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Q.

A string of length l = 1 m is fixed at one end and carries a mass of 100gm at other end. The string makes 5/π revolutions per second about a vertical axis passing through its second end. What is the angle of inclination of the string with the vertical ? Take g = 10 m/s2

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a

30∘

b

45∘

c

60∘

d

75∘

answer is C.

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Detailed Solution

The different forces are shown in fig. (10). From figureTsin⁡θ=mv2r=mω2r=mω2lsin⁡θ           ....(1)and     Tcos⁡θ=mgFrom eq. (1),T=mω2l=mω2 (∵l=1m)From eq. (2) mω2cos⁡θ=mg∴ cos⁡θ=gω2=g4π2n2=104π25/π2or         cos⁡θ=0.5 i.e., θ=60∘
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