A) \[mgl\text{ }sin\theta \]
B) \[mgl\text{ }(1+sin\theta )\])
C) \[mgl(1+cos\theta )\]
D) \[mgl\text{ }(1-\cos \theta )\]
Correct Answer: D
Solution :
\[h=l-l\cos \theta \] The energy of pendulum at point A will be completely potential energy at point B will be kinetic energy. \[\Rightarrow \] \[\frac{1}{2}m{{v}^{2}}=mgh\] \[\frac{1}{2}m{{v}^{2}}=mgl(1-\cos \theta )\]You need to login to perform this action.
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