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JCECE Engineering JCECE Engineering Solved Paper-2008

  • question_answer
    The period of oscillation of a simple pendulum is given by\[T=2\pi \sqrt{\frac{l}{g}}\], where \[l\] is about \[100\,\,cm\]  and is known to have \[1\,\,mm\] accuracy. The period is about \[2s\]. The time of \[100\] oscillations is measured by a stop watch of least count\[0.1s\]. The percentage error in \[g\] is

    A) \[0.1%\]                              

    B) \[1%\]

    C) \[0.2%\]                              

    D)  \[0.8%\]

    Correct Answer: C

    Solution :

                    \[T=2\pi \sqrt{\frac{l}{g}}\] \[\Rightarrow \]               \[{{T}^{2}}=4{{\pi }^{2}}\frac{l}{g}\] \[\Rightarrow \]               \[g=\frac{4{{\pi }^{2}}l}{{{T}^{2}}}\] Here \[%\] error in                 \[l=\frac{1\,\,mm}{100\,\,cm}\times 100=\frac{0.1}{100}\times 100=0.1%\] and \[%\] error in\[T=\frac{0.1}{2\times 100}\times 100=0.05%\] \[\therefore \]\[%\]error in\[g=%\]error\[l+2(%\]\[\text{error}\,\,\text{in}\,\,T\text{)}\]                             \[=0.1+2\times 0.05\]                             \[=0.2%\]


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