Question

A load of 4 kg is suspended from a ceiling through a steel wire of length 20 m and radius 2 mm. It is found that the length of the wire increases by 0.031 mm, as equilibrium is achieved. If g = 3.1xxpi ms(-2), the value of young's modulus of the material of the wire (in Nm^(-2)) is

Answer 1

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Answer 2

Given :

Load = 4 kg

length of steel = 20 m

Radius of wire = 2mm

Change in length = 0.031mm

To Find :

Young modulus of wire is 2×10¹² Nm^-²

• On application of stress , strain is produced which leads to change in length of wire.

•Force applied (F) = mg = 4 × 3.1 ×

3.4 N

•Length of steel bar = 20 m

•Cross-sectional area of wire = (3.14)(2×10^-³)² m²

• ∆L = 0.031 mm = 3.1 × 10^-5 m

•Also , Relation between stress and strain is such that , stress is Young's modulus times strain

i.e. Stress = Y . Strain

where , stress is Force per unit Area

strain is change in length per unit length

We get,

F/A = (Y∆L)/L

Y = (F.L)/A.∆L

where ∆L is elongation of wire

Now putting the given values in equation .

Y = (4×3.1×3.14×20)/

(3.14)(2×10^-³)².3.1 × 10^-5

Y = 20 /10^-¹¹

Y = 2×10¹² Nm^-²

Value of young modulus of material of wire is 2×10¹² Nm^-²