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JEE Main & Advanced Sample Paper JEE Main Sample Paper-41

  • question_answer
    The period of oscillation of a simple pendulum is given by \[T=2\pi \sqrt{\frac{l}{g}},\]where I 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 0.1 s. The percentage error in g is

    A)  0.1%                          

    B)  1%

    C)  0.2%                          

    D)  0.8%

    Correct Answer: C

    Solution :

    \[T=2\pi \sqrt{\frac{l}{g}}\] \[{{T}^{2}}=4{{\pi }^{2}}\frac{l}{g}\]\[\Rightarrow \]\[g=\frac{4{{\pi }^{2}}l}{{{T}^{2}}}\] Here percentage 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 in I + 2 (% error in T) = 0.1+ 2 x 0.05 = 0.2%


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