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NEET Physics Mathematical Tools, Units & Dimensions Question Bank Errors of Measurement

  • 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 1mm 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 :

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


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