The length of a simple pendulum is increased four times to its...

The time period of a simple pendulum is given by $T = 2\pi\sqrt{\dfrac{l}{g}}$.

The time period is independent of the mass of the bob.

Given that the new length is $l_2 = 4l_1$.

The new time period is $T_2 = 2\pi\sqrt{\dfrac{l_2}{g}} = 2\pi\sqrt{\dfrac{4l_1}{g}} = 2 \times 2\pi\sqrt{\dfrac{l_1}{g}} = 2T_1$.

The ratio of the new and previous time period is $\dfrac{T_2}{T_1} = \dfrac{2}{1}$.

Answer: 2 : 1