    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

$0.25m$

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B

$0.5mm$

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C

$1mm$

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D

$2mm$

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Solution

## The correct option is C $1mm$Step 1: GivenMaximum 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 lawHooke'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 deformationsForm 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 deformationLet the deformation after breaking be $∆{L}_{2}$$\frac{∆{L}_{1}}{∆{L}_{2}}=\frac{{L}_{1}}{{L}_{2}}\phantom{\rule{0ex}{0ex}}\frac{2}{∆{L}_{2}}=\frac{L}{\frac{L}{2}}\phantom{\rule{0ex}{0ex}}∆{L}_{2}=1mm$Therefore, option C is the correct option.  Suggest Corrections  1      Explore more