Question

dimensions of density =3g/4pie radius of gravitational ​

Answer 1

Answer:

given expression,  

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

dimension of  = [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 = dimension of {g/rG}

LHS = dimension of  = [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.

Explanation: