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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2000

  • question_answer
    A steel wire of 1 m long and 1 mm2 cross section area is hanged from rigid end when weight of 1 kg is hung from it then change in length will be (for wire Youngs coefficient \[Y=2\times {{10}^{11}}\,N/{{m}^{2}}\]):

    A)  0.5 mm               

    B)         0.25 mm

    C)  0.05 mm             

    D)         5 mm

    Correct Answer: C

    Solution :

    Youngs coefficient of elasticity \[Y=\frac{mgL}{\pi {{r}^{2}}\Delta L}\] substituting the values we get \[\Delta L=0.05\] mm


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