Solveeit Logo
Login

Question

Question: The dimension of universal constant of gravitation G is A. \(\left[ M{{L}^{2}}{{T}^{-1}} \right]\)...

The dimension of universal constant of gravitation G is
A. [ML2T−1]\left[ M{{L}^{2}}{{T}^{-1}} \right]
B. [M−1L3T−2]\left[ {{M}^{-1}}{{L}^{3}}{{T}^{-2}} \right]
C. [M−1L2T−2]\left[ {{M}^{-1}}{{L}^{2}}{{T}^{-2}} \right]
D. [ML3T−2]\left[ M{{L}^{3}}{{T}^{-2}} \right]

Explanation

Solution

Hint : The gravitational force between two objects of mass M and m are given by F=GMmr2F=\dfrac{GMm}{{{r}^{2}}}, where G is the gravitational constant, by putting the dimensions of each quantity that is present in the formula we can find the dimensional formula for gravitational constant.

Formula Used:
F=GMmr2F=\dfrac{GMm}{{{r}^{2}}}
Where:
F is the Gravitational force acting on two objects
M, m are mass of two objects and,
R is the distance between the objects.

Complete step by step answer:
The gravitational force acting on two particles placed at a distance apart is given by F=GMmr2F=\dfrac{GMm}{{{r}^{2}}}. This implies that gravitational force is directly proportional to the product of mass of two objects and inversely proportional to the square of the distance between them.
We know that the dimension of mass is [M][M] dimension of length is [L][L] and the dimension of time is [T][T].
We know that force acting on the particle is given by F=maF=ma, therefore first we find the dimensional formula for force.
F=ma F=mvt F=mlt2 F=[M][L][T2] F=[MLT−2] \begin{aligned} & F=ma \\\ & F=m\dfrac{v}{t} \\\ & F=m\dfrac{l}{{{t}^{2}}} \\\ & F=\dfrac{\left[ M \right]\left[ L \right]}{\left[ {{T}^{2}} \right]} \\\ & F=\left[ ML{{T}^{-2}} \right] \\\ \end{aligned}
With the help of this we can find the dimensions of gravitational constant:

F=GMmr2 G=Fr2Mm G=[MLT−2][L2][M2] G=[M−1L3T−2] \begin{aligned} & F=\dfrac{GMm}{{{r}^{2}}} \\\ & G=\dfrac{F{{r}^{2}}}{Mm} \\\ & G=\dfrac{\left[ ML{{T}^{-2}} \right]\left[ {{L}^{2}} \right]}{\left[ {{M}^{2}} \right]} \\\ & G=\left[ {{M}^{-1}}{{L}^{3}}{{T}^{-2}} \right] \\\ \end{aligned}
Hence, the dimensional formula for gravitational constant is G=[M−1L3T−2]G=\left[ {{M}^{-1}}{{L}^{3}}{{T}^{-2}} \right].

Therefore option B is the correct answer.

Note : the value of gravitation constant is 6.67×10−11Nm2kg−26.67\times {{10}^{-11}}N{{m}^{2}}k{{g}^{-2}}. Acceleration due to gravity in terms of gravitation constant is given by g=GMR2g=\dfrac{GM}{{{R}^{2}}} 9.8ms−29.8m{{s}^{-2}}
Where:
g is acceleration due to gravity. The value of g is 9.8ms−29.8m{{s}^{-2}}
M is mass of earth which is 6×1024kg6\times {{10}^{24}}kg
R is radius of earth which is 6.4×106m6.4\times {{10}^{6}}m