ytsimple34
Answered

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correctness the relation Where pressure=3accleration due to gravity/4pi G​

Sagot :

worlongyangya

Explanation:

given expression, \rho=\frac{3g}{4rG}ρ=

4rG

3g

where \rhoρ is density , g is acceleration due to gravity , r is radius and G is universal gravitational constant.

dimension of \rhoρ = [ML^-3]

dimension of g = [LT^-2]

dimension of r = [L]

dimension of G = [M^-1L^3T^-2]

expression will be dimensionally correct,

dimension of \rhoρ = dimension of {g/rG}

LHS = dimension of \rhoρ = [ML^-3]

RHS = dimension of {g/rG} = dimension of g/dimension of r × dimension of G

= [LT^-2]/[L][M^-1L^3T^-2]

= [ML^-3]

here, LHS = RHS

so, expression is dimensionally correct.

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