JEE Main & Advanced JEE Main Paper (Held on 09-4-2019 Morning)

  • question_answer A simple pendulum oscillating in air has period T. The bob of the pendulum is completely immersed in a non-viscous liquid. The density of the liquid is \[\frac{1}{16}\]th of the material of the bob. If the bob is inside liquid all the time, its period of oscillation in this liquid is : [JEE Main 9-4-2019 Morning]

    A) \[4T\sqrt{\frac{1}{15}}\]

    B)   \[2T\sqrt{\frac{1}{10}}\]

    C) \[4T\sqrt{\frac{1}{14}}\]                   

    D) \[2T\sqrt{\frac{1}{14}}\]

    Correct Answer: A

    Solution :

    For a simple pendulum \[T=2\pi \sqrt{\frac{L}{{{g}_{eff}}}}\] situation 1 : when pendulum is in air \[\to {{g}_{eff}}=g\] situation 2 : when pendulum is in liquid \[\to {{g}_{eff}}=g\left( 1-\frac{{{\rho }_{liquid}}}{{{\rho }_{body}}} \right)=g\left( 1-\frac{1}{16} \right)=\frac{15g}{16}\] So,\[\frac{T'}{T}=\frac{2\pi \sqrt{\frac{L}{15g/16}}}{2\pi \sqrt{\frac{L}{g}}}\Rightarrow T'=\frac{4T}{\sqrt{15}}\] Option [a]


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