Question

Q If force (F), velocity (V) and acceleration (A) are taken as the fundamental units instead of mass, length and time; then which of the following gives the
dimensions of the Gravitational constant in terms of F, V, and A?
​

Answer 1

Explanation:

This will be the dimensions of mass if force, velocity and time are taken asfundamental quantities. Hence, if force(F), velocity (V) and time (T) are taken asfundamental units, the dimensions ofmass are [FV−1T]. So, the correct answer is “Option D”.

Answer 2

Answer:

The correct answer is equal to $[F]^{-1}[V]^{8/3}[A]^{4/3}$.

Explanation:

By the equation,

$F= \frac{GM_{1} M_{2}}{r^{2} }$

[F] denoted as the dimensional formula of F.

[$M_{1}$] denoted as the dimensional formula of $M_{1}$.

[$M_{2}$] denoted as the dimensional formula of $M_{2}$.

[$r^{2}$] denoted as the dimensional formula of $r^{2}$.

[G] denoted as the dimensional formula of G.

[V] denoted as the dimensional formula of V.

[A] denoted as the dimensional formula of A.

[G]=[$r^{2}$]×[F]/[$M_{1}$]×[$M_{2}$]

[G]=$L^{2}$×$MLT^{-2}$/M×M

[G]=$M^{-1} L^{3}T^{-2}$

The dimensional formula of the gravitational constant in terms of F, V, and A.

[G]=$[F]^{a}$$[V]^{b}$$[A]^{c}$

$M^{-1} L^{3}T^{-2}=(MLT^{-2})^{a}(LT)^{b}(LT^{-2})^{c}$

a=-1

b=8/3

c=4/3

So, the dimensional formula of the gravitational constant in terms of F, V, and A is equal to $[F]^{-1}[V]^{8/3}[A]^{4/3}$.