Question 41: Units , Dimensions and Error Analysis

Question

Dimensions of universal gravitational constant ($G$) in terms of Planck's constant ($h$), distance ($L$), mass ($M$) and time ($T$) are _______.

Accepted Answer

Using the dimensional formula of gravitational constant:
$ [G] = \frac{[F]\,[r^2]}{[m_1][m_2]} $
Since,
$ [F] = [M L T^{-2}] $
So,
$ [G] = \frac{[M L T^{-2}] [L^2]}{[M][M]} = [M^{-1} L^3 T^{-2}] $
Now, we need to express this in terms of $h$, $L$, $M$, and $T$.
We know PlanckÃ¢ÂÂs constant has dimensions:
$ [h] = [M L^2 T^{-1}] $
Assume
$ [G] = [h^a L^b M^c T^d] $
Then,
$ [h^a L^b M^c T^d] = [M^a L^{2a} T^{-a}] [L^b] [M^c] [T^d] = [M^{a+c} L^{2a+b} T^{-a+d}] $
Compare with
$ [G] = [M^{-1} L^3 T^{-2}] $
So,
$ a+c = -1 $
$ 2a+b = 3 $
$ -a+d = -2 $
Now check options.
Option A:
$ [h T L M^{-2}] $
Dimensions:
$ [h][T][L][M^{-2}] = [M L^2 T^{-1}] [T] [L] [M^{-2}] = [M^{-1} L^3 T^0] $
Not correct.
Option B:
$ [h T^{-1} L M^{-2}] $
Dimensions:
$ [h][T^{-1}][L][M^{-2}] = [M L^2 T^{-1}] [T^{-1}] [L] [M^{-2}] = [M^{-1} L^3 T^{-2}] $
This matches $[G]$.
So, the correct answer is:
$ \boxed{[h T^{-1} L M^{-2}]} $
Hence, Option B is correct.

Answer

$[h T L M^{-2}]$

Answer

$[h T^{-1} L M^{-2}]$

Answer

$[h T L^2 M^{-2}]$

Answer

$[h^{-1} T^{-1} L M^{-2}]$

Units , Dimensions and Error Analysis

Dimensions of universal gravitational constant (G) in terms of Planck's constant (h), distance (L), mass (M) and time (T) are _______.

Choose the correct answer: