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A solid uniform ball having volume V and density \mathrm{\rho;}  floats at the interface of two immiscible liquid as shown. The densities of upper and lower liquid are \mathrm{ \rho_1} and \mathrm{ \rho_2}  respectively such that  \mathrm{ \rho_1<\rho<\rho_2}. What fraction of the ball will be in the lower liquid

Option: 1

\mathrm{\frac{\rho_1-\rho_2}{\rho_2}}


Option: 2

\mathrm{\frac{\rho_1-\rho}{\rho_1-\rho_2}}


Option: 3

\mathrm{\frac{\rho_1}{\rho_2-\rho_1}}


Option: 4

\mathrm{\frac{\rho_2}{\rho_2-\rho_1}}


Answers (1)

best_answer

\mathrm{\begin{aligned} & V_L\left(\rho_2-\rho\right) g=V_u\left(\rho-\rho_1\right) g \\\\ & \frac{V_L}{V_u}=\frac{\rho-\rho_1}{\rho_2-\rho} \\\\ & \frac{V_L}{V_L+V_u}=\frac{\rho-\rho_1}{\rho_2-\rho_1} \end{aligned}}

Posted by

avinash.dongre

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