Units and Measurement

Kindly go through the solution 

Angular momentum is an axial vector along axis of rotation.

Angular momentum is an axial vector along axis of rotation.

Angular momentum is an axial vector along axis of rotation.

We cannot call the trailing zeroes as significant numbers. If you remember from the rules, it needs to have a decimal point somewhere to be considered as one. What is recommended is that one can use the measure as to the power of 10.  

%error= $\left(\frac{dV}{V}+\frac{dR}{R}\right)*100$  

$=\left(\frac{5}{200}+\frac{0.2}{20}\right)*100$                          

= 3.5

$S=\frac{F}{L}$  

[S] = [MT^–2]

$F=nA\frac{dv}{dx}$  

$\eta =\frac{ML{T}^{-2}\cdot T}{L^2}=M{L}^{-1}{T}^{-1}$

$\overrightarrow{L}=\overrightarrow{r}*\overrightarrow{p}$

L º [ML²T^–1]

$KE=\frac{1}{2}l{\omega }^2$  

KE º ML²T⁻²

$T=2\pi \sqrt[]{\frac{I}{g}}$

$g=\frac{4{\pi }^{2}I}{T^2}$

$\frac{\Delta g}{g}=\frac{\Delta I}{I}+\frac{2\Delta T}{T}$

$=\frac{0.2}{20}+2\left(\frac{1}{40}\right)$

= 6%

[A] = L²

$B=\frac{x^2}{tE}\equiv \frac{L^2}{TM{L}^2{T}^{-2}}=\frac{1}{M{T}^{-1}}$            

[B] = M^–1T

[AB] = [M^–1L²T]