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InterviewSolution
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A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1m, is whirled in a vertical circle with an angular velocity of `2 rev.//s` at the bottom of the circle. The cross-sectional area of the wire is `0.065 cm^(2)` . Calculate the elongaton of the wire when the mass is at the lowest point of its path `Y_(steel) = 2 xx 10^(11) Nm^(-2)`.A. 0.52 mmB. 1.52 mmC. 2.52 mmD. 3.52 mm |
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Answer» Correct Answer - C Here, m = 15 kg , r = L = 1 m `v = 2 "rev s"^(-1), A = 0.05 cm^(2) = 0.05 xx 10^(-4) m^(2)` When the mass is the lowest point of the vertical circle, the stretching force is `F = mg + mromega^(2) = mg + mr (2piv)^(2)` `= 15 xx 10 + 15 xx (1)xx (2pixx2)^(2) = 2516 N` As ` Y= (F//A)/(DeltaL//L) therefore DeltaL = (FL)/(AY)` `DeltaL = (2526N xx 1 m)/(0.05 xx 10^(-4)m^(2) xx 2 xx 10^(11))= 2.52 xx 10^(-3)m = 2.52 mm` |
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