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)`.

Options

A. 0.52 mm
B. 1.52 mm
C. 2.52 mm
D. 3.52 mm

Answer

C

Explanation

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`