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CMC Medical CMC-Medical Ludhiana Solved Paper-2013

  • question_answer
    A wire 4m long and area of cross-section 12 \[c{{m}^{2}}\] is stretched by a force of\[4.8\times {{10}^{3}}N\]. If Youngs elasticity coefficient of material of wire is \[1.2\times {{10}^{11}}N\text{/}{{m}^{2}},\]then increases in length will be

    A)  1.33 mm                             

    B)  1.33 cm

    C)  2.66 cm                               

    D)  0.55 cm

    Correct Answer: A

    Solution :

                    Young elasticity coefficient \[Y=\frac{F\cdot L}{A\cdot I}-\] \[I=\frac{F\cdot L}{AY}\] \[=\frac{4.8\times {{10}^{3}}\times 4}{1.2\times {{10}^{-4}}\times 1.2\times {{10}^{m}}}\] \[=\frac{4}{3}\times {{10}^{-3}}m\] \[=1.33\,mm\]


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