A wire 2.5 m long with and cross-section area of 10⁻⁵ m² is stretched 1.5 mm by a force of 100 N in the elastic region. Calculate (i) the strain (ii) Young's modulus (iii) the energy stored in the wire. (Ans : 6.02 x 10⁻⁴, 66 x 10¹⁰  Pa, 7.5 x 10⁻² J)



Given Data:

Length of wire = L =  2.5 m 

Area = A = 10⁻⁵ m²

Increased Length = ΔL=  1.5 mm = 1.5ｘ10⁻³ m

Radius steel rod = r = 1 cm = 0.01 m

Force = F = 100 N



To Find:


(a). Tensile Strain = Ɛ = ?

(b). Young's modulus of steel  = Y = ?

(c). Energy stored = Work done = ?



Solution:


(a). Tensile Strain = Ɛ = ?

The formula for tensile strain Ɛ is

Ɛ = `\frac {ΔL}{L}`

putting values

Ɛ = `\frac {1.5ｘ10⁻³ m}{2.5 m}`

Ɛ = 0.6 ｘ10⁻³

Ɛ = 6.0 ｘ10⁻⁴  ----------Ans.1


(b). Young's modulus of steel  = Y = ?

Young's Modulus is defined as :

Y = `\frac {Stress}{Strai n}` ---------(1)


Where Stress = `\frac {F}{A}` 

Stress = `\frac {100 N}{ 10⁻⁵ m²}`

Stress = 10⁷ N m⁻² 

putting the values of stress and strain in equation (1)

Y = `\frac {10⁷ N m⁻² }{0.6ｘ10⁻³ }`

by simplifying we get

Y = 1.667 ｘ10¹⁰ N m⁻² ------------Ans. (2)



(c). Energy stored = Work done = ?

The formula for work done is

 Work done = Energy stored in the wire = `\frac {1}{2}` F ｘ ΔL

Energy stored in the wire= `\frac {1}{2}` 100 N ｘ  1.5ｘ10⁻³ m

Energy stored = 50 N ｘ  1.5ｘ10⁻³ m

Energy stored = 75ｘ10⁻³ N m

or

Energy stored = 7.5ｘ10⁻² J ------------Ans.3




Similar Questions:

A force of 1.5ｘ10⁴ N causes a strain of 1.4ｘ10⁻⁴ in a steel cable of cross-sectional area 4.8ｘ10⁴ m². 

(a) What is the Young's Modulus of the steel cable? 

(b) The stress strain graph is linear for this cable. Calculate the strain energy per unit volume stored in the cable when the cable has a strain of 1ｘ10⁻⁴.  Answer((a) 2.2ｘ10¹¹ N m⁻², (b) 1.5 ｘ 10³  Jm⁻³)



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