Units & Measurement

Mean of the five observations,

$\mu=\frac{80.0+80.5+81.0+81.5+82}{5}=\frac{405.0}{5}=81$

$\therefore$ Mean error

$=\frac{|80-\mu|+|80.5-\mu|+|81.0-\mu|+|81.5-\mu| +|82-\mu|}{5}$

$=\frac{|80-81|+|80.5-81|+|81.0-81| +|81.5-81|+|82-81|}{5}$

$\begin{gathered} =\frac{1+0.5+0+0.5+1}{5}=\frac{3}{5}=0.6 \\ \therefore \quad \text { Mean % error }=\frac{0.6}{81} \times 100 \%=0.74 \% \end{gathered}$

According to question, $[X]=[C]=\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^2 \mathrm{Q}^2\right]$ and $[Z]=[B]=\left[\mathrm{MT}^{-1} \mathrm{Q}^{-1}\right]$

$\begin{aligned} \therefore \quad & {[Y]=\frac{[X]}{\left[Z^2\right]} } \\ & {[Y]=\frac{\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^2 \mathrm{Q}^2\right]}{\left[\mathrm{M}^2 \mathrm{~T}^{-2} \mathrm{Q}^{-2}\right]}=\left[\mathrm{M}^{-3} \mathrm{~L}^{-2} \mathrm{~T}^4 \mathrm{Q}^4\right] } \end{aligned}$