Answer
Conversions:
1 in. $=25.4 \mathrm{~mm}$
Calculate the number of lead balls that together weighs $1 \mathrm{lb}$.
$n \frac{1}{12} \mathrm{lb}=1 \mathrm{lb}$
Here, $\frac{1}{12} \mathrm{lb}$ is the weight of single lead ball.
$n=12$
Therefore, the number of lead balls that fit in the barrel is 12 .
Calculate the diameter of the lead ball.
$$
n \times \frac{\pi}{6} d^{3} \rho=1 \mathrm{lb}
$$
Here, $\frac{\pi}{6} d^{3}$ is the volume of the lead ball and $\rho$ is the density of the lead ball.
Substitute 12 for $n$ and $0.411 \mathrm{lb} / \mathrm{in}^{3}$ for $\rho$
$$
\begin{array}{l}
12 \times \frac{\pi}{6} d^{3} \times 0.411=1 \mathrm{lb} \\
d^{3}=0.38743 \\
d=0.729 \mathrm{in} .
\end{array}
$$
$=0.729 \times 25.4 \mathrm{~mm}$
$$
=18.51 \mathrm{~mm}
$$
Therefore, the diameter of the lead ball is $18.51 \mathrm{~mm}$.
Find the diameter of the lead ball in inches by using conversion formula.
$$
\begin{aligned}
d &=18.51 \mathrm{~mm} \times \frac{1 \mathrm{in}}{25.4 \mathrm{~mm}} \\
&=0.729 \mathrm{in}
\end{aligned}
$$
Therefore, the diameter of the lead ball is $0.729 \mathrm{in}$.
Work Step by Step
Conversions:
1 in. $=25.4 \mathrm{~mm}$
Calculate the number of lead balls that together weighs $1 \mathrm{lb}$.
$n \frac{1}{12} \mathrm{lb}=1 \mathrm{lb}$
Here, $\frac{1}{12} \mathrm{lb}$ is the weight of single lead ball.
$n=12$
Therefore, the number of lead balls that fit in the barrel is 12 .
Calculate the diameter of the lead ball.
$$
n \times \frac{\pi}{6} d^{3} \rho=1 \mathrm{lb}
$$
Here, $\frac{\pi}{6} d^{3}$ is the volume of the lead ball and $\rho$ is the density of the lead ball.
Substitute 12 for $n$ and $0.411 \mathrm{lb} / \mathrm{in}^{3}$ for $\rho$
$$
\begin{array}{l}
12 \times \frac{\pi}{6} d^{3} \times 0.411=1 \mathrm{lb} \\
d^{3}=0.38743 \\
d=0.729 \mathrm{in} .
\end{array}
$$
$=0.729 \times 25.4 \mathrm{~mm}$
$$
=18.51 \mathrm{~mm}
$$
Therefore, the diameter of the lead ball is $18.51 \mathrm{~mm}$.
Find the diameter of the lead ball in inches by using conversion formula.
$$
\begin{aligned}
d &=18.51 \mathrm{~mm} \times \frac{1 \mathrm{in}}{25.4 \mathrm{~mm}} \\
&=0.729 \mathrm{in}
\end{aligned}
$$
Therefore, the diameter of the lead ball is $0.729 \mathrm{in}$.
