2. Solved Papers
  3. JEE Main & Advanced

JEE Main & Advanced JEE Main Paper (Held on 7 May 2012)

  • question_answer
    A structural steel rod has a radius of 10 mm and length of 1.0 m. A 100 kN force stretches it along its length. Young's modulus of structural steel is\[2\times {{10}^{11}}N{{m}^{-2}}.\]The percentage strain is about   JEE Main Online Paper (Held On 07 May 2012)

    A) 0.16%     

    B)                                                        0.32%

    C)                        0.08%                                   

    D)                        0.24%

    Correct Answer: A

    Solution :

    Given: \[F=100kN={{10}^{5}}N\] \[Y=2\times {{10}^{11}}N{{m}^{-2}}\] \[{{\ell }_{0}}=1.0m\] radius r = 10mm= 10-2m From formula, \[\text{Y=}\frac{\text{Stress}}{\text{Strain}}\] \[\Rightarrow \]Strain \[\text{=}\frac{\text{Stress}}{Y}=\frac{S}{AY}\] \[\text{=}\frac{{{10}^{5}}}{\pi {{r}^{2}}Y}=\frac{{{10}^{5}}}{3.14\times {{10}^{-4}}\times 2\times {{10}^{11}}}\] \[=\frac{1}{628}\] Therefore % strain \[=\frac{1}{628}\times 100=0.16%\]


You need to login to perform this action.
You will be redirected in 3 sec spinner