A brass wire of diameter 1 mm and length 2 m is stretched by applying a force of 20 N. If the increase in length is 0.51 mm, find i. the stress ii. the strain and iii. Young's modulus of the wire
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Given, diameter of brass wire, d = 1mm
radius of the brass wire, r = 5 × 10^-4 m
force applied , F = 20N
Length of brass wire, L = 2m
Change in length of wire, ∆L = 0.51mm = 5.1 × 10^-4 m
i. stress is the ratio of force to the cross sectional area.
e.g., stress = force/area
= F/πr²
= 20/{3.14 × (5 × 10^-4)²}
= 2.546 × 10^7 N/m²
ii. Strain = change in length/original length
= 5.1 × 10^-4/2
= 2.55 × 10^-4
iii. Young's modulus = stress/strain
= 2.546 × 10^7/2.55 × 10^-4
= 9.984 × 10^10 N/m²
radius of the brass wire, r = 5 × 10^-4 m
force applied , F = 20N
Length of brass wire, L = 2m
Change in length of wire, ∆L = 0.51mm = 5.1 × 10^-4 m
i. stress is the ratio of force to the cross sectional area.
e.g., stress = force/area
= F/πr²
= 20/{3.14 × (5 × 10^-4)²}
= 2.546 × 10^7 N/m²
ii. Strain = change in length/original length
= 5.1 × 10^-4/2
= 2.55 × 10^-4
iii. Young's modulus = stress/strain
= 2.546 × 10^7/2.55 × 10^-4
= 9.984 × 10^10 N/m²
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Explanation:
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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