JEE Main & Advanced Physics Elasticity JEE PYQ-Elasticity

  • question_answer
    Two steel wires having same length are suspended from a ceiling under the same load. If the ratio of their energy stored per unit volume is 1 : 4, the ratio of their diameters is [JEE MAIN Held on 09-01-2020 Evening]

    A)  \[\sqrt{2}:1\]    

    B)       \[2:1\]

    C)  \[1:\sqrt{2}\]    

    D)       \[1:2\]

    Correct Answer: A

    Solution :

    [a] Energy density \[=\frac{1}{2}\,stress\times strain\]
    Energy density\[=\frac{1}{2}\frac{F}{A}\times \frac{F}{AY}\]
    \[\frac{{{u}_{1}}}{{{u}_{2}}}={{\left( \frac{{{d}_{2}}}{d{{}_{1}}} \right)}^{4}}\]
    \[\frac{{{d}_{1}}}{{{d}_{2}}}={{\left( 4 \right)}^{1/4}}\]
    \[\frac{{{d}_{1}}}{{{d}_{2}}}=\sqrt{2}:1\]


You need to login to perform this action.
You will be redirected in 3 sec spinner