A mass of 6.5 kg is hanging from the end of
a 60 cm long steel wire (Y = 2 x 10^11 Pa) with
area of cross-section 0.05cm2. When it is
revolving in a vertical circel it has an angular
velocity of 2 revolutions per second, at the
bottom of the circle. Approximate elongation
of the wire (in meters) when the mass is at its
lowest point of the trajectory is:
Answers
Approximate elongation of the wire is 8 x 10^-4 m.
Explanation:
Tension in the string on crossing the mean position.
T = mg + mv^2/r
= mg + mv^2/l
But V = √2gl(1 - cosθ)
Given θ = 90°
Therefore
V = √2gl
T = = mg + m/l x 2gl = 3 mg
Young's modulus "Y" = FL/Ae
e / L = F / AY
3 x 1 x 1 /3 x 10^-6 x 10^11 = 8 x 10^-4 m
Approximate elongation of the wire is 8 x 10^-4 m
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For a given material the youngs modulus is 2.4 times its modulus of rigidity. its poisons ratio is ?
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Answer:
8 x 10^-4
Explanation:
Tension in the string on crossing the mean position.
T = mg + mv^2/r
= mg + mv^2/l
But V = √2gl(1 - cosθ)
Given θ = 90°
Therefore
V = √2gl
T = = mg + m/l x 2gl = 3 mg
Young's modulus "Y" = FL/Ae
e / L = F / AY
3 x 1 x 1 /3 x 10^-6 x 10^11 = 8 x 10^-4 m
Approximate elongation of the wire is 8 x 10^-4 m