2. Questions Bank
  3. JEE Main & Advanced
  4. Physics
  5. Elasticity

JEE Main & Advanced Physics Elasticity Question Bank Youngs Modulus and Breaking Stress

  • question_answer
    Young's modulus of rubber is \[{{10}^{4}}\,N/{{m}^{2}}\] and area of cross-section is \[2\,c{{m}^{2}}\]. If force of \[2\times {{10}^{5}}\]dynes is applied along its length, then its initial length l  becomes

    A) 3L          

    B)                 4L

    C)                 2L          

    D)                 None of the above

    Correct Answer: C

    Solution :

        \[Y={{10}^{4}}N/{{m}^{2}},A=2\times {{10}^{-4}}{{m}^{2}},F=2\times {{10}^{5}}dyne=2N\] \[l=\frac{FL}{AY}=\frac{2\times L}{2\times {{10}^{-4}}\times {{10}^{4}}}=L\] Final length = initial length + increment = 2L


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