when tension 'F' is applied the elongation produced in uniform wire of length l,radius r is e; when the tension 2F is applied the elongation produced in another uniform wire of length 2l and radius 2r made of same material is
ans:
1.)1.0 e
To find the elongation produced in the second wire when a tension of 2F is applied, we can use the concept of stress and strain.
We know that the stress on a wire is given by the formula:
Stress = Force/Area
The area of a wire is given by the formula:
Area = π * (radius)^2
When a tension F is applied to the first wire, the stress on the wire is F/(π * r^2). Let's call this stress σ1.
Now, the elongation produced in the wire is given by Hooke's Law:
Elongation = (Stress * Length) / (Young's Modulus)
Let's call the elongation produced in the first wire as e1.
So, e1 = (σ1 * l) / (Young's Modulus)
When a tension of 2F is applied to the second wire, the stress on the wire is 2F / (π * (2r)^2) = F / (π * r^2). Let's call this stress σ2.
Now, we can find the elongation produced in the second wire, let's call it e2:
e2 = (σ2 * 2l) / (Young's Modulus)
To find the ratio of e2 to e1, we can substitute the values of σ1 and σ2:
e2 / e1 = [(F / (π * r^2)) * 2l] / [(F / (π * r^2)) * l]
Simplifying this expression, we get:
e2 / e1 = 2l / l
e2 / e1 = 2
Therefore, the elongation produced in the second wire when a tension of 2F is applied is 2 times the elongation produced in the first wire.