JEE Main & Advanced Physics Simple Harmonic Motion JEE PYQ-Simple Harmonic Motion

  • question_answer
    A simple pendulum is being used to determine the value of gravitational acceleration g at a certain place. The length of the pendulum is 25.0 cm and a stopwatch with 1 s resolution measures the time taken for 40 oscillation to be 50 s. The accuracy in g is [JEE MAIN Held on 08-01-2020 Evening]

    A)  3.40% 

    B)       2.40%

    C)  5.40% 

    D)       4.40%

    Correct Answer: D

    Solution :

    [d] l = 25.0 cm
    Time of 40 oscillation is 50 sec
    \[\therefore \,\,\,\,\,\,\,\,g=\frac{4{{\pi }^{2}}l}{{{T}^{2}}}\]            \[\Rightarrow \,\,\,\,\,\,\,\frac{\Delta g}{g}=\frac{\Delta l}{l}+\frac{2\Delta T}{T}\]
    \[\Rightarrow \,\,\,\,\,\,\,\frac{\Delta g}{g}=\left( \frac{0.01}{25.0} \right)+2\left( \frac{1}{50} \right)\]
    \[\Rightarrow \,\,\,\,\,\,\,\left( \frac{\Delta g}{g}\times 100 \right)=4.4%\]


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