# In an experiment, brass and steel wires of length 1 m each with areas of cross section 1 mm2 are used . the wires are connected in series and one end of the combined wire is conneced to a rigid  support and other end is subjected to elongation The stress required to produce a net elongation of 0.2mm is, [Given ,the young 's Modulus for steel and brass are respectively ,$120\times 10^{9}N/m^{2}\, \, and\, \, 60\times 10^{9}N/m^{2}$] Option 1) $1.2\times 10^{6}N/m^{2}$ Option 2) $8\times 10^{6}N/m^{2}$ Option 3) $1.8\times 10^{6}N/m^{2}$ Option 4) $0.2\times 10^{6}N/m^{2}$

Work Done in Stretching Wire / Elastic P.E. -

$\frac{1}{2}Kx^2=\frac{1}{2}\frac{YAl^{2}}{L}=\frac{1}{2}Fl$

$(K=\frac{YA}{L})$

- wherein

L - Length of wire

$l$ - increase in length

$k_{1}=\frac{y_{1}A_{1}}{l_{1}}=\frac{120\times 10^{9}\times A}{1}$

$k_{2}=\frac{y_{2}A_{2}}{l_{2}}=\frac{60\times 10^{9}\times A}{1}$

$K_{eq}=\frac{K_{1}K_{2}}{K_{1}+K_{2}}=\frac{120\times 60}{180}\times 10^{9}\times A$

$\Rightarrow K _{eq} =40\times 10^{9}\times A$

$F=k_{eq} \left (x \right )$

$F=\left ( 40\times 10^{9} \right )A_{0}\left ( 0.2\times 10^{-3} \right )$

$\Rightarrow \frac{F}{A}=8\times 10^{6}N/m^{2}$

Option 1)

$1.2\times 10^{6}N/m^{2}$

Option 2)

$8\times 10^{6}N/m^{2}$

Option 3)

$1.8\times 10^{6}N/m^{2}$

Option 4)

$0.2\times 10^{6}N/m^{2}$

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