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Question

If work done in stretching a wire by 1 mm is 2J. Then the work necessary for stretching another wire of same material but with double the radius and half the length by 1 mm in joule is

A
1/4
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B
4
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C
8
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D
16
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Solution

The correct option is D 16
The stretching force F=YAΔll
where Y=Young's modulus, A=Area of cross-section of wire, l=actual length of wire, Δl=increase in length.
F=Yπr2Δll
As the material is same Y does not change.
F1F2=r21Δl1l1r22Δl2l2
Here Δl1=1mm
Δl2=1mm
l2=12l1
r2=2r1
F1F2=r21×1mml14r21×1mm12l1
F1F2=18
The work done in stretching wire by amount Δl is W=12FΔl
Hence W1W2=F1F2=18
As F1=2
F2=2×8=16
Hence the correct option is (D).

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