Savitha, a XI standard student, while conducting an experiment to determine the effective length of a simple pendulum $L$, notes down the data of time taken to complete 30 oscillations as 60 s and hence calculates the length of the simple pendulum as: (Take $\pi^2=9.8$, and $g=9.8 ext{ m/s}^2$)

Question

Savitha, a XI standard student, while conducting an experiment to determine the effective length of a simple pendulum $L$, notes down the data of time taken to complete 30 oscillations as 60 s and hence calculates the length of the simple pendulum as: (Take $\pi^2=9.8$, and $g=9.8 	ext{ m/s}^2$)

Accepted Answer

D. 1. The time period $T$ of a pendulum is the time for **one** complete oscillation. Since 30 oscillations take 60 s, dividing gives the time for one oscillation:

$$
T = \frac{60}{30} = 2 	ext{ s}
$$

2. The relationship between the time period and the length of a simple pendulum is given by the standard formula:

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

3. Rearranging this formula to isolate $L$ (squaring both sides and solving) gives:

$$
L = \frac{gT^2}{4\pi^2}
$$

4. Substituting $g = 9.8 	ext{ m/s}^2$, $T = 2	ext{ s}$, and $\pi^2 = 9.8$:

$$
L = \frac{9.8 	imes (2)^2}{4 	imes 9.8} = \frac{9.8 	imes 4}{39.2} = 1 	ext{ m}
$$

Hence, the answer is **D. $1 	ext{ m}$**.