Question

A hollow metallic sphere the outer and inner diameter of which are $d_1$ and $d_2$ floats on the surface of the liquid. The density of the metal is ${\rho }_1$ and the density of the liquid is ${\rho }_2$. What weight must be added inside the sphere in order for it to float below the level of liquid?

• A. $\pi \left[d_1^3\right({\rho }_2-{\rho }_1)+d_2^3{\rho }_1]g$
• B. $\frac{\pi }{6}\left[d_1^3\right({\rho }_2-{\rho }_1)+d_2^3{\rho }_1]g$
• C. $\frac{4\pi }{3}\left[d_1^3\right({\rho }_2-{\rho }_1)+d_2^3{\rho }_1]g$
• D. $\frac{4\pi }{3}{d}_1^3({\rho }_2-{\rho }_1)g$

Answer 1

The correct option is B $\frac{\pi }{6}\left[d_1^3\right({\rho }_2-{\rho }_1)+d_2^3{\rho }_1]g$

Weight of liquid displaced $=\frac{4}{3}\pi {\left(\frac{d_1}{2}\right)}^3{\rho }_{2}g$
Weight of metal sphere $=[\frac{4}{3}\pi {\left(\frac{d_1}{2}\right)}^3-\frac{4}{3}\pi {\left(\frac{d_2}{2}\right)}^3]{\rho }_{1}g$
Using the law of floatation
$\text{weight of sphere + extra weight}=\text{weight of liquid displaced}$
$\Rightarrow [\frac{4}{3}\pi {\left(\frac{d_1}{2}\right)}^3-\frac{4}{3}\pi {\left(\frac{d_2}{2}\right)}^3]{\rho }_{1}g+x=\frac{4}{3}\pi {\left(\frac{d_1}{2}\right)}^3{\rho }_{2}g$
by solving the above equation we get
$x=\frac{\pi }{6}\left[d_1^3\right({\rho }_2-{\rho }_1)+d_2^3{\rho }_1]g$