A neutron star has a density equal to that of nuclear matter (2.8x 10¹⁷ kg m⁻³). Assume the star to be spherical, find the radius of the neutron star whose mass is 4.0 x 10¹⁰ kg (twice the mass of the sun).

The correct option is 15 km.

Problem Analysis:
We are given the following values for the neutron star:
1. Density of the neutron star (ρ), which is equal to the density of nuclear matter:
$\rho =2.8\times {10}^{17}\mathrm{kg}{m}^{-3}$
2. Mass of the neutron star ($M$). The problem text mentions "twice the mass of the sun" which is:
$M=4.0\times {10}^{30}\mathrm{kg}$
Note: The exponent in the question's mass is written as 10¹⁰ kg, which is a typo for 10³⁰ kg, as confirmed by the description "twice the mass of the sun" (since the mass of the Sun is approximately 2.0 × 10³⁰ kg).

Step-by-Step Derivation:

1. Formula for Density:
Assuming the neutron star is a perfect sphere, its volume ($V$) is given by the formula:
$V=\frac{4}{3}\pi R^3$
where $R$ is the radius of the star.

2. Relating Density, Mass, and Volume:
Density is defined as mass per unit volume:
$\rho =\frac{M}{V}=\frac{M}{\frac{4}{3}\pi R^3}=\frac{3M}{4\pi R^3}$

3. Solving for Radius ($R$):
Rearranging the formula to solve for $R^3$:
$R^3=\frac{3M}{4\pi \rho }$

4. Substituting the values:
Let us substitute the values into the equation:
$R^3=\frac{3\times (4.0\times {10}^{30})}{4\times 3.1416\times (2.8\times {10}^{17})}$
Simplify the numerator and the denominator:
$R^3=\frac{12\times {10}^{30}}{35.1858\times {10}^{17}}$
$R^3\approx 0.341\times {10}^{13}{m}^3$
$R^3\approx 3.41\times {10}^{12}{m}^3$

5. Calculating the Cube Root:
Taking the cube root of both sides to find the radius $R$:
$R=(3.41\times {10}^{12}{)}^{\frac{1}{3}}$
$R=(3.41{)}^{\frac{1}{3}}\times {10}^{4}m$
Since ${1.5}^3=3.375$, we have $(3.41{)}^{\frac{1}{3}}\approx 1.5$:

$R\approx 1.5\times {10}^{4}m$
$R\approx 15,000m=15\mathrm{km}$

Thus, the radius of the neutron star is approximately 15 km.