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1a. Define Hooke's law. b. A wire of length 10m and a uniform area of 0.1 mm² is made into a resistor. If the resistivity of the wire is 50µcm, calculate the resistance. 2. If 2 kilogram-force is applied to the end of a wire of length L= 2m and diameter 0.64mm produces an extension of 0.60mm, calculate the Young modulus for the wire.​

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Answer:

a. Hooke's law is a principle in physics that states that the force required to extend or compress a spring or elastic material is directly proportional to the displacement of its end or deformation.

b. The formula for resistance of a wire is given by:

R = (ρL)/A

where R is resistance, ρ is resistivity, L is length, and A is cross-sectional area.

Given L = 10m, A = 0.1 mm² = 0.1 x 10^-6 m², and ρ = 50 µcm = 50 x 10^-8 Ωm, we have:

R = (50 x 10^-8 x 10) / (0.1 x 10^-6)

R = 5 Ω

Therefore, the resistance of the wire is 5 Ω.

The formula for Young's modulus is given by:

Y = F L / A ΔL

where Y is Young's modulus, F is force, L is length, A is cross-sectional area, and ΔL is the change in length.

Given F = 2 kgf = 2 x 9.81 N, L = 2 m, d = 0.64 mm = 0.64 x 10^-3 m, and ΔL = 0.60 mm = 0.60 x 10^-3 m, we have:

A = πd²/4 = π(0.64 x 10^-3)²/4 = 3.21 x 10^-7 m²

Y = (2 x 9.81 x 2) / (3.21 x 10^-7 x 0.60 x 10^-3)

Y ≈ 1.24 x 10^11 N/m²

Therefore, the Young's modulus of the wire is approximately 1.24 x 10^11 N/m².

Step-by-step explanation:

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