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Question

A copper rod of length L and radius ris suspended from the ceiling by one of its ends. What will be elongation of the rod due to its own weight when rho blank a n d blank Y are the density and Young’s modulus of the copper respectively?

  1. fraction numerator rho to the power of 2 end exponent g L to the power of 2 end exponent over denominator 2 Y end fraction    
  2. fraction numerator rho g L to the power of 2 end exponent over denominator 2 Y end fraction    
  3. fraction numerator rho to the power of 2 end exponent g to the power of 2 end exponent L to the power of 2 end exponent over denominator 2 Y end fraction    
  4. fraction numerator rho g L over denominator 2 Y end fraction    

The correct answer is: fraction numerator rho g L over denominator 2 Y end fraction


    The weight of the rod can be assumed to act at its mid-point.
    Now, the mass of the rod is
    M equals V rho
    M equals A L rho
    Here, A equals a r e a blank o f blank c r o s s minus s e c t i o n s comma
    L= length of the rod.
    Now, we know that the Young’s modulus

    Y equals fraction numerator fraction numerator M g L over denominator 2 end fraction over denominator A bullet l blank end fraction blank left parenthesis H e r e comma blank L equals fraction numerator L over denominator 2 end fraction comma l equals e x t e n s i o n right parenthesis
    l equals fraction numerator fraction numerator M g L over denominator 2 end fraction over denominator A Y end fraction
    or l equals fraction numerator M g L over denominator 2 A Y end fraction
    On putting the value of M from Eq.(i), we get
    l equals fraction numerator A L rho bullet g L over denominator 2 A Y end fraction
    orl equals fraction numerator rho g L to the power of 2 end exponent over denominator 2 Y end fraction

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