In an experiment, brass and steel wires of length 1 m each with areas of cross section 1 mmare 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}

     

     

     

     

     

     

     

     

 

Answers (1)

 

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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