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JEE Main & Advanced JEE Main Paper (Held on 08-4-2019 Afternoon)

  • 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 :

    \[T=\frac{30\sec }{20}\]              \[\Delta T=\frac{1}{20}\sec .\] \[L=55cm\]                 \[\Delta L=1mm=0.1cm\] \[g=\frac{4{{\pi }^{2}}L}{{{T}^{2}}}\] 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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