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JEE Main & Advanced JEE Main Solved Paper-2015

  • question_answer
    A pendulum made of a uniform wire of cross sectional area A has time period T. When an additional mass M is added to its bob, the time period change to \[{{T}_{M}}.\]If the Young's modulus of the material of the wire is Y then\[\frac{1}{Y}\]is equal to : (g = gravitational acceleration) [JEE Main Solved Paper-2015 ]

    A)  \[\left[ 1-{{\left( \frac{{{T}_{M}}}{T} \right)}^{2}} \right]\frac{A}{Mg}\]

    B)  \[\left[ 1-{{\left( \frac{T}{{{T}_{M}}} \right)}^{2}} \right]\frac{A}{Mg}\]

    C) \[\left[ {{\left( \frac{{{T}_{M}}}{T} \right)}^{2}}-1 \right]\frac{A}{Mg}\] 

    D) \[\left[ {{\left( \frac{{{T}_{M}}}{T} \right)}^{2}}-1 \right]\frac{Mg}{A}\]

    Correct Answer: C

    Solution :

                    \[T=2\pi \sqrt{\frac{\ell }{g}}\] \[{{T}_{M}}=2\pi \sqrt{\frac{\ell }{g}}\] \[{{T}_{M}}=2\pi \sqrt{\frac{\frac{{{\ell }_{0}}\left( 1+\frac{F}{AY} \right)}{g}}{{}}}\] \[{{T}_{M}}={{T}_{0}}\sqrt{\left( 1+\frac{F}{AY} \right)}\] \[{{\left( \frac{{{T}_{M}}}{{{T}_{0}}} \right)}^{2}}=1+\frac{F}{AY}\] \[\frac{F}{AY}={{\left( \frac{{{T}_{M}}}{{{T}_{0}}} \right)}^{2}}-1\] \[\frac{1}{Y}=\frac{A}{F}{{\left( 1-\frac{{{T}_{M}}}{{{T}_{0}}} \right)}^{2}}\] \[\frac{1}{Y}=\frac{A}{mg}\left( {{\left( \frac{{{T}_{M}}}{{{T}_{0}}} \right)}^{2}}-1 \right)\]


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