Physics, asked by ShaniKat6096, 1 year ago

Two wires made up of the same material are subjected to forces in the ratio of 1:4. Their lengths are in the ratio of 8:1 and diameter in the ratio of 2:1. Find the ratio of their extensions.

Answers

Answered by ajay6287
37
since both wires are of same material there modules of Elasticity is same
Attachments:
Answered by Dhruv4886
17

The ratio of their extensions is 1 / 2 or 1 : 2.

Both the wires are made of same material, therefore the modulus of Elasticity is equal.

E = Stress / Strain

Stress = Force / Area

Strain = change in length(Extension) / original length

Let the force for first wire be F

∴ Force for second wire is 4F

Diameters are in ratio 2:1

Areas are in ratio 4:1 ratio ( As Area is directly proportional to d²)

Area of first wire is 4A. Area of second wire is A.

Lengths are in ratio 8:1

Length of first wire is 8L. Length of second wire is L.

Let the extension in first wire be Δl1

Let the extension in second wire be Δl2

E for first wire= (F / 4A) / (Δl1 / 8L)

= F×8L / 4A×Δl1

E for second wire= (4F / A) / (Δl2 / L)

= 4F×L / A×Δl2

We know, E for first wire = E for second wire.

⇒ F×8L / 4A×Δl1 = 4F×L / A×Δl2

⇒ Δl1/ Δl2 = 1 / 2 or 1 : 2

Similar questions