Question

check the correctness of the relation, p=3g/4πRG, where the letters have their usual meaning.​

Answer 1

Answer:

Explanation:

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]

Answer 2

Answer:

The equation p=3g/4πRG is correct.

Explanation:

Given: The equation p=3g/4πRG

To find: To prove that the given equation is dimensionally correct

Solution:

we have

ρ = $(\frac{3g}{4.r.G} )$

ρ  is the density of the planet

g is the acceleration due to gravity

R is the radius of the planet

G is the universal constant

Here,

The dimension of LHS is  = $[ML^{-3} ]$

The dimension of RHS is = $\frac{[LT^{-2}] }{[L].[M^{-1} L^{3} T^{-2} ]}. }$

                                        = $[ML^{-3} ]$

Therefore the dimension of LHS = dimension of RHS

The given equation is dimensionally correct

Final answer:

The equation  ρ = $(\frac{3g}{4.r.G} )$ is dimensionally correct and its satisfy the statement.

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