Question

When a certain wire stretches 0.90cm when outward force with magnitude F are applied to each end. Thr same force are applied to a wire of same material but with three times the diameter and three times the length. The second wire stretches-

Answer 1

The second wire stretches 0.30cm.

Given,

Force (F₁) =F

Initial length = L

Extension in length (ΔL₁) =0.90 cm

Length (L₂) = 3L

Diameter (D₂) = 3D₁

Force (F₂)= F

Same material

To find,

Extension in length (ΔL₂) =?

Solution,

According to Hooke's law of elasticity:

Young's Modulus(Y) =stress/strain =FL₀ /A(ΔL)

Now, the wire is of same material so Young's Modulus will be same.

∴Y₁ = Y₂

FL₁/ A₁(ΔL₁) = FL₂ /A₂(ΔL₂)

FL/A₁ * 0.9 = F*3L / 9A₁ *(ΔL₂)

(ΔL₂) = 0.3 cm

Hence, the second wire stretches 0.30 cm.

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