Question

If force F, acceleration A and time T are taken as fundamental physical quantity then the dimension of length in this new system with be

Answer 1

Answer:-

$\mathsf{L = AT^2}$

Explanation:-

The dimensional formula of physical quantity:-

Force → $[MLT^{-2}]$

Acceleration → $[ LT^{-1}]$

Time → T

Now,

• This quantities are taken as fundamental physical quantity of length.

→$\mathsf{L = [F]^a [A]^b [T]}$

→$\mathsf{L = [MLT^{-2}]^a [LT^{-2}]^b [T]^c }$

→$\mathsf{L = M^a L^a T^{-2a} \times L^b T^{-2b}\times T^c }$

→$\mathsf{[M^0 L^1 T^0]= M^a L^{a +b}T^{-2a -2b +c}}$

• On equating with powers on L. H. S side.

→$M^0 = M^a$

→$a = 0$

→$L^1 = L^{a + b}$

→$1 = a + b$

→$b = 1$

→$T^0 = T^{-2a-2b+c}$

→$0 = -2a -b + c$

→$0 = -2\times 0 -2 +c$

→$0 = -2 + c$

→$c = 2$

Dimension of length is :-

$\mathsf{L = F^0 A^1 T^2}$

$\mathsf{L = AT^2}$