The period of a simple pendulum is given by:

$$ T = 2\pi \sqrt{\frac{L}{g}} $$

where:

• T is the period,
• L is the length of the pendulum, and
• g is the acceleration due to gravity.

Since g is constant, we see that the period T is proportional to the square root of the length L.

If L is increased by 21%, the new length L' is L + 21%L = 1.21L. The new period T' is then:

$$ T' = 2\pi \sqrt{\frac{L'}{g}} = 2\pi \sqrt{\frac{1.21L}{g}} = \sqrt{1.21}T \approx 1.1T $$

The percentage increase in the time period is then:

$$ \frac{T' - T}{T} \times 100\% = (\sqrt{1.21} - 1) \times 100\% \approx 10\% $$

Therefore, the percentage increase in the time period of the pendulum of increased length is approximately 10%.