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Question: What is the density of a neutron star? A. \[6.87 \times {10^6}kg{m^{ - 3}}\] B. \(5.46 \times {1...

What is the density of a neutron star?
A. 6.87×106kgm36.87 \times {10^6}kg{m^{ - 3}}
B. 5.46×1012kgm35.46 \times {10^{12}}kg{m^{ - 3}}
C. 2.42×1017kgm32.42 \times {10^{17}}kg{m^{ - 3}}
D. None

Explanation

Solution

Before we proceed, let us first get a basic understanding of neutron stars. Neutron stars are dense phenomena that exist throughout the cosmos. They are formed of neutrons and have a mass bigger than the sun, but just a few km in diameter.

Complete step by step answer:
Because a neutron star is almost entirely made up of neutrons, let's first define a neutron. Except for hydrogen, neutrons can be found in the nuclei of all other particles. A neutron is an atom's fundamental particle with roughly the same mass as a proton. Along with protons, it is found in the nucleus. The mass of a neutron is equal to 1,8381,838 electrons and 1.00141.0014 protons.

Now, coming to the question;we have, angular frequency and rotational kinetic energy of the body;
ω=2πTand K=12Iω2\omega = \dfrac{{2\pi }}{T}{\mkern 1mu} and{\mkern 1mu} \\\ \Rightarrow K = \dfrac{1}{2}I{\omega ^2}
K=2π2IT2K = \dfrac{{2{\pi ^2}I}}{{{T^2}}}
dKdt=4π2IT2dTdt\Rightarrow \dfrac{{dK}}{{dt}} = - \dfrac{{4{\pi ^2}I}}{{{T^2}}}\dfrac{{dT}}{{dt}}
This is the rate at which energy is being lost.Now, there is a moment of inertia.

\Rightarrow I = \dfrac{{\left( {5 \times {{10}^{31}}} \right){{\left( {0.0331} \right)}^3}}}{{4{\pi ^2}}}\dfrac{1}{{4.22 \times {{10}^{ - 13}}}} \\\ \Rightarrow I = 1.09 \times {10^{38}}kg{m^2} $$ Now, the neutron star's radius is $R = \sqrt {\dfrac{{5I}}{{2M}}} \\\ \Rightarrow R= \sqrt {\dfrac{{5\left( {1.09 \times {{10}^{38}}} \right)}}{{2\left( {1.4} \right)\left( {1.99 \times {{10}^{30}}} \right)}}} \\\ \Rightarrow R= 9.9 \times {10^3}m$ The velocity is now expressed as $v = \dfrac{{2\pi R}}{T} \\\ \Rightarrow v= \dfrac{{2\pi \left( {9.9 \times {{10}^3}} \right)}}{{0.0331}} \\\ \Rightarrow v= 1.9 \times {10^6}m{s^{ - 1}}$ The density is now expressed as $$\rho = \dfrac{M}{V} \\\ \Rightarrow \rho = \dfrac{M}{{\dfrac{{4\pi }}{{3{R^3}}}}} \\\ \therefore \rho = 6.8 \times {10^6}kg{m^{ - 3}}$$ **So, the correct option is A.** **Note:** High temperatures and pressures are required for the formation of a neutron star. Such pressures and temperatures are only found in the aftermath of a supernova, which occurs when enormous stars collapse. The neutron star that results is almost entirely made up of neutrons. Neutron stars are the densest known objects in the cosmos, with a diameter about equal to a city but a mass of nearly $$500,000$$ Earths or $$2$$ suns.