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 $4 L$ and radius $4 r$ is pulled by a force $4 F$ under same conditions The increase in length of this wire is ______$cm$

Answer 1

Correct Answer: 5

To calculate the increase in length of the second wire, we use the formula for linear expansion under tension, given by the relation \(\Delta L = \frac{FL}{A \cdot Y}\), where \(F\) is the force applied, \(L\) is the original length, \(A\) is the cross-sectional area of the wire, and \(Y\) is Young's modulus of the material. The cross-sectional area \(A\) is determined by \(A = \pi r^2\).
For the first wire, we have:
\[ \Delta L_1 = \frac{F \times L}{\pi r^2 \cdot Y} = 5 \, \text{cm} \]
For the second wire, the length is \(4L\), the radius is \(4r\), and the force is \(4F\).
The cross-sectional area \(A_2\) of the second wire is \(\pi (4r)^2 = 16 \pi r^2\).
The increase in length \(\Delta L_2\) is calculated as:
\[ \Delta L_2 = \frac{4F \times 4L}{16 \pi r^2 \cdot Y} \]
Simplifying:
\[ \Delta L_2 = \frac{16FL}{16 \pi r^2 \cdot Y} = \frac{FL}{\pi r^2 \cdot Y} = 5 \, \text{cm} \]
Therefore, the increase in length of the second wire is \(5\) cm, which matches the given expected range of \(5,5\).