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Question
Answer the following.
A wire of mild steel having an initial length 1.5 m and diameter 0.60 mm gets extended by 6.3 mm when a certain force is applied to it. If Young’s modulus of mild steel is 2.1 × 1011 N/m2, calculate force applied.
Solution
Given: L = 1.5 m, d = 0.60 mm,
r = `"d"/2 = 0.30`mm = 3 × 10-4 m,
Y = 2.1 × 1011 N/m2,
l = 6.3 mm = 6.3 × 10-3 m
To find: Force (F)
Formula: Y = `"FL"/("A"l)`
Calculation: From formula,
F = `("YA"l)/"L"`
`= ("Y" pi "r"^2l)/"L"`
`= (2.1 xx 10^11 xx 3.142 xx (3 xx 10^-4)^2 xx 6.3 xx 10^-3)/1.5`
`= (2.1 xx 3.142 xx 9 xx 6.3 xx 10^11 xx 10^-8 xx 10^-3)/1.5`
= 2.1 × 3.142 × 6 × 6.3
= antilog [log 2.1 + log 3.142 + log 6 + log 6.3]
= antilog [0.3222 + 0.4972 + 0.7782 + 0.7993]
= antilog [2.3969]
= 2.494 × 102
≈ 250 N
The force applied on wire is 250 N.
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