Q140 / 180 · NEET 2026 (Cancelled) · Physics
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 $g = 9.8$ m/s$^2$)
(Take $\pi^2 = 9.8$, and $g = 9.8$ m/s$^2$)
- A$0.75$ m
- B$1.5$ m
- C$2$ m
- Correct answerD$1$ m
Solution
Time period $T = \dfrac{60}{30} = 2$ s.
The formula for the time period of a simple pendulum is $T = 2\pi \sqrt{\dfrac{L}{g}}$.
Squaring both sides, we get $T^2 = 4\pi^2 \dfrac{L}{g}$.
Substituting the given values $T = 2$ s, $\pi^2 = 9.8$, and $g = 9.8$ m/s$^2$:
$(2)^2 = 4 \times 9.8 \times \dfrac{L}{9.8}$
$4 = 4L$
$L = 1$ m.
Answer: $1$ m
The formula for the time period of a simple pendulum is $T = 2\pi \sqrt{\dfrac{L}{g}}$.
Squaring both sides, we get $T^2 = 4\pi^2 \dfrac{L}{g}$.
Substituting the given values $T = 2$ s, $\pi^2 = 9.8$, and $g = 9.8$ m/s$^2$:
$(2)^2 = 4 \times 9.8 \times \dfrac{L}{9.8}$
$4 = 4L$
$L = 1$ m.
Answer: $1$ m
Hence, the correct answer is option: 'D'.