A) \[\frac{E}{4}\]
B) \[\frac{3E}{4}\]
C) \[\frac{\sqrt{3}}{4}E\]
D) \[{{K}_{1}}\]
Correct Answer: D
Solution :
Let bob velocity be v at point B where it makes an angle of 60o with the vertical, then using conservation of mechanical energy \[K{{E}_{A}}+P{{E}_{A}}\]\[=K{{E}_{B}}+P{{E}_{B}}\] Þ \[\frac{1}{2}m\times {{3}^{2}}=\frac{1}{2}m{{v}^{2}}+mgl(1-\cos \theta )\] \[\therefore \frac{{{T}_{2}}}{{{T}_{1}}}=\sqrt{\frac{{{M}_{2}}}{{{M}_{1}}}}=\sqrt{\frac{4M}{M}}=2\]You need to login to perform this action.
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