The pressure caused by gravitational pull inside earth at a distance a measured from its centre, when its mass,
density and radius are M, `rho` and R respectively , is given by

Options

A. `(8)/(3)("GM"^(2))/(pi R^(3))(1-(a^(2))/(R^(2)))`
B. `(3)/(8)"GM"^(2)(1-(a^(2))/(R^(2)))`
C. `(3)/(8)("GM"^(2))/(pi R^(4))(1-((a)/(r))^(2))`
D. `(8)/(3)("GM"^(2))/(R^(3))(1-(a)/(R))`

Answer

C

Explanation

Here pressure = `("Gravitational Pull")/("area")=("GMm")/(r^(2) xx A)`
Then pressure of a small element da is given by `dp=(-G(4pi)/(3)a^(3)rho" dS dap")/(a^(2)dS)`
Where dS is area of element of width d a
`=-(4 pi)/(3) G rho^(2)"ada"`
Integrating `P=(4)/(3) pi G rho^(2) underset(a)overset(R)int "a da"`
`P=(2 pi)/(3)G rho^(2)(R^(2)-a^(2))=(2 pi)/(3)G rho^(2)R^(2)(1-(a^(2))/(R^(2)))`
Substituting `(4 pi)/(3)R^(3) rho=M,` `P=(3)/(8)("GM"^(2))/(pi R^(4))(1-(a^(2))/(R^(2)))`