# Answer the following. Calculate the work done in stretching a steel wire of length 2 m and cross-sectional area 0.0225 mm2 when a load of 100 N is slowly applied to its free end. - Physics

Sum

Calculate the work done in stretching a steel wire of length 2 m and cross-sectional area 0.0225 mm2 when a load of 100 N is slowly applied to its free end. (Young’s modulus of steel = 2 × 1011 N/m2)

#### Solution

Given: L = 2 m, F = 100 N, A = 0.0225 mm2 = 2.25 × 10-8 m2, Y = 2 × 1011 N/m2

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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#### APPEARS IN

Balbharati Physics 11th Standard Maharashtra State Board
Chapter 6 Mechanical Properties of Solids
Exercises | Q 5. (v) | Page 113