Physics - Fluid Question with Solution | TestHub

Let $V$ be the volume of an object. When the object is floating in the liquid, then Weight of the object $=$ Weight of the liquid displaced
Divide (i) by (ii), we get $1=\frac{0.9{\rho }_{\text{water }}}{0.7{\rho }_{\text{oil }}}$ or $\frac{{\rho }_{\text{oil }}}{{\rho }_{\text{water }}}=\frac{0.9}{0.7}=\frac{9}{7}$ Specific gravity of the oil $={\rho }_{\text{oil }}=\frac{9}{7}\times {\rho }_{\text{water }}=1.3$