BHU PMT BHU PMT (Mains) Solved Paper-2011

  • question_answer
    In a simple pendulum the breaking strength of the string is double the weight of the bob. The bob is released from rest when the string is horizontal. The string breaks when it makes an angle\[\theta \]With the vertical

    A)  \[\theta ={{\cos }^{-1}}\left( \frac{1}{3} \right)\]             

    B)  \[\theta ={{\cos }^{-1}}\left( \frac{2}{3} \right)\]

    C)  \[\theta ={{60}^{o}}\]                  

    D) \[\theta =zero\]

    Correct Answer: B

    Solution :

                      By law of conservation of energy loss in PE = gain in KE \[\Rightarrow \]               \[mgl(1-\cos \theta )=\frac{1}{2}m{{v}^{2}}\] \[\Rightarrow \]               \[\frac{m{{v}^{2}}}{l}=2\,mg(1-\cos \theta )\] At point of breaking \[mg\cos \theta =\frac{m{{v}^{2}}}{l}\] \[\therefore \]  \[mg\cos \theta =2mg(1-\cos \theta )\]                 \[2mg=3mg\cos \theta \] \[\cos \theta =\frac{2}{3}\] \[\theta ={{\cos }^{-1}}\left( \frac{2}{3} \right)\]              


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