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

  • question_answer
    A uniform plank of Young?s modulus Y is moved over  a smooth horizontal surface by a constant horizontal force F. The area of cross section of the plank is A. The compressive strain on the plank in the direction of the force is [Kerala PET 2002]

    A)             \[F/AY\]

    B)                      \[2F/AY\]

    C)             \[\frac{1}{2}(F/AY)\]

    D)                      \[3F/AY\]

    Correct Answer: A

    Solution :

                    \[Y=\frac{F/A}{\text{Strain}}\Rightarrow \text{strain}=\frac{F}{Ay}\]


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