A 20 kN weight is suspended by two wires as shown in the figure. The length of each wire is 2 meters. The steel wire (E value 200 GPa) has a cross-sectional area of 60 × 10^{-6} m^{2} and the aluminum wire (E value 70 GPa) has a cross-sectional area of 120 × 10^{-6 }m^{2}. The stress in aluminium wire is

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ISRO Scientist ME 2014 Paper

Option 4 : \(\frac{7}{102}~GPa\)

CT 1: Ratio and Proportion

2672

10 Questions
16 Marks
30 Mins

**Concept:**

The both wires having equal deformation due to load of 20 kN.

Therefore, δ_{Al} = δ_{St}

And the deformation is given as, \(\delta=\frac{PL}{AE}\)

**Calculation:**

**Given:**

L_{Al} = L_{St} = 2 m, A_{Al} = 120 × 10^{-6} m^{2}, E_{Al} = 70 GPa, A_{St }= 60 × 10^{-6} m^{2}, E = 200 GPa

\( \Rightarrow \frac{{{P_{Al}}{L_{Al}}}}{{{A_{Al}}{E_{Al}}}} = \frac{{{P_{st}}{L_{st}}}}{{{A_{st}}{E_{st}}}}\)

\(\frac{{{P_{Al}}}}{{{P_{st}}}} = \frac{{{L_{st}}}}{{{L_{Al}}}} × \frac{{{A_{Al}}}}{{{A_{st}}}} × \frac{{{E_{Al}}}}{{{E_{st}}}}\)

\(\frac{P_{Al}}{P_{St}} = \frac{2}{2} × \frac{{120 × {{10}^{ - 6}}}}{{60 × {{10}^{ - 6}}}} × \frac{{70}}{{200}}=\frac{7}{10}\)....................(1)

But we know that, P_{Al} + P_{St }= 20..........................(2)

From equation (1), \(P_{St}=\frac{10}{7}P_{Al}\)

Therefore, from equation (1) and (2), we will get,

\(P_{Al}+\frac{10}{7}P_{Al}=20\)

\({P_{Al}} = \frac{{140}}{{17}}~kN\)

\({\sigma _{Al}} = \frac{{{P_{Al}}}}{{{A_{Al}}}} = \frac{{140 × {{10}^3}}}{{17 × 120 × {{10}^{ - 6}}}}\)

\(\sigma _{Al} = \frac{{14}}{{12 × 17}} × {10^9}\frac{N}{{{m^2}}}\)

\({\sigma _{Al}} = \frac{7}{{6 × 17}}~GPa = \frac{7}{{102}}~GPa\)