Question

If force[f], velocity [v],and time [t] are taken as fundamental units then the dimensions of mass are

Answer 1

Explanation:

force/velocity×time

Answer 2

ANSWER:

If force[f], velocity [v], and time [t] are taken as fundamental units then the dimensions of mass are $\text { (Force) }^{1}(\text { Velocity })^{-1}(\text { Time })^{1}$

EXPLANATION:

Let the formula for mass be: $M=(f)^{a} \times(v)^{b} \times(t)^{c}$

=> $\mathrm{M}=\left(\mathrm{MLt}^{-2}\right)^{\mathrm{a}} \times\left(\mathrm{Lt}^{-1}\right)^{\mathrm{b}} \times(\mathrm{t})^{\mathrm{c}}$

=> $\mathrm{M}^{1} \mathrm{L}^{0} \mathrm{t}^{0}=\mathrm{M}^{\mathrm{a}} \times \mathrm{L}^{\mathrm{a}+\mathrm{b}} \times \mathrm{t}^{\mathrm{c}-2 \mathrm{a}-\mathrm{b}}$

Comparing the powers

=> a = 1   and  a + b = 0  => b = -1  and  c - 2a - b = 0 => c = 2 -1 = 1

So, the dimensions of mass will be $\text { (Force) }^{1}(\text { Velocity })^{-1}(\text { Time })^{1}.$