A brass wire of cross-sectional area 2 mm² is suspended from a rigid support and a body of volume 100 cm³ is attached to its other end. If the decrease in the length of the wire is 0.11 mm, when the body is completely immersed in water, find the natural length of the wire (Ybrass = 0.91 × 10¹¹ N m⁻², ρwater = 10³ kg m⁻³)
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Given, cross sectional area , A = 2mm² = 2 × 10^-6 m²
Volume of the body , V = 100cm³ = 100 × 10^-6 m³ = 10^-4 m³
decrease in the length of wire, ∆L = 0.11mm = 11 × 10^-5 m
Young's modulus of brass, Y = 0.91 × 10¹¹N/m²
And density of water , d = 10³ kg/m³
Body completely immersed in water . So, a upward force acts on the body. e.g., buoyancy force .
Buoyancy force, F = volume of body × density of water × acceleration due to gravity
= 10^-4 × 10³ × 9.8
= 0.98 N
Now, Young's modulus , Y = FL/A∆L
so, L = YA∆L/F
= 0.91 × 10¹¹ × 2 × 10^-6 × 11 × 10^-5/0.98
= 2.043m
Hence, length of brass wire = 2.043 m
Volume of the body , V = 100cm³ = 100 × 10^-6 m³ = 10^-4 m³
decrease in the length of wire, ∆L = 0.11mm = 11 × 10^-5 m
Young's modulus of brass, Y = 0.91 × 10¹¹N/m²
And density of water , d = 10³ kg/m³
Body completely immersed in water . So, a upward force acts on the body. e.g., buoyancy force .
Buoyancy force, F = volume of body × density of water × acceleration due to gravity
= 10^-4 × 10³ × 9.8
= 0.98 N
Now, Young's modulus , Y = FL/A∆L
so, L = YA∆L/F
= 0.91 × 10¹¹ × 2 × 10^-6 × 11 × 10^-5/0.98
= 2.043m
Hence, length of brass wire = 2.043 m
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Explanation:
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