JEE Main & Advanced Physics Simple Harmonic Motion Question Bank Simple Pendulum

  • question_answer
    The length of the second pendulum on the surface of earth is 1 m. The length of seconds pendulum on the surface of moon, where g is 1/6th value of g on the surface of earth, is [CPMT 1971]

    A)            1 / 6 m                                    

    B)            6 m

    C)            1 / 36 m                                  

    D)            36 m

    Correct Answer: A

    Solution :

                       \[T=2\pi \sqrt{\frac{l}{g}}\] Þ \[\sqrt{\frac{l}{g}}\]= constant                               \[\Rightarrow l\propto g;\] Þ \[\frac{{{l}_{m}}}{1}=\frac{1}{6}\frac{g}{g}\Rightarrow {{l}_{m}}=\frac{1}{6}m\]


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