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Question

Breaking stress for steel is F/A, where A is the area of cross-section of steel wire. A body of mass M is tied at the end of the steel wire of length L and whirled in a horizontal circle. The maximum number of revolution it can make per second is:

A
FA4π2ML
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B
F4π2MLA
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C
F4π2ML
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D
4π2MLA
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Solution

The correct option is C F4π2ML
Breaking stress for steel is FA. On rotating the mass M, F=mω2Lw=FML Revolutions per second =f=w2π=F4π2ML

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