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The acceleration due to gravity g and density of the earth P are related by which of the following relations? (Here G is the gravitational constant and R is the radius of the earth).(A) $p = \dfrac{{4\pi }}{{3GRd}}$(B) $p = \dfrac{{3g}}{{4\pi GR}}$(C) $p = \dfrac{{3G}}{{4\pi GR}}$ (D) $p = \dfrac{{4\pi GR}}{{3G}}$

Last updated date: 20th Jun 2024
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Hint Write down the equation for acceleration due to gravity. Rearrange the equation and find the formula for mass. Then calculate the volume of earth. Finally substitute, mass and volume in the equation which states that the ratio of mass to volume gives us density.

Complete step-by-step solution
Acceleration due to gravity on earth is given as,
$g = \dfrac{{GM}}{{{R^2}}}\,\,\, \ldots (1)$
Where, g: Acceleration due to gravity
G: Gravitational constant
M: Mass of earth
R: Radius of earth

Now we know, density is given as,
$density = \dfrac{{mass}}{{volume}}$
$\therefore p = \dfrac{M}{V}\,\,\, \ldots (2)$
Volume of Earth is
$V = \dfrac{4}{3}\pi {R^3}\,\, \ldots (3)$
Now, rearranging equation.(1) gives us,
${\text{M = }}\dfrac{{{\text{g}}{{\text{R}}^{\text{2}}}}}{{\text{G}}}\,\,\,{\ldots (4)}$
By substituting equation.(3) and equation.(4) in equation.(2) we get,
$p=\dfrac{{{\text{g}}{{\text{R}}^{\text{2}}}}}{ \dfrac{4}{3}\pi {R^3}}$
$\therefore {\text{p = }}\dfrac{{{\text{3g}}}}{{{\text{4}\pi \text{GR}}}}$

So, the relation between density and acceleration due to gravity is given as $p = \dfrac{{3g}}{{4\pi GR}}$.
Acceleration due to gravity is proportional to mean density of earth and its radius.

Hence, the correct answer is option (B).

${\text{F = G}}\dfrac{{{\text{Mm}}}}{{{{\text{R}}^{\text{2}}}}}\,\,{ \ldots (1)}$
${\text{F = mg}}\,\,{ \ldots (2)}$
${\text{mg = G}}\dfrac{{{\text{Mm}}}}{{{{\text{R}}^{\text{2}}}}}$
$\therefore {\text{g = }}\dfrac{{{\text{GM}}}}{{{{\text{R}}^{\text{2}}}}}\,\,$