Q2

A standard baseball has a circumference of apoximately 23cm. If a baseball had the same mass per unit volume as a neutron or a p

BigorU [14]

Baseball, neutrons, and protons are all perfect spheres.

Circumference of the baseball, $L_{\text {ball }}=23 \mathrm{~cm}$ .

mass of the proton or neutron, $m_{\mathrm{p}}=10^{-27} \mathrm{~kg}$

diameter of the proton or neutron, $r_{0}=10^{-15} \mathrm{~m}$

Mass per unit volume is the definition of a material's density. The density equation is

$\rho=\frac{M}{V} .$

Here, M is the mass of the material, V is the volume, and $\rho$ is the density of the material.

We know that formula for the volume of a sphere having diameter d is $V=\frac{\pi d^{3}}{6}$.$

You may obtain this result by plugging this number into the proton/neutron density equation:

$\rho_{0}=\frac{m_{0}}{\left(\frac{\pi d_{0}^{3}}{6}\right)}$

By replacing the variables in the aforementioned equation with their values, you will obtain

$\begin{aligned}\rho_{\circ} &=\frac{10^{-27}}{\left(\frac{\pi\left(10^{-15}\right)^{3}}{6}\right)} \\&=1.91 \times 10^{18} \mathrm{kgm}^{-3}\end{aligned}$

The formula for calculating a circle's circumference with diameter d is $L=\pi d$.$

Using this, you get the diameter of the baseball:

$\begin{aligned}d_{\text {ball }} &=\frac{L_{\text {ball }}}{\pi} \\&=\frac{23}{\pi} \\&=7.32 \mathrm{~cm} \\&=0.0732 \mathrm{~m}\end{aligned}$

The volume of the baseball may be calculated using the formula for the volume of a sphere:

$\begin{aligned}V_{\text {ball }} &=\frac{\pi d_{\text {ball }}^{3}}{6} \\&=\frac{\pi \times(0.0732)^{3}}{6} \\&=2.054 \times 10^{-4} \mathrm{~m}^{3}\end{aligned}$

The mass of the baseball is determined by the concept of density if you suppose that its density is equal to that of a proton or neutron.

$m_{\text {ball }}=\rho_{\circ} \cdot V_{\text {ball }}$

Here, $\rho_{\circ}$ is the density of the proton/neutron.

By replacing the variables in the aforementioned equation with their values, you will obtain

$\begin{aligned}m_{\text {ball }} &=\left(1.91 \times 10^{18}\right) \cdot\left(2.054 \times 10^{-4}\right) \\&=3.92 \times 10^{14} \mathrm{~kg}\end{aligned}$