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JEE Main & Advanced JEE Main Paper Phase-I Held on 07-1-2020 Morning

  • question_answer
    Speed of transverse wave on a straight wire (mass 6.0 g, length 60 cm and area of cross section\[1.0\,\,m{{m}^{2}}\]) is \[90\text{ }m{{s}^{1}}\]. If the Young?s modulus of wire is \[16\times {{10}^{11}}\text{ }N{{m}^{2}}\], the extension of wire over its natural length is: [JEE MAIN Held on 07-01-2020 Morning]

    A) 0.01 mm

    B) 0.02 mm

    C) 0.04 mm

    D) 0.03 mm

    Correct Answer: D

    Solution :

    [d] \[l=60\text{ }cm,\text{ }m=6\text{ }g,\text{ }A=1\text{ }m{{m}^{2}},\text{ }v=90\text{ }m/s\] \[v=\sqrt{\frac{T}{m}\times l}\Rightarrow T=\frac{m{{v}^{2}}}{l}\] \[\therefore \Delta L=\frac{Tl}{YA}=\frac{m{{v}^{2}}\times l}{l(YA)}\] \[=\frac{6\times {{10}^{-3}}\times {{90}^{2}}}{16\times {{10}^{11}}\times {{10}^{-6}}}=3\times {{10}^{-4}}m\] \[=0.03\,\,mm\]


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