MHTCET 14 May 2023 Shift 1 Solved Paper

To derive an expression for the acceleration due to gravity ' g ' at the surface of the Earth, in terms of its radius 'R', its uniform density ' ρ ', and the gravitational constant ' G ', we can start by calculating the mass ' M ' of the Earth in terms of its density and volume.
The volume ' V ' of a sphere is given by
V=‌

4

3

πR3
For a sphere with uniform density ' ρ ', the mass ' M ' can be calculated by multiplying the volume by the density:
M=ρV=ρ(‌

4

3

πR3)
Now that we have the mass, we can use Newton's law of universal gravitation to find the force ' F ' exerted on a mass 'm' at the surface of the Earth:
F=‌

GMm

R2

The acceleration ' g ' due to gravity at the Earth's surface is simply the force per unit mass 'm': square
g=‌

F

m

=‌

GM

R2

Substituting the expression for 'M' in terms of ' ρ ' and ' R ', we get:
g=‌

G(ρ( 4 3 πR3))

R2

When we simplify this expression, we have:
g=‌

Gρ 4 3 πR3

R2

g=‌

4

3

πGρR
Therefore, the correct answer is:
Option C
g=‌

4

3

πρGR