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Manipal Medical Manipal Medical Solved Paper-2010

  • question_answer
    A force of\[6\times {{10}^{6}}N{{m}^{-2}}\] required for breaking a material. The density p of the material is\[3\times {{10}^{3}}kg\text{ }{{m}^{-3}}\]. If the wire is to break under its own weight, the length of the wire made of that material should be (take\[g=10\text{ }m{{s}^{-2}}\])

    A)  20 m          

    B)  200 m

    C)  100 m         

    D)  2000 m

    Correct Answer: B

    Solution :

     When a wire is pulled it stretches (undergoes strain) upto a certain limit the amount it stretches is proportional to the load divided by the cross-sectional area of the wire. \[Stress=\frac{Force}{Area}=\frac{mg}{A}=\frac{V\rho g}{A}=\frac{LA\rho g}{A}\] \[\therefore \] Stress \[=L\rho g\] Given, stress\[=6\times {{10}^{6}}N{{m}^{-2}}\] \[\rho =3\times {{10}^{3}}kg\text{ }{{m}^{-3}}\] \[g=10m{{s}^{-2}}\] \[L=\frac{stress}{\rho g}\] \[=\frac{6\times {{10}^{6}}}{3\times {{10}^{3}}\times 10}\] \[=2\times {{10}^{2}}=200m\]


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