Correct Answer is Option (D)

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:

$T^{\prime }=2\pi \sqrt{\frac{L^{\prime }}{g}}=2\pi \sqrt{\frac{1.21L}{g}}=\sqrt{1.21}T\approx 1.1T$

The percentage increase in the time period is then:

$\frac{T^{\prime }-T}{T}\times 100\mathrm{\%}=(\sqrt{1.21}-1)\times 100\mathrm{\%}\approx 10\mathrm{\%}$

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

Correct Answer is Option (B)

KEY CONCEPT : The time period $T=2\pi \sqrt{\frac{\ell }{g}}$ where

$\ell$ $=$ distance between the point of suspension and the center of mass of the child. This distance decreases when the child stands

$\therefore$ $T^{\prime }<T$ i.e., the period decreases.