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A copper wire  y=10^{11}\, N/m^{2} of length 8 m and a steel wire  of  y=2\times 10^{11}\, N/m^{2}  length 4 m, each of 0.5 cm2 cross -section are fastened end to end and stretched with a tension of 500 N.

 

Option: 1

Elongation in copper wire is 0.8 mm.


Option: 2

Elongation in steel is  (\frac{1}{4}) th  the elongation in copper wire.

 


Option: 3

Total elongation is 1.0 mm.


Option: 4

All of the above


Answers (1)

best_answer

\begin{aligned} & (\Delta l)_c=\left(\frac{\mathrm{Fl}}{\mathrm{AY}}\right)_c=\frac{500 \times 8}{0.5 \times 10^{-4} \times 10^{11}} \\ \end{aligned}

(\Delta l)_c=0.8 \times 10^{-3} \mathrm{~m}=0.8 \mathrm{~mm} \text {. } \\
{ (\Delta l })_S=\left(\frac{F l}{A Y}\right)_S \\
=\frac{500 \times 4}{0.5 \times 10^{-4 }\times 2 \times 10^{11}}=0.2 \times 10^{-3} \mathrm{~m} \\
                                                     =0.2 \mathrm{~mm} \text {. } \\

(\Delta l)_s =\frac{1}{4}(\Delta l)_c \\

\Delta l =0.9+0.2=1.00 \mathrm{~mm}
 

Posted by

vishal kumar

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