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  • A composite wire of a uniform cross-section of <img alt="5.5 \times 10^{-5} \mathrm{~m}^2" src="https://entrancecorner.oncodecogs.com/gif.latex?5.5%20%5Ctimes%2010%5E%7B-5%7D%20%5Cmathrm%7B%7Em%7D%

A composite wire of a uniform cross-section of 5.5 \times 10^{-5} \mathrm{~m}^2 consists of a steel wire of length 1.5 \mathrm{~m} and a copper wire of length 2.0 \mathrm{~m}. By how much will it stretch when it is loaded with a mass of \mathrm{200 \mathrm{~kg}} ? Young's modulus of steel is \mathrm{2 \times 10^{11} \mathrm{Nm}^{-2}} and that of copper is 1 \times 10^{11} \mathrm{Nm}^{-2}. Take g=10 \mathrm{~ms}^{-2}.

Option: 1

1 mm


Option: 2

2 mm


Option: 3

3 mm


Option: 4

4 mm


Answers (1)

best_answer

Since the steel and copper parts of the wire are of the same cross-sectional area (A) and are loaded with the same weight \mathrm{F=M g, } they are under the same stress \mathrm{ (F / A)}. Therefore,

\mathrm{ \begin{aligned} \frac{Y_s l_s}{L_s} & =\frac{Y_c l_c}{L_c}=\frac{F}{A}=\frac{M g}{A} \\\\ l_s & =\frac{M g}{A} \cdot \frac{L_s}{Y_s} \quad \text { and } \quad l_c=\frac{M g}{A} \cdot \frac{L_c}{Y_c} \end{aligned} }
or
\mathrm{ l_s=\frac{M g}{A} \cdot \frac{L_s}{Y_s} \quad \text { and } \quad l_c=\frac{M g}{A} \cdot \frac{L_c}{Y_c} }
\therefore \quad  Extension of the composite wire is

                                          \mathrm{ l=l_s+l_c=\frac{M g}{A}\left(\frac{L_s}{Y_s}+\frac{L_c}{Y_c}\right) }
Substituting the given values, we find that \mathrm{l=10^{-3} \mathrm{~m}=1 \mathrm{~mm}}

. Hence the correct choice is (a).

Posted by

Ritika Harsh

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