Question

A wire of length L and radius r is clamped rigidly at one end. When the other end of the wire is pulled by a force F, its length increases by 5 cm. Another wire of the same material of length 4L and radius 4r is pulled by a force 4F under same conditions. The increase in length of this wire is ___ cm.

Answer 1

Correct Answer: 5

To solve this problem, we apply the formula for elongation in a wire under tension: ΔL = (F×L)/(A×Y), where ΔL is the change in length, F is the force, L is the original length, A is the cross-sectional area, and Y is the Young's modulus of the material.
Since both wires are made of the same material, their Y value is identical.
The cross-sectional area A of a wire with radius r is given by A=πr².
For the first wire, we know:
1. F = F
2. L = L
3. A = πr²
4. ΔL = 5 cm
Now, applying the formula for the first wire:
5 = (F×L)/(πr²×Y)(1)
For the second wire:
1. F = 4F
2. L = 4L
3. A = π(4r)² = 16πr²
Let's find ΔL' for the second wire using the formula:
ΔL' = (4F×4L)/(16πr²×Y) = (16F×L)/(16πr²×Y)
This simplifies to:
ΔL' = (F×L)/(πr²×Y)(2)
Comparing (1) and (2), we see that ΔL' = 5 cm. Therefore, the increase in length of the second wire is confirmed to be 5 cm, which fits within the expected range of 5 to 5 cm.