JEE Main & Advanced Sample Paper JEE Main Sample Paper-28

  • question_answer
    The time period of oscillation of a simple pendulum is given by \[T=2\pi \sqrt{\frac{\ell }{g}}\]. Find percentage error in calculation g while \[\ell =10\pm -0.1\,cm\] and \[T=0.5\pm 0.02\,cm\]

    A)  \[9%\]                                 

    B)  \[7%\] 

    C)  \[4%\]                                 

    D)  \[2%\]

    Correct Answer: A

    Solution :

    \[g=\frac{4{{\pi }^{2}}\ell }{{{T}^{2}}}\] \[\frac{\Delta g}{g}=\frac{\Delta \ell }{l}+\frac{2\Delta T}{T}\]each term in rhs is positive because error always increases in each step of calculation \[\frac{\Delta g}{g}\times 100=\frac{0.1}{10}\times 100+2\times \frac{0.02}{0.5}\times 100\] \[=1+8=9%\]


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