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 figureTsinθ=mv2r=mω2r=mω2lsinθ ....(1)and Tcosθ=mgFrom eq. (1),T=mω2l=mω2 (∵l=1m)From eq. (2) mω2cosθ=mg∴ cosθ=gω2=g4π2n2=104π25/π2or cosθ=0.5 i.e., θ=60∘