A steel wire of radius 0.4 x l0⁻³ m and length 1 m is tightly clamped between points A and B which are separated by 1 m and in the same horizontal plane. A mass is hung from the middle point of the wire such that the middle point sags by 2 cm from the original position. Compute the mass of body
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As given the figure :
AD = BD = 0.50 m
In ∆ADC,
Let AC = x
∴ x =√( 50² + 2 ²) =√ (2500 + 4 )=√( 2504)
= 50.03 cm = 50.03 × 10⁻³m
Elongation, l = x – 0.50 = 50.03 – 50 = 0.03 cm
Resolve T as shown in figure
2T cos (90 – θ) = mg
∴ 2T sin θ = mg
∴ 2T × 2x 10⁻²/x = mg
∴ 2T × 2 x10⁻²/ 50.03 x10⁻² = mg
∴ 2 × Y × Ax l x 2/ (50 10 –2 × 50.03) = m × 9.8
2 × Y × πr ² (x0.03 x10⁻² / 50x 10 ⁻²) × (2/ 50.03) = m × 9.8
∴ m = 2x20x10¹⁰x3.14x(0.14x10⁻³)2x0.03x2/50x50.03x9.8
∴ m = 0.492 kg
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Answer:
As given the figure :
AD = BD = 0.50 m
In ∆ADC,
Let AC = x
∴ x =√( 50² + 2 ²) =√ (2500 + 4 )=√( 2504)
= 50.03 cm = 50.03 × 10⁻³m
Elongation, l = x – 0.50 = 50.03 – 50 = 0.03 cm
Resolve T as shown in figure
2T cos (90 – θ) = mg
∴ 2T sin θ = mg
∴ 2T × 2x 10⁻²/x = mg
∴ 2T × 2 x10⁻²/ 50.03 x10⁻² = mg
∴ 2 × Y × Ax l x 2/ (50 10 –2 × 50.03) = m × 9.8
2 × Y × πr ² (x0.03 x10⁻² / 50x 10 ⁻²) × (2/ 50.03) = m × 9.8
∴ m = 2x20x10¹⁰x3.14x(0.14x10⁻³)2x0.03x2/50x50.03x9.8
∴ m = 0.492 kg
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