Question

If force (F), velocity (u) and time (T) are taken
as fundamental quantities, then dimension of
length is​

Answer 1

Answer:

[ u T ]  

Explanation:  

$\large \text{Let L \propto F^a \ V^b \ T^c \ ...(i) }\\\\\\\large \text{We know dimension of Force = \left[MLT^{-2}\right] }\\\\\\\large \text{Velocity = \left[LT^{-1}\right] }\\\\\\\large \text{Time= \left[T^1 \right] }}\\\\\\\large \text{We putting all values }\\\\\\\large \left[M^0LT^0\right]= k\left( \left[MLT^{-2}\right] \right )^a \left( \left[LT^{-1} \right] \right)^b\left(\left[T^1 \right]\right)^c$

$\large \left[M^0LT^0\right]= k\left( \left[M^\right] \right )^a \left( \left[L\right] \right)^{a+b}\left(\left[T\right]\right)^{-2a-b+c}$

Now comparing on both side we have :  

a = 0  

a + b = 1

b = 1 - a

b = 1

-2 a - b + c = 0

- b + c = 0  

c = b  

c = 1  

Now put all these value in ( i ) we get

$\large L \propto F^a \ V^b \ T^c \ ...(i)\\\\\\\large L= k \ F^0 \ V^1 \ T^{1} \ ..\\\\\\\large L= k \ VT\\\\\\\large \text{Given V = velocity = u}\\\\\\\large \text{So , we can write it as L = [ u T ]}$

Thus , we get Dimension of Length is [ u T ] .

Answer 2

We're given that force (F), velocity (u) and time (T) are taken as fundamental quantities. We have to derive length (L) in terms of these three.

Well, a simple method is given below.

Among the given three, first we consider the derivation of a derived quantity in terms of the others, including L, in real situation.

In real situation, T is a fundamental quantity, so it can't be considered.

Consider F.

We know  $[F]=[ma],\ \ \ [a]=\left[\dfrac{u}{T}\right]$  and  $[u]=\left[\dfrac{L}{T}\right]$

So,  $[F]=\left[\dfrac{mL}{T^2}\right]\ \implies\ [L]=\left[\dfrac{FT^2}{m}\right]$

But this can't be accepted, because it's not given in the question that mass is taken as a fundamental quantity!

So this can't be taken.

Finally, consider u.

We had  $[u]=\left[\dfrac{L}{T}\right]$

This implies  $[L]=[u\ T]$

Well, this is enough!

Including F, we can rewrite it as,

$\Large\boxed{[L]=F^0\ u^1\ T^1}$