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Question

The area of cross section of a steel wire of Young's modulus Y=2.0×1011 N/m2 is 0.1 cm2. Find the force required to double its length.
(Assume area of cross section remains same)
  1. 2×104 N
  2. 2×1011 N
  3. 2×1010 N
  4. 2×1012 N


Solution

The correct option is A 2×104 N
Given,
Young's modulus (Y)=2.0×1011 N/m2
Area of cross-section(A)=0.1 cm2
Let l be the length of steel wire
When the length of wire is doubled, then  strain =Δll=1

Y=Stressstrain=FA

Force=Y×A=2×1011×0.1×104=2×106 N

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