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Question

A 6-kg weight is fastened to the end of a steel wire of unstreched length 60 cm. It is whirled in a vertical circle and has an angular velocity of 2 revolution per second at the bottom of the circle. The area of cross-section of the wire is 0.05cm2. The elongation of the wire when the weight is at the lowest point of the path is (Ysteel=2×1011Pa)


A

3.77×103m

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B

3.77×102m

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C

3.77 m

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D

3.77×104m

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Solution

The correct option is D

3.77×104m


Let l be the length of the at the bottommost point. l=l0+Δl.

We have,

Δl=Wl0YA. Also TA=YAΔll0

In the rest frame of the weight, we have at the bottommost position

T=mg+mω2l

YAΔll0=mg+mω2(l0+Δl)

On solving and substituting,

Δl=mg+mω2l0YAl0mω2

=3.77×104m


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