In a simple-pendulum experiment to find g, the length L is measured as 100.0 cm with a metre scale of least count 0.1 cm, and the time for 20…

The period is $T=t/20$ and $g=4\pi^2L/T^2=4\pi^2L\,(20)^2/t^2$, so the maximum fractional error is $\dfrac{\Delta g}{g}=\dfrac{\Delta L}{L}+2\dfrac{\Delta t}{t}$. Dividing the total time by the number of oscillations does not change the fractional error in $T$ since the count of oscillations is exact. Here $\Delta L/L=0.1/100.0=1.0\times10^{-3}$ and $\Delta t/t=0.1/40.0=2.5\times10^{-3}$. Therefore $\Delta g/g=1.0\times10^{-3}+2\times2.5\times10^{-3}=6.0\times10^{-3}$, so $N=6$. The result shows that the timing error dominates, which is why counting many oscillations is advised in practice. Since the fractional error multiplies to $6\times10^{-3}$, i.e. $0.6\%$, the integer asked for is $N=6$ exactly.