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प्रश्न
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. (Young’s modulus of steel = 2 × 1011 N/m2)
उत्तर
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.
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