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Question: \([{M^{ - 1}}{L^3}{T^{ - 2}}]\) are the dimensions of: A. Acceleration due to gravity B. Gravita...

[M1L3T2][{M^{ - 1}}{L^3}{T^{ - 2}}] are the dimensions of:
A. Acceleration due to gravity
B. Gravitational constant
C. Gravitational force
D. Gravitational potential energy

Explanation

Solution

Hint There is a very simple way to solve these types of questions. Consider each option separately and write their formula. Now replace all the masses with MM, all the lengths or distances with LL and all the times with TT. Solve it and you will get the result.

Complete step-by-step solution :For option B: Gravitational constant
Formula for gravitational constant is
G=Fr2m1m2G = \dfrac{{F{r^2}}}{{{m_1}{m_2}}}---(1)
Where F=F = Force exerted between two bodies
R=R = distance between centre of two bodies
m1,m2{m_1},{m_2}Are masses of the bodies.

F=maF = ma
a=vt=st2a = \dfrac{v}{t} = \dfrac{s}{{{t^2}}}
To write dimensional formula, replace all masses with “MM”, replace all the lengths and distances with “LL” and all the times with “TT”.
So dimensional formula of force will be [M1l1t2][{M^1}{l^1}{t^{ - 2}}]
Dimensional formula for r2{r^2}will be [M0L2T0][{M^0}{L^2}{T^0}]
Dimensional formula of m1m2 will be[M2L0T0][{M^2}{L^0}{T^0}]
Now putting these 3 in equation 1, we get:
G=[M1L1T2]×[M0L2T0]/[M2L0T0] G=[M1L1T2]×[L2]/[M2]  G = [{M^1}{L^1}{T^{ - 2}}] \times [{M^0}{L^2}{T^0}]/[{M^2}{L^0}{T^0}] \\\ G = [{M^1}{L^1}{T^{ - 2}}] \times [{L^2}]/[{M^2}] \\\
Take all the denominator values in numerator, we get:
G=[M1L1T2]×[L2]×[M2]G = [{M^1}{L^1}{T^{ - 2}}] \times [{L^2}] \times [{M^{ - 2}}]
Adding up all the powers we get:
G=[M1L3T2]G = [{M^{ - 1}}{L^3}{T^{ - 2}}]

Note:- Make sure, while expanding the formula, and expand it to its simplest form and try to for every term in the formula try to write all in terms ofM,L,TM,L,T. And make sure while taking the denominator to the numerator change the power from positive to negative of only the denominator variables.