The time period $T$ of a pendulum of length $l$ is given by $T=2\pi\sqrt{\frac{l}{g}}$. What is the percentage error in $T$ when the error in length is $1\text{%}$.
Attempt-
$\displaystyle T=2\pi\sqrt{\frac{l}{g}}$ and $\displaystyle\frac{\Delta l}{l}=0.01$
By using differentials we get
$\displaystyle\Delta T\approx\frac{dT}{dl}\Delta l=\frac{\pi}{\sqrt{gl}}0.01$
Therefore $\displaystyle\frac{\Delta T}{T}*100=\frac{1}{2l}$ which is wrong dimensionally by an extra $l$ in denominator.
What is the problem here?
${\Delta l} = 0.01*l$ . You have missed out an extra $l$
Therefore Percentage error in $\displaystyle T=\frac{dT}{dl}\Delta l=\frac{dT}{dl}*0.01*l$
Now the $l$ in denominator cancels
To avoid such mistakes you can use logarithmic differentiation also.