Question

# Two wire of equal cross section, but one made up of steel and the other copper are joined end to end. when the combination is kept under tension, the elongations in the two wires are found to be equal if $${Y_{steel}} = 2.0 \times {10^{11}}N{m^2}\,and\,{Y_{copper}} = 1.1 \times {10^{11}}N{m^2}$$, the ratio of the length of the two wires is

A
20:11
B
11:20
C
5:4
D
4:5

Solution

## The correct option is A 20:11$$\begin{array}{l}Y_{\text {Steel }}=2 \times 10^{11}\mathrm{~N} / \mathrm{m}^{2} \\Y_\text { copper }=1.1\times 10^{11} \mathrm{~N} / \mathrm{m}^{2}\end{array}$$$$\begin{array}{l}\text { As they are joined end to end tenspon will be Same. } \\\text { Given Area of cross section is also same } \\\text { that means Stress is also Same. }\end{array}$$$$\begin{array}{l}\text { let } L_1,L_2 \text { be length of steel, copper wires respectively } \\\text { and } \Delta l_{1}, \Delta l_{2} \text { elongations. }\end{array}$$$$Y_{S t{e e l}}= \frac{s t{r e s s}}{\frac{\Delta l_{1}}{l_{1}}} \text { and } Y_{\text {copper }}=\frac{\text { stress }}{\frac{\Delta l_{2}}{l_{2}}}$$$$\begin{array}{l}\frac{Y_{\text {steel }}}{Y_\text { copper }}=\frac{\text { stress }}{\Delta l_{1}} \times \frac{l_{1}}{l_{2}} \times \frac{\Delta x_{2}}{\text { stress }}\\\frac{l_{1}}{l_{2}}=\frac{2}{1.1}\end{array}$$$$l_{1}: l_{2}=20: 11$$Physics

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