If $T=2\pi \sqrt{\frac{l}{g}},$ then relative errors in $T$ and $I$ are in the ratio 

Solution:

We have,

$\begin{array}{l}T=2\pi \sqrt{\frac{l}{g}}\\ \Rightarrow \log T=\log 2\pi +\frac{1}{2}\log l-\frac{1}{2}\log 8\\ \Rightarrow \frac{1}{T}dT=0+\frac{1}{2l}dl-0\\ \Rightarrow \frac{dT}{T}=\frac{1}{2}\frac{dl}{l}\\ \Rightarrow \frac{\mathrm{\Delta }T}{T}=\frac{1}{2}\frac{\mathrm{\Delta }l}{l} [\because dT\cong \mathrm{\Delta }T\text{ and }dl\cong \mathrm{\Delta }l]\\ \Rightarrow \frac{\mathrm{\Delta }T}{T}\frac{\mathrm{\Delta }l}{l}=\frac{1}{2}\\ \Rightarrow l\end{array}$