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JEE Main & Advanced Physics Mathematical Tools, Units & Dimensions Question Bank Self Evaluation Test - Physical World, Units and Measurements

  • question_answer
    The frequency (f) of a wire oscillating with a length \[\ell \], in p loops, under a tension T is given by \[f=\frac{P}{2\ell }\sqrt{\frac{T}{\mu }}\] where \[\mu =\] linear density of the wire. If the error made in determining length, tension and linear density be 1%, -2% and 4%, then find the percentage error in the calculated frequency              

    A) - 4%

    B)  - 2%

    C)  - 1%

    D)  - 5%

    Correct Answer: A

    Solution :

    [a] Given \[f=\frac{\rho }{2\ell }\sqrt{\frac{T}{\mu }}\] Taking log of both sides \[\log \,f=\log \left( \frac{p}{2} \right)-\log \ell +\frac{1}{2}\log T-\frac{1}{2}\log \mu \] Differentiating partially on both sides, \[\frac{df}{f}=0-\frac{d\ell }{\ell }+\frac{1}{2}\frac{dT}{T}-\frac{1}{2}\frac{d\mu }{\mu }\] or \[\frac{df}{f}\times 100=\left( -\frac{d\ell }{\ell }\times 100 \right)+\left( \frac{1}{2}\frac{dT}{T}\times 100 \right)-\left( \frac{1}{2}\frac{d\mu }{\mu }\times 100 \right)\]\[=(-1)+\frac{1}{2}(-2)-\frac{1}{2}(4)=-1-1-2=-4%\]


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