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JEE Main & Advanced JEE Main Paper (Held On 9 April 2014)

  • question_answer
    An experiment is performed to obtain the value of acceleration due to gravity g by using a simple pendulum of length L. In this experiment time for 100 oscillations is measured by using a watch of 1 second least count and the value is 90.0 seconds. The length L is measured by using a meter scale of least count 1 mm and the value is 20.0 cm. The error in the determination of g would be:   [JEE Main Online Paper ( Held On 09 Apirl  2014  )

    A) 1.7%

    B) 2.7%

    C) 4.4%                     

    D) 2.27%

    Correct Answer: B

    Solution :

    According to the question. \[T=(90\pm 1)\]or,\[\frac{\Delta t}{t}=\frac{1}{90}\] \[l=(20\pm 0.1)\]or\[\frac{\Delta l}{l}=\frac{0.1}{20}\] \[\frac{\Delta g}{g}%=?\] As we know,\[t=2\pi \sqrt{\frac{l}{g}}\]\[\Rightarrow \]\[g=\frac{4{{\pi }^{2}}l}{{{t}^{2}}}\] or\[\frac{\Delta g}{g}=\pm \left( \frac{\Delta l}{l}+2\frac{\Delta t}{t} \right)\] \[=\left( \frac{0.1}{20}+2\times \frac{1}{90} \right)\]       \[=0.027\] \[\therefore \]\[\frac{\Delta g}{g}%=2.7%\]


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