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Sum

**Answer the following.**

Calculate the work done in stretching a steel wire of length 2 m and cross-sectional area 0.0225 mm^{2} when a load of 100 N is slowly applied to its free end. (Young’s modulus of steel = 2 × 10^{11} N/m^{2})

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#### Solution

**Given:** L = 2 m, F = 100 N, A = 0.0225 mm^{2} = 2.25 × 10^{-8} m^{2}, Y = 2 × 10^{11} N/m^{2}

**To find:** Work (W)

**Formula: **W = `1/2 xx "F" xx l`

**Calculation:** Since, Y = `"FL"/("A"l)`

∴ l = `"FL"/"AY"`

From formula,

W = `1/2 xx ("F" xx "FL")/"AY" = 1/2 ("F"^2"L")/"AY"`

`= 1/2 xx (100 xx 100 xx 2)/(2.25 xx 10^-8 xx 2 xx 10^11)`

`= (10^4 xx 10^-11 xx 10^8)/(2.25 xx 2)`

`= 10/4.5`

= antilog [log 10 – log 4.5]

= antilog [1.0000 – 0.6532]

= antilog [0.3468]

∴ W = **2.222 J**

The work done in stretching the steel wire is **2.222 J**.

Concept: Strain Energy

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