Answer :
The mass of a black hole needs to be in order for its mass divided by its volume to be equal to the density of water (1g/cm³) is 7x10²¹ solar masses or 1.45 x 10⁵⁵g
A black hole is a collapsed star where all of the mass is concentrated into a single point in space. As it is a point, there is no volume. Therefore the density of the singularity is therefore infinite regardless of mass which is given by,
Density = Mass/Volume
Density =Mass/0
Density=Infinite
However, it is said that black holes have an event horizon, which is said to be the point where light is "captured" by the black hole. If we assume this event horizon is a spherical boundary for the black hole, then we can use its volume for our density calculation instead of the singularity.
Effectively, we could able to calculate the "average" density within the event horizon. The radius of the event horizon is called the Schwarzschild Radius, and can be found using the following formula,
R = 2MG/c²
where M is the mass of gravity, G is the gravitational coefficient and c is the speed of light. The volume of the spherical horizon could be computed as,
V = πR²
On substituting the value of R to this formula we get,
V = π(4M²G²)/c⁴
Density = c⁴/π(4MG²)
M = c⁴/4πPG²
If we consider the density to be the density of water that is 1g/cm², then mass could ve computed as,
[tex]M = \frac{(3*10^{8})^{4} }{3.14*4*1*(6.6743*10^{-11})^{2} } \\M = \frac{(3*10^{8})^{4}}{12.56*44.52*10^{-22}} \\M = \frac{81*10^{32}}{559.2833*10^{-22}}} \\[/tex]
M = 0.14482 x 10⁵⁴
M = 1.45 x 10⁵⁵ g
In more meaningful terms, this could be approximated to 7×10²¹ solar masses, within the range of stellar black holes.
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