Question

A cable breaks if stretched by more than $2mm$. It is cut into two equal parts. How much either part can be stretched without breaking?

• A. $0.25m$
• B. $0.5mm$
• C. $1mm$
• D. $2mm$

Answer 1

The correct option is C

$1mm$

Step 1: Given

Maximum elongation in the original cable, $∆L_1=2mm$

The original length of the cable $L$

Length of the cables after breaking $\frac{L}{2}$

Step 2: Hooke's law

Hooke's law states that the deformation of a body is directly proportional to the applied load.

$∆L=\frac{PL}{AE}$

(Where $∆L$ is the deformation, $P,L,A,E$ are the applied load, the length, the area, and Young's modulus respectively.)

Step 3: Determine the relationship between the lengths and the deformations

Form the above formula, $∆L\propto L$

From the above proportionality, we can write

$\frac{∆L_1}{∆L_2}\propto \frac{L_1}{L_2}$

Step 4: Calculate the maximum deformation

Let the deformation after breaking be $∆L_2$

$$\begin{gathered} \frac{∆L_1}{∆L_2}=\frac{L_1}{L_2} \\ \frac{2}{∆L_2}=\frac{L}{{\displaystyle \frac{L}{2}}} \\ ∆L_2=1mm \end{gathered}$$

Therefore, option C is the correct option.