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An object floats in water with $10 \%$ of its volume outside and in oil $30 \%$ of its volume outside. The specific gravity of the oil is
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$1.3$
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
$1=\frac{0.9 \rho_{\text {water }}}{0.7 \rho_{\text {oil }}} \quad$ or $\quad \frac{\rho_{\text {oil }}}{\rho_{\text {water }}}=\frac{0.9}{0.7}=\frac{9}{7}$
Specific gravity of the oil $=\rho_{\text {oil }}=\frac{9}{7} \times \rho_{\text {water }}=1.3$
Divide (i) by (ii), we get
$1=\frac{0.9 \rho_{\text {water }}}{0.7 \rho_{\text {oil }}} \quad$ or $\quad \frac{\rho_{\text {oil }}}{\rho_{\text {water }}}=\frac{0.9}{0.7}=\frac{9}{7}$
Specific gravity of the oil $=\rho_{\text {oil }}=\frac{9}{7} \times \rho_{\text {water }}=1.3$
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