Q123 / 180 · NEET 2026 (Cancelled) · Physics
The sum of kinetic energy and potential energy of a simple pendulum bob is $0.02$ joule. The speed of the simple pendulum bob at equilibrium position is approximately :
(Consider mass of the bob $= 20$ g)
(Consider mass of the bob $= 20$ g)
- A$0.2$ m/s
- Correct answerB$1.41$ m/s
- C$14.1$ m/s
- D$2.0$ m/s
Solution
Total energy of the simple pendulum is given as $E = 0.02$ J.
Mass of the bob, $m = 20$ g $= 0.02$ kg.
At the equilibrium position, the potential energy is zero and the total energy is purely kinetic.
$E = \dfrac{1}{2} m v^2$
$0.02 = \dfrac{1}{2} \times 0.02 \times v^2$
$v^2 = 2$
$v = \sqrt{2} \approx 1.41$ m/s.
Answer: $1.41$ m/s
Mass of the bob, $m = 20$ g $= 0.02$ kg.
At the equilibrium position, the potential energy is zero and the total energy is purely kinetic.
$E = \dfrac{1}{2} m v^2$
$0.02 = \dfrac{1}{2} \times 0.02 \times v^2$
$v^2 = 2$
$v = \sqrt{2} \approx 1.41$ m/s.
Answer: $1.41$ m/s
Hence, the correct answer is option: 'B'.