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CMC Medical CMC-Medical VELLORE Solved Paper-2015

  • question_answer
    The length of an elastic rope is 4.5 m and radius 3mm is fixed to a free-limb on an end. A monkey weighing 100 N jumps to catch the free end and stays there. If Youngs modulus for the rope is \[4.8\times {{10}^{11}}N{{m}^{-2}}\] and Poissons ratio is 0.2, then change in diameter corresponding to elongation would be

    A)  \[8.6\times {{10}^{-5}}\,m\]                      

    B)  \[7.9\times {{10}^{-9}}\,m\]

    C)  \[8.2\times {{10}^{-7}}\,m\]                      

    D)  \[8.6\times {{10}^{-8}}\,m\]

    E)  \[8.8\times {{10}^{-9}}\,m\]

    Correct Answer: E

    Solution :

                    As \[Y=\frac{\text{stress}}{\text{strain}}=\frac{T/A}{l/L}\] \[\Rightarrow \]               \[l=\frac{TL}{AY}\]   \[=\frac{100\times 4.5}{\pi \times 9\times {{10}^{-6}}\times 4.8\times {{10}^{11}}}\]   \[=3.32\times {{10}^{-6}}m\] Again Poisson ratio,                 \[=\frac{\Delta d/d}{l/L}=\frac{\Delta d\times 4.5}{3.32\times {{10}^{-5}}\times 6\times {{10}^{-3}}}\] \[\Rightarrow \]               \[\Delta d=\frac{0.2\times 6\times 3.32\times {{10}^{-8}}}{4.5}\]                       \[=8.8\times {{10}^{-9}}m\]


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