A simple pendulum is being used to determine the value of gravitational acceleration  $\left(g\right)$  at a certain place the length of the pendulum is  $25.0cm$  and a stop watch with  $1s$  resolution measures the time taken for 40 oscillations to be  $50s$ , the accuracy in  $g$ is  $x\left(1.10\right)\%$ . Then the value of  x  is _________

Given length of pendulum $l=25.0cm$
So, there is an uncertainity of 0.1 cm  in measurement of length 
Resolution of stop watch is 1 sec, so uncertainity in measurement of time is 1 sec
Now using  $T=2\pi \sqrt{\frac{l}{g}}$   (or)  $g=4{\pi }^2\left(\frac{l}{g}\right)$
We have  $\frac{\Delta g}{g}\times 100=[\frac{\Delta l}{l}+\frac{2\left(\Delta T\right)}{T}]\times 100\%$
Accuracy in measurement of g  is $$\begin{gathered} \frac{\Delta g}{g}\times 100\%=[\frac{0.1}{25}+\frac{2\left(1\right)}{50}]\times 100\% =(0.004+0.04)\times 100\% =4.40\%=4\left(1.10\right)\% \\ ;x=4 \end{gathered}$$