Question

The dimensional formula for gravitational constant G is
a) [ML³T⁻²]
b) [M⁻¹L³T⁻²]
c) [M⁻¹L⁻³T⁻²]
d) [ML⁻³T²]

Answer 1

Correct option (b) -  M⁻¹L³T⁻²

The dimensional formula for gravitational constant is M⁻¹L³T⁻²

Explanation :-

We know that the formula of gravitational Force 

$F = Gm_{1}m_{2} /r^2$

Gravitational constant  $G = Fr^2/m_{1} m_{2}$  ----→(i)

Now we know the gravitational formula for Force
 F = ma 
 F = M¹L¹T⁻²

Now in the equation (i)
$G = \frac{M^1 L^1 T^-2 ( L^2 )}{M^1 M^1}$ 

G = M¹L³T⁻²/M²
G = M⁻¹L³ T⁻²


Hence the Dimension of Gravitational constant G is  M⁻¹L³ T⁻² .



Hope it Helps. :-)

Answer 2

The Universal Gravitational Constant can be seen in Newton's Universal Law of Gravitation:

$\boxed{F = G\frac{m_1m_2}{r^2}}$

Here,
$F$ = Gravitational Force
$m_1 \, \, and \, \, m_2$ = Masses
$r$ = Distance between centers of masses

We need the Dimension of Force.

$F = ma$
So, Unit of Force = $kg \, m/s^2$
So, dimensions are:

$[\, F \, ] = [ \, M \, L \, T^{-2} \, ]$


We have:

$F = G \frac{m_1m_2}{r^2} \\ \\ \\ \implies G = \frac{Fr^2}{m_1m_2} \\ \\ \\ \implies [ \, G \, ] = \frac{[ \, F \, ] \, \, [ \, r^2 \, ]}{[ \, m_1 \, ] \, \, [ \, m_2 \, ]} \\ \\ \\ \implies [ \, G \, ] = \frac{[ \, M \, L \, T^{-2} \, ] \, \, [ \, L^2 \, ]}{[ \, M \, ] \, \, [ \, M \, ]} \\ \\ \\ \implies [ \, G \, ] = \frac{[ \, M \, L^3 \, T^{-2} \, ]}{[ \, M^2 \, ]} \\ \\ \\ \implies \boxed{[ \, G \, ] = [ \, M^{-1} \, L^3 \, T^{-2} \, ]}$

Thus, the answer is Option b) $[ \, M^{-1} \, L^3 \, T^{-2} \, ]$