Physics

If length L, mass M and force F are taken as fundamental quantities then the dimensions of time will be

$M^{\frac{1}{2}}L^{\frac{1}{2}}F^{-\frac{1}{2}}$

SOLUTION
We are  given that force $F$, length $L$, and mass $M$ are the fundamental quantities of time $T$ and hence we put,
$T=F^{x}L^{y}M^{z}$
Substituting their dimensions we have
$\left[T\right]=\left[M^xL^xT^{-2x}\right]L^{y}M^{z}$
Applying principle of homogeneity of dimensions we have
$1=-2x$ or $x=\dfrac{-1}{2}$,
$y= \dfrac{1}{2}$,
$z= \dfrac{1}{2}$,
$\therefore$ dimension of time becomes $M^{\frac{1}{2}} L^{\frac{1}{2}}F^{\frac{-1}{2}}$

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Single Correct Medium Published on 18th 08, 2020
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