Physics-
General
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Question

A cube of aluminium of sides 0.1 m is subjected to a sharing force of 100 N. The top face of the cube is displaced through 0.02 cm with respect to the bottom face. The shearing strain would be

  1. 0.02    
  2. 0.1    
  3. 0.005    
  4. 0.002    

The correct answer is: 0.002


    Shearing strain = fraction numerator 0.02 blank cross times 10 to the power of negative 2 end exponent over denominator 0.1 blank end fraction equals 0.002

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