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Question 76 of 104

Q.A steel wire having cross sectional area 1.5 mm² when stretched by a load produces a lateral strain 1.5×10−51.5\times10^{-5}. Calculate the mass attached to the wire. (Ysteel=2×1011Y_{steel} = 2\times10^{11} N/m², Poisson's ratio σ=0.291\sigma = 0.291, g = 9.8 m/s²)

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2016Subjective· 3mImportance★★★★★
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Use Poisson's ratio to get the longitudinal strain from the given lateral strain, then Young's modulus to get the stress, and finally the force (weight).

Poisson's ratio σ\sigma relates lateral strain to longitudinal strain:

σ=lateral strainlongitudinal strain\sigma=\frac{\text{lateral strain}}{\text{longitudinal strain}}

So the longitudinal strain is

longitudinal strain=lateral strainσ=1.5×10−50.291≈5.155×10−5\text{longitudinal strain}=\frac{\text{lateral strain}}{\sigma}=\frac{1.5\times10^{-5}}{0.291}\approx 5.155\times10^{-5}

Young's modulus relates stress to longitudinal strain:

Y=stresslongitudinal strain ⇒ stress=Y×longitudinal strainY=\frac{\text{stress}}{\text{longitudinal strain}}\ \Rightarrow\ \text{stress}=Y\times\text{longitudinal strain}

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