\=\frac{1.5}{1.5-0.9}*\frac{1.2*10^3}{1*10^3}=3.0$

so 3.0 is the answer.

Answer 2

Answer:

Weight of body in air (W1) = 1•5 kgf

Weight of body in liquid (W2) =0•9 kgf

RD of solid = W1 / W1 - W2 × RD of liquid

RD of solid = 1•5 / 1•5 - 0•9 × 1•2

= 1•5 / 0•6 × 1•2

= 3•0

Relative density has no unit as It's a perfect ration ....

Hope it helps