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

  • question_answer
    In a simple pendulum experiment for determination of acceleration due to gravity (g), time taken for 20 oscillations is measured by using a watch of 1 second least count. The mean value of time taken comes out to be 30 s. The length of pendulum is measured by using a meter scale of least count 1 mm and the value obtained is 55.0 cm. The percentage error in the determination of g is close to:-                                                  [JEE Main 8-4-2019 Afternoon]

    A)  0.7%   

    B)       0.2%

    C)  3.5%   

    D)       6.8%

    Correct Answer: D

    Solution :

    [d] \[T=\frac{30\sec }{20}\]                 \[\Delta T=\frac{1}{20}\sec .\]
                \[L=55cm\]                    \[\Delta L=1mm=0.1cm\]
                \[g=\frac{4{{\pi }^{2}}L}{{{T}^{2}}}\]
    percentage error in g is
                \[\frac{\Delta g}{g}\times 100%=\left( \frac{\Delta L}{L}+\frac{2\Delta T}{T} \right)100%\]
    \[=\left( \frac{0.1}{55}+\frac{2\left( \frac{1}{20} \right)}{\frac{30}{20}} \right)100%\simeq 6.8%\]


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