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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2010

  • question_answer
    On all the six surfaces of a unit cube, equal tensile force of F is applied. The increase in length of each side will be (Y = Youngs modulus, \[\rho \]= Poissions ratio)

    A)  \[\frac{F}{Y(1-\sigma )}\]            

    B)         \[\frac{F}{Y(1+\sigma )}\]

    C)  \[\frac{F(1-2\sigma )}{Y}\]         

    D)         \[\frac{F}{Y(1+2\sigma )}\]

    Correct Answer: C

    Solution :

    Tensile strain on each face\[=\frac{F}{Y}\] Lateral strain due to the other two forces acting on perpendicular faces\[=\frac{-2\sigma F}{Y}\] Total increase in length\[=(1-2\sigma )\frac{F}{Y}\]


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