Question

A glass ball of radius $\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathit{c}\mathit{m}$sits at the bottom of a container of milk that has a density of $\mathbf{1}\mathbf{.}\mathbf{03}\mathbf{}\mathit{g}\mathbf{/}\mathit{c}{\mathit{m}}^{\mathbf{3}}$. The normal force on the ball from the container’s lower surface has magnitude $\mathbf{9}\mathbf{.}\mathbf{48}\mathbf{}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathit{N}$. What is the mass of the ball?

Short answer

The mass of the ball is $0.0442kg$.

Step-by-step solution

Listing the given quantities

• Radius of the ball$\left(r\right)=2\times {10}^{-2}m$.
• Normal force, $N=9.48\times {10}^{-2}N$.
• The density of milk,

$\begin{array}{rcl}\left({\rho }_m\right)& =& 1.03g/c{m}^3\\ & =& 1030kg/m^3\\ & & \end{array}$

Understanding the concept of buoyant force

There are three forces acting on the bubble –buoyant force$\left(F_b\right)$,weight,and normal reaction.As the ball is stationary, the net force acting on the ball is zero. Using this concept, we can find

Formula:

$$\begin{gathered} F_{net}=0 \\ {F}_b=M_{b}g \\ ={\rho }_b{V}_{b}g \\ W=M_{b}g \\ V=\frac{4}{3}\pi r^3 \end{gathered}$$

(a) Calculation of mass of the ball

The volume of the milk displaced by the ball(Vmilk) = Volume of the ball (Vb)

$\begin{array}{rcl}{V}_{milk}& =& \frac{4}{3}\pi r^3\\ & =& \frac{4}{3}\times 3.14\times {\left(2\times {10}^{-2}\text{m}\right)}^3\\ & =& 33.472\times {10}^{-6}{\text{m}}^3\\ & & \end{array}$

Using Newton’s second law,

$$\begin{gathered} N+F_b--W=0 \\ N+{\rho }_m{V}_{b}g--M_{b}g=0 \end{gathered}$$

Rearranging the equation,

$M_b=\frac{N+{\rho }_m{V}_{b}g}{g}$

Substitute the values in the above equation.

$=\frac{9.48\times {10}^{-2}\text{N+}\left(1030{\text{kg/m}}^3\text{×}\left(33.472\text{×}{10}^{-6}{\text{m}}^3\right)\text{×}9.8{\displaystyle \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.}\right)}{9.8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.}$

Therefore, the mass of the ballis $0.0442kg$.