The period of a pendulum is given as $$T = 2\pi \sqrt{\frac{l}{g}}$$​ where g = 9.81 m/s² and π = 3.1416. The period of a pendulum of length 1 m correct to the first place of decimal in seconds is

Correct option is C

​$$\begin{aligned}&\text{Given:} \\&\quad T = 2\pi \sqrt{\frac{l}{g}},\quad l = 1\, \text{m},\quad g = 9.81\, \text{m/s}^2,\quad \pi = 3.1416 \\\\&\text{Step 1: Substitute the values into the formula} \\&\quad T = 2 \times 3.1416 \times \sqrt{\frac{1}{9.81}} \\\\&\text{Step 2: Calculate the square root} \\&\quad \sqrt{\frac{1}{9.81}} = \sqrt{0.10193} \approx 0.319 \\\\&\text{Step 3: Multiply} \\&\quad T = 6.2832 \times 0.319 \approx 2.003 \\\\&\text{Final Answer: } \boxed{2.0 \text{ seconds (to one decimal place)}}\end{aligned}$$​​