Arrange the following units of length in the increasing order of their magnitude

i) Tm       (ii) pm             (iii)  $\mu m$              (iv) Pm

Given units of length are

(i) $1\text{Tm}={10}^{12}\text{m\hspace{0.17em}\hspace{0.17em}}(\because 1\text{T}={10}^{12})$

$\left(\text{ii}\right)1\text{pm\hspace{0.17em}}={10}^{-12}\text{m\hspace{0.17em}\hspace{0.17em}}(\because 1\text{p\hspace{0.17em}}={10}^{-12})$

$\text{(iii)\hspace{0.17em}}1\mu \text{m\hspace{0.17em}}={10}^{-6}\text{m\hspace{0.17em}\hspace{0.17em}(}\because \text{\hspace{0.17em}1}\mu {\text{\hspace{0.17em}=\hspace{0.17em}10}}^{-6})$

$\left(\text{iv}\right)1\text{Pm\hspace{0.17em}}={10}^{15}\text{m\hspace{0.17em}}(\because 1\text{P\hspace{0.17em}}={10}^{15})$

We get the above units in the increasing order as Pm(ii), $\mu$m(iii),Tm(i),Pm(iv)