Average density of the earth-
A) does not depend on \[g\]
B) is a complex function of \[g\]
C) is directly proportional to \[g\]
D) is inversely proportional to \[g\]

Hint: In this question, use the concept of the universal law of gravitation that is the force applied between the bodies is directly proportional to the product of the masses and inversely proportional to square of the distance between the bodies.

Complete step by step solution:
As we know that the force acting on a body due to gravity is given by,
$F = m \times g$
Where, $F$ is the force acting on the body, $g$ is the acceleration due to gravity, $m$ is mass of the body.
As we know according to the universal law of gravitation,
\[F = \dfrac{{GmM}}{{{{\left( {r + h} \right)}^2}}}\]
Where, $F$ is the force between two bodies, \[G\] is the universal gravitational constant, \[m\] is the mass of the object, \[M\] is the mass of the earth, \[r\] is the radius of the earth, and \[h\] is the height at which the body is from the surface of the earth.
As the height \[\left( h \right)\]is negligibly small compared to the radius of the earth we can-frame the equation as follows,
$F = \dfrac{{GmM}}{{{r^2}}}$
Now we are equating both the expressions of the forces,
$ \Rightarrow mg = \dfrac{{GmM}}{{{r^2}}}$
\[ \Rightarrow {\text{ }}g{\text{ }} = \dfrac{{GM}}{{{r^2}}}\]
Therefore, the formula of acceleration due to gravity is given by,
\[\therefore g{\text{ }} = \dfrac{{GM}}{{{r^2}}}......(i)\]
Now, we obtain the expression for the mass of the earth as,
$M = \rho V$
Here, the average density of the Earth is $\rho $ and the volume of the earth is $V$
As we know that, even though the earth is not exactly a sphere shaped, but it is close enough to that. So, the equation for the volume of the sphere is given by:
\[V = \dfrac{4}{3} \times \pi \times {r^3}\]
Now, we substitute the value of the volume in the mass equation as,
$ \Rightarrow M = \rho \left( {\dfrac{4}{3}\pi {r^3}} \right)$
Now, we substitute the mass equation in equation \[\left( i \right):\] as
\[ \Rightarrow g = \dfrac{{G\left( {\rho \left( {\dfrac{4}{3}\pi {r^3}} \right)} \right)}}{{{r^2}}}\]
\[\therefore g = \dfrac{4}{3}\rho \pi G\]
Hence, from the above equation we can clearly see that $g$ is directly proportional to average density.

So, option (C) is the correct answer of the provided question.

Note:Acceleration due to gravity depends on the mass and radius of the earth. While calculating the average density of the earth, we have to keep in mind the shape of the earth. Mistakes often occur during calculation of Volume of the earth.