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Chapter 7: Problem 971

When \(100 \mathrm{~N}\) tensile force is applied to a rod of $10^{-6} \mathrm{~m}^{2}\( cross-sectional area, its length increases by \)1 \%$ so young's modulus of material is \(\ldots \ldots \ldots \ldots\) (A) \(10^{12} \mathrm{~Pa}\) (B) \(10^{11} \mathrm{~Pa}\) (C) \(10^{10} \mathrm{~Pa}\) (D) \(10^{2} \mathrm{~Pa}\)

Short Answer

Expert verified
The Young's modulus of the material is \(10^{10} \mathrm{~Pa}\).

Step by step solution

01

Find the change in length ΔL

We are given that the length increases by 1% when a tensile force is applied. So, ΔL = L0 × 0.01
02

Rearrange the formula for Young's modulus

Young's modulus (Y) = (Tensile force (F) × Original length(L0)) / (Area (A) × Change in length (ΔL))
03

Substitute the values into the formula

Substitute the values given in the problem: Tensile force (F) = 100 N, Area (A) = 10^{-6} m², and the change in length (ΔL) = L0 × 0.01 into the formula: Y = (100 × L0) / (10^{-6} × L0 × 0.01) You can notice that 'L0' on the numerator and denominator can be canceled out.
04

Simplify the formula

After canceling out L0, we have: Y = (100) / ( 10^{-6} × 0.01) Now, let's simplify: Y = (100) / (10^{-8}) Y = 100 × 10^{8} Y = 10^{10} Pa So, the Young's modulus of the material is \(10^{10} \mathrm{~Pa}\). The correct answer is option (C).

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