Q192 / 200 · WBJEE 2013 · Physics
A cylindrical block floats vertically in a liquid of density $\rho_{1}$ kept in a container such that the fraction of volume of the cylinder inside the liquid is $x_{1}$. Then some amount of another immiscible liquid of density $\rho_{2}\left(\rho_{2} < \rho_{1}\right)$ is added to the liquid in the container so that the cylinder now floats just fully immersed in the liquids with $x_{2}$, fraction of volume of the cylinder inside the liquid of density $\rho_{1}$. The ratio $\rho_{1} / \rho_{2}$ will be
- Correct answerA$\frac{1-x_{2}}{x_{1}-x_{2}}$
- B$\frac{1-x_{1}}{x_{1}+x_{2}}$
- C$\frac{x_{1}-x_{2}}{x_{1}+x_{2}}$
- D$\frac{x_{2}}{x_{1}}-1$
Solution
The given situation is shown as 
As the Bouyant force in both the cases are same
$\therefore$
$$
\rho_{1} x_{1} g=\rho_{1} x_{2} g+\rho_{2}\left(1-x_{2}\right) g
$$
On solving $\quad \frac{\rho_{1}}{\rho_{2}}=\frac{\left(1-x_{2}\right)}{\left(x_{1}-x_{2}\right)}$
As the Bouyant force in both the cases are same
$\therefore$
$$
\rho_{1} x_{1} g=\rho_{1} x_{2} g+\rho_{2}\left(1-x_{2}\right) g
$$
On solving $\quad \frac{\rho_{1}}{\rho_{2}}=\frac{\left(1-x_{2}\right)}{\left(x_{1}-x_{2}\right)}$Hence, the correct answer is option: 'A'.