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JCECE Medical JCECE Medical 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 V  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 s. 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 \[g=%\,\,\text{error}\,\text{in}\,l+2(%\,\text{error}\,\text{in}\,\text{T})\] \[=0.1+2\times 0.05\] \[=0.2%\]


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