Question

A wire of cross section A is stretched horizontally between two clamps located ‘2l’ m apart. A weight W kg is suspended from the mid-point of the wire. If the mid-point sags vertically through a distance x << l the strain produced is

A) 2x² / l²
B) x² / l²
C) x² / 2l²
D) x / 2l²

Answer 1

The strain produced in a wire of length 2l with a midpoint sag of x for the condition x << l is calculated using an approximation from the Pythagorean theorem to be 2x² / l².

Step-by-step explanation:

The question is asking for the strain produced in a wire of length 2l when a weight is suspended from the mid-point causing a sag of x. The calculation of strain involves the relationship between the amount of deformation experienced by an object and the original dimensions of the object. For small deformations, where x is much less than l, we can use the geometric relation in a right triangle where half of the wire (length l) and the vertical sag (x) form the sides of the triangle. The length of the stretched wire can be approximated using the Pythagorean theorem.

Using the approximation (A + x) ≈ √l² + x², the additional length (Δl) can be expressed as the difference between the original length (l) and approximated stretched length (√l² + x²), yielding Δl = √l² + x² - l. For small deformations, the strain (ε) which is strain = Δl / l can be approximated as:

1. Square both sides of the triangular relationship: l² + x² = (l + Δl)²
2. Assuming Δl << l, approximate (l + Δl)² as l² + 2lΔl.
3. Divide by l² to obtain the strain: (Δl/ l) = (2lΔl / l²) = (2x² / l²).

Therefore, the correct option is A) 2x² / l², which is the strain produced in the wire due to the sag.