AMU Medical AMU Solved Paper-2004

  • question_answer
    The diameter of brass road is 4 mm. Youngs modulus of brass is 9 xl010 N/m2. The force required to stretch 0.1% of its length is

    A)  360 \[\pi \]N                     

    B)  36 N

    C) \[\text{36}\pi \times \text{1}{{0}^{5}}\text{ N}\]             

    D)  \[\text{14471}\times \text{1}{{0}^{\text{3}}}\text{ N}\]

    Correct Answer: A

    Solution :

                     The ratio of stress to strain is known as Youngs modulus of wire                 \[Y=\frac{F/A}{l/L}\] \[\Rightarrow \]               \[F=\frac{YAl}{L}\] Given,   \[Y=9\times {{10}^{10}}N/{{m}^{2}}\],                 \[A=\pi {{r}^{2}}=\pi {{(2\times {{10}^{-3}})}^{2}}\],                 \[l=0.1%L\] \[\therefore \]  \[F=\frac{9\times {{10}^{10}}\times \pi \times {{(2\times {{10}^{-3}})}^{2}}\times 0.1}{100}\] \[\Rightarrow \]               \[F=360\,\pi \,N\]


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