Question

if the unit of length be doubled then the numerical value of Universal gravitational constant G will become ​

Answer 1

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Answer 2

Answer:

The numerical value of universal gravitational constant will be 1/8  times of  G'

Explanation:

The gravitational force is defined as:

$F = \dfrac{Gm_{1}m_{2}}{r^2}$

$G= \dfrac{Fr^2}{m_{2}m_{2}}$

Where, m₁ = mass of the first object

m₂ = mass of the second object

r = distance

G = gravitational constant

The dimension formula of the gravitational constant is

$G = [MLT^{-2}]\times[L^2]\times[M^{-2}]$

$G = [M^{-1}L^{3}T^{-2}]$

If the unit of length be doubled then,

The unit of new gravitational constant

$G' =  [M^{-1}(2L)^{3}T^{-2}]$

$G'= 8\times[M^{-1}L^{3}T^{-2}]$

$G'=8G$

The unit of gravitational constant G' becomes 8 times of G.

$G = \dfrac{G'}{8}$

Hence, The numerical value of universal gravitational constant will be 1/8  times of G'