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.

Answer 1

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.