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A cylindrical bar of L meters deforms by I cm. The strain in bar is
1. \(\frac{l}{L}\)
2. \(\frac{0.1\ l}{L}\)
3. \(\frac{0.01\ l}{L}\)
4. \(\frac{100\ l}{L}\)

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Correct Answer - Option 3 : \(\frac{0.01\ l}{L}\)

Concept:

Whenever the bar is subjected to the axial tensile load, there will be an increase in the length of the bar along the direction of the loading.

The longitudinal strain is defined as the ratio of increase in the length of the bar in the direction of applied load to that of the original length.

\(Strain~=~\frac{Change~in~length}{Original~length}~=~\frac{\Delta L}{L}\)

Calculation:

Given:

Length of cylinder = L meters, deformation = l cm

deformation = ΔL

\(Strain= \frac{{{\rm{Δ }}L}}{L} \)

\(Strain= \frac{l}{L}\frac{{cm}}{m}\)

\(Strain= \frac{l}{{L \times 100}}\frac{{m}}{{\;m}}\)

\(Strain= \frac{0.01\ l}{L}\)

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