
The dimensional formula of the universal gravitational constant is …………………
$
(a){\text{ }}\left[ {{L^0}{M^0}{T^0}} \right] \\
(b){\text{ }}\left[ {{L^2}{M^1}{T^0}} \right] \\
(c){\text{ }}\left[ {{L^{ - 1}}{M^1}{T^{ - 2}}} \right] \\
(d){\text{ }}\left[ {{L^3}{M^{ - 1}}{T^{ - 2}}} \right] \\
$
Answer
600.6k+ views
Hint – In this question use the dimensional formula of force, distance and mass, as gravitational constant G can be expressed in terms of Force, distance between two bodies and the mass of the bodies that is $F = G\dfrac{{m.M}}{{{r^2}}}$.
Formula used: $F = G\dfrac{{m.M}}{{{r^2}}}$
Complete step-by-step solution -
As we know force (F) between two masses (m) and (M) is
$F = G\dfrac{{m.M}}{{{r^2}}}$, where G is called a gravitational constant and r is the distance between them.
So, the formula of universal gravitational constant G is
$ \Rightarrow G = \dfrac{{F.{r^2}}}{{m.M}}$
Now as we know according to Newton’s second law of motion, force acting on the moving body is the product of the mass (M) and the acceleration (a)
Therefore, F = (M. a).
Now as we know that the dimension of mass (M) is ($M^1$).
And we know the S.I unit of acceleration (a) is m/$s^2$.
The dimension of meter is ($L^1$) and the dimension of second (s) is ($T^1$).
So the dimension of acceleration is ($L^1 T^{-2}$).
Therefore, the dimension of force (F) is $[M^1 L^1 T^{-2}]$.
And we all know distance is measured in meters so the dimension of r is [$L^1$].
Therefore, the dimension of G is
$ \Rightarrow G = \dfrac{{\left[ {{M^1}{L^1}{T^{ - 2}}} \right]{{\left[ {{L^1}} \right]}^2}}}{{\left[ {{M^2}} \right]}}$
Now on simplifying we have,
$ \Rightarrow G = \left[ {{M^{ - 1}}{L^3}{T^{ - 2}}} \right]$
So this is the required dimension of universal gravitational constant (G).
Hence option (D) is the correct answer.
Note – Dimension formula is the expression for the unit of a physical quantity in terms of the fundamental quantities. The fundamental quantities are mass (M), Length (L) and time (T). A dimensional formula is expressed in terms of power of M, L and T.
Formula used: $F = G\dfrac{{m.M}}{{{r^2}}}$
Complete step-by-step solution -
As we know force (F) between two masses (m) and (M) is
$F = G\dfrac{{m.M}}{{{r^2}}}$, where G is called a gravitational constant and r is the distance between them.
So, the formula of universal gravitational constant G is
$ \Rightarrow G = \dfrac{{F.{r^2}}}{{m.M}}$
Now as we know according to Newton’s second law of motion, force acting on the moving body is the product of the mass (M) and the acceleration (a)
Therefore, F = (M. a).
Now as we know that the dimension of mass (M) is ($M^1$).
And we know the S.I unit of acceleration (a) is m/$s^2$.
The dimension of meter is ($L^1$) and the dimension of second (s) is ($T^1$).
So the dimension of acceleration is ($L^1 T^{-2}$).
Therefore, the dimension of force (F) is $[M^1 L^1 T^{-2}]$.
And we all know distance is measured in meters so the dimension of r is [$L^1$].
Therefore, the dimension of G is
$ \Rightarrow G = \dfrac{{\left[ {{M^1}{L^1}{T^{ - 2}}} \right]{{\left[ {{L^1}} \right]}^2}}}{{\left[ {{M^2}} \right]}}$
Now on simplifying we have,
$ \Rightarrow G = \left[ {{M^{ - 1}}{L^3}{T^{ - 2}}} \right]$
So this is the required dimension of universal gravitational constant (G).
Hence option (D) is the correct answer.
Note – Dimension formula is the expression for the unit of a physical quantity in terms of the fundamental quantities. The fundamental quantities are mass (M), Length (L) and time (T). A dimensional formula is expressed in terms of power of M, L and T.
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