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
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