MCQ
The bob of a pendulum of length $1$ is pulled aside from its equilibrium position through an angle $\theta$ and then released. The bob will then pass through its equilibrium position with a speed $v,$ where $v$ equals:
  • A
    $\sqrt{2\text{gl}(1-\sin\theta)}$
  • B
    $\sqrt{2\text{g}(1-\cos\theta)}$
  • $\sqrt{2\text{gl}(1-\cos\theta)}$
  • D
    $\sqrt{2\text{gl}(1+\sin\theta)}$

Answer

Correct option: C.
$\sqrt{2\text{gl}(1-\cos\theta)}$
The height of fall of pendulum, $\text{h = l}(1-\cos\theta)$ and $\text{v}=\sqrt{2\text{gh}}$

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