Question

If force (F), velocity (V) and mass (M) are chosen as fundamental quantities, then the dimensions of length

in terms of these quantities would be

Answer 1

Explanation:

F = ma

= kg M/s^2

v = M/s

m= kg

$M = (kg \: M \: {s}^{ - 2} )^{a} \times (M \: {s}^{ - 1} )^{b} \times (kg)^{c}$

$M^{1} = (kg)^{a + c} \times M^{a + b} \times {s}^{ - 2a - b}$

Equating powers on both sides

a+c= 0

a+b=1

-2a-b=0

We get :-

a = -1

b = 2

c = 1

So, the dimensions of length in terms of these quantities would be

$F ^{ - 1} V^{2} M^{1}$