Two intervals of time are measured as Δ t_1=(2.00 ± 0.02) s and Δ...

Here, $\Delta t_1=(2.00 \pm 0.02) \mathrm{s}$ and $\Delta t_2=(4.00 \pm 0.02) \mathrm{s}$ Given, $T=\sqrt{\left(\Delta t_1\right)\left(\Delta t_2\right)}=\sqrt{(2.00)(4.00)}=2.828427 \mathrm{~s}$ Now, according to the relative error of product \pm $\frac{\Delta T}{T}= \pm \left(\frac{1}{2} \frac{\Delta t_1}{t_1}+\frac{1}{2} \frac{\Delta t_2}{t_2}\right)$Since, $\Delta t_1$ and $\Delta t_2$ have 3 significant figures, then $T$ also has 3 significant figures. Hence, after rounding off, $$\begin{array}{rlrl} \text{ and } \therefore \text{ and } T \text{ and } =2.83 \mathrm{~s} \\ \text{ and } \text { Now, } \text{ and } \Delta T \text{ and } = \pm \frac{1}{2}\left(\frac{0.02}{2.00}+\frac{0.02}{4.00}\right) \times2.8284 \\ \Delta T \text{ and } =0.02121 \mathrm{~s} \end{array}$$So, after rounding off, $$\begin{aligned} \DeltaT & =0.02 \\ T & =(2.83 \pm 0.02) \mathrm{s} \end{aligned}$$Hence, So, most matched answer is (b).