$\text{ The maximum error in }\mathrm{T}\text{ due to possible errors upto }1\mathrm{\%}\text{ in }l\text{ and }2.5\mathrm{\%}\text{ in }\mathrm{g}\text{ where period }\mathrm{T}$$\text{ of a simple pendulum is }T=2\pi \sqrt{lg}$

$\begin{array}{l}\frac{\delta l}{l}\times 100=1;\frac{\delta g}{g}\times 100=2.5;\Rightarrow \frac{\delta T}{T}\times 100=1.75\\ T=2\pi \sqrt{lg}\\ \log T=\log 2\pi +\frac{1}{2}\left(\log l+\log g\right)\\ \frac{\delta T}{T}=0+\frac{1}{2}\left(\frac{\delta l}{l}+\frac{\delta g}{g}\right)\\ \frac{\delta T}{T}\times 100=\frac{1}{2}\left(\frac{\delta l}{l}\times 100+\frac{\delta g}{g}\times 100\right)\\ =\frac{1}{2}\left(1+2.5\right)\\ =1.75\\ \text{ Hence option (A) is correct. }\end{array}$