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BCECE Engineering BCECE Engineering Solved Paper-2011

  • question_answer
    A man measures the period of simple pendulum inside a stationary lift and finds it to be T second. If the lift accelerates  downwards with acceleration of \[\frac{g}{4},\] the period of oscillation will be

    A) \[T\times \frac{\sqrt{3}}{2}s\]                                   

    B) \[T\times \frac{2}{\sqrt{3}}s\]

    C)  \[\frac{T}{2}s\]                

    D)        \[\sqrt{T}s\]

    Correct Answer: B

    Solution :

    Effective acceleration due to gravity \[g=g-\frac{g}{4}=\frac{3g}{4}\] Time period   \[T=2\pi \sqrt{\left( \frac{l}{g} \right)}\]     ?(i) \[T=2\pi \sqrt{\left( \frac{4l}{3g} \right)}=2\pi \times \frac{2}{\sqrt{3}}\sqrt{\left( \frac{l}{g} \right)}\]    ?(ii) Now dividing Eq. (ii) by Eq. (i), we get \[\frac{T}{T}=\frac{2}{\sqrt{3}}\]or \[T=T\times \frac{2}{\sqrt{3}}\]


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