The length of the pendulum which is displaying SHM is increased by 21%. Calculate the percentage increase in the periodic time of the pendulum

The correct option is 10%.

Step-by-Step Explanation:

The time period of a simple pendulum executing Simple Harmonic Motion (SHM) is given by the formula:
$$T=2\pi \sqrt{\frac{L}{g}}$$ where:
T is the periodic time,
L is the length of the pendulum, and
g is the acceleration due to gravity (which remains constant).

From this formula, we can see that the time period T is directly proportional to the square root of the length L:
$$T\propto \sqrt{L}$$

Let the initial length of the pendulum be L₁ = L. Therefore, the initial time period is T₁ ∝ L^1/2.

The length of the pendulum is increased by 21%. Thus, the new length L₂ becomes:
$$L_2=L+21\%\text{ of }L=L+0.21L=1.21L$$

The new time period T₂ is proportional to the square root of the new length L₂:
$$T_2\propto \sqrt{L_2}\propto \sqrt{1.21L}$$

We can write the ratio of the new time period to the initial time period as:
$$\frac{T_2}{T_1}=\sqrt{\frac{L_2}{L_1}}=\sqrt{\frac{1.21L}{L}}=\sqrt{1.21}$$

Since the square root of 1.21 is 1.1, we have:
$$\frac{T_2}{T_1}=1.1\Rightarrow T_2=1.1T_1$$

Now, we calculate the percentage increase in the periodic time:
$$\text{Percentage Increase}=\left(\frac{T_2-T_1}{T_1}\right)\times 100\%$$
$$\text{Percentage Increase}=\left(\frac{1.1T_1-T_1}{T_1}\right)\times 100\%$$
$$\text{Percentage Increase}=\left(\frac{0.1T_1}{T_1}\right)\times 100\%=0.1\times 100\%=10\%$$

Therefore, the percentage increase in the periodic time of the pendulum is 10%.