Question

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

(a) 1.3 (b) 0.33

(c) 2 (d) 7
​

Answer 1

Given:

An object floats in water with 10% of its volume outside and in oil 30% of its volume outside.

To find:

Specific gravity of oil.

Calculation:

Let density of object be $\rho$ and that of oil be $\sigma$.

In 1st case ;

Volume of object inside water = 100-10 = 90%.

$\therefore \: \sf{ \dfrac{ \rho}{1} = \dfrac{V_{inside}}{V_{total}} }$

$= > \: \sf{ \dfrac{ \rho}{1} = 90\% }$

$= > \: \sf{ \dfrac{ \rho}{1} =0.9}$

$= > \: \sf{ \rho =0.9}$

In 2nd case ;

Volume of object inside oil = 100-30 = 70%.

$\therefore \: \sf{ \dfrac{ \rho}{ \sigma} = \dfrac{V_{inside}}{V_{total}} }$

$= > \: \sf{ \dfrac{ \rho}{ \sigma} =70 \%}$

$= > \: \sf{ \dfrac{ \rho}{ \sigma} =0.7}$

$= > \: \sf{ \dfrac{ 0.9}{ \sigma} =0.7}$

$\sf{ = > \: \sigma = 1.28}$

$\sf{ = > \: \sigma \approx 1.3}$

So, final answer is:

$\boxed{ \red{ \sf{ \: \sigma \approx 1.3}}}$