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JEE Main & Advanced JEE Main Online Paper (Held on 9 April 2013)

  • question_answer
                                    Two simple pendulums of length 1m and 4mrespectively are both given small d is placement in the same direction at the same instant. They will be again in phase after the shorter pendulum has completed number of oscillations equal to:    JEE Main Online Paper (Held On 09 April 2013)           

    A)                 2                             

    B)                 7                                                             

    C)                 5                                                             

    D)                 3                

    Correct Answer: A

    Solution :

                                                    Let \[{{T}_{1}}\] and \[{{T}_{2}}\] be the time period of shorter length and larger length pendulums respectively. According to question,                 \[n{{T}_{1}}=(n-1){{T}_{2}}\]                 So, \[n2\pi =\sqrt{\frac{1}{g}}=(n-1)\,2\pi \sqrt{\frac{4}{g}}\]                 or \[n=(n-1)2\]                 \[n=2n-2\Rightarrow \,n=2\]                


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