Question

In a particular system the units of length, mass and time are chosen to be $10cm,10g$ and $0.1s$ respectively. The unit of force in this system will be equal to

Answer 1

$\because n_1{u}_1=n_2{u}_2\therefore n_2=\left(\begin{array}{c}\frac{u_1}{u_2}\end{array}\right)n_1=\left(\begin{array}{c}\frac{M_1{L}_1{T}_1^{-2}}{M_2{L}_2{T}_2^{-2}}\end{array}\right)\left(n_1\right)$

$\Rightarrow n_2=\left[\begin{array}{c}\frac{\left(1000g\right)\left(100cm\right)(1{)}^{-2}}{\left(10g\right)\left(10cm\right)(0.1{)}^{-2}}\end{array}\right]\left[\begin{array}{c}1\end{array}\right]=10$

$1N=10\times$ Unit of force in new system

So, unit of force in new system $=0.1N$

As $\left[F\right]=ML{T}^{-2}$

So, unit of force $=\left(10g\right)\left(10cm\right(0.1s{)}^{-2}$

$=\left({10}^{-2}kg\right)\left({10}^{-1}m\right)100(s{)}^{-2}$

$=0.1\left(kg\right)\left(m\right)(s{)}^{-2}=0.1N$