A body of mass 1 kg is fastened to one end of a steel wire of cross- section area 3 xx 10^(-6)m^(2) and is rotated in horizantal circle of radius 20 cm with a constant speed 2 m//s. Find the elongation of the wire (Y= 2 xx 10^(11) N//m^(2))
Answers
Given : A body of mass 1 kg is fastened to one end of a steel wire of cross sectional area 3 × 10^-6 m² and is rotated in horizontal circle of radius 20 cm with a constant speed 2 m/s.
To find : The elongation of the wire.
solution : using formula,
Young's modulus, Y = F/A(l/L)
⇒l = FL/AY
here, here r = 20 cm = 0.2 m , v = 2m/s, m = 1 kg
F = mv²/r = 1kg × (2m/s)²/(0.20 m) = 20 N
A = 3 × 10^-6 m²
Y = 2 × 10¹¹ N/m²
now l = (20 N × 0.2 m)/(3 × 10^-6 m² × 2 × 10¹¹ N/m²)
= 4/(6 × 10^5) m
= 0.67 × 10^-5 m
Therefore elongation of the wire is 0.67 × 10^-5 m.
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Explanation:
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