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Chapter 7: Problem 851

A steel wire of cross-sectional area \(3 \times 10^{-6} \mathrm{~m}^{2}\) can with stand a maximum strain of \(10^{-3}\) Young's modulus of steel is $2 \times 10^{11}\left(\mathrm{~N} / \mathrm{m}^{2}\right)$. The maximum mass the wire can hold is..... $\left[\mathrm{g}=10\left(\mathrm{~m} / \mathrm{s}^{2}\right)\right]$ (A) \(40 \mathrm{~kg}\) (B) \(60 \mathrm{~kg}\) (C) \(80 \mathrm{~kg}\) (D) \(100 \mathrm{~kg}\)

Short Answer

Expert verified
The maximum mass the wire can hold is 60 kg, which corresponds to option (B).

Step by step solution

01

Calculate Stress

Calculate the stress (σ) using the Young's Modulus (Y) and the maximum strain (strain_max): \[σ = Y × strain\_max\]
02

Find the maximum force

Use the calculated stress (σ) and the cross-sectional area (A) to find the maximum force (F_max) the wire can withstand: \[F\_max = σ × A\]
03

Calculate the maximum mass

Now, use the formula for the force due to gravity to find the maximum mass (m_max) the wire can hold: \[F\_max = m\_max × g\] Solving for m_max, we get: \[m\_max = \frac{F\_max}{g}\] Now, let's plug in the given values and find the maximum mass.
04

Plug in values

Use the given values from the problem: A = \(3 × 10^{-6} m^2\) strain_max = \(10^{-3}\) Y = \(2 × 10^{11}(N/m^2)\) g = 10 m/s²
05

Calculate σ and F_max

Calculate the stress (σ) and the maximum force (F_max): σ = \((2 × 10^{11})(10^{-3}) = 2 × 10^8 N/m^2\) F_max = \((2 × 10^8)(3 × 10^{-6}) = 600 N\)
06

Compute m_max

Finally, calculate the maximum mass (m_max): m_max = \(\frac{600}{10} = 60 kg\) Thus, the maximum mass the wire can hold is 60 kg, which corresponds to option (B).

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