The adjacent graph shows the extension ( $Δ l$ ) of a wire of length $1 m$ suspended from the top of a roof at one end and with a load w connected to the other end. If the cross-sectional area of the wire is $10^{- 6} m^{2}$ , calculate from the graph the Young's modulus of the material of the wire.

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$1.0 \times 10^{11} N / m^{2}$

$2.0 \times 10^{11} N / m^{2}$

$4.0 \times 10^{11} N / m^{2}$

$6.0 \times 10^{11} N / m^{2}$

answer is B.

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Detailed Solution

$Δ l = (\frac{l}{Y A}) \cdot w \Rightarrow Δ l \propto w$

i.e. $Δ l$ versus w graph is a straight line passing through origin (as shown in question also), the slope of which is $\frac{l}{Y A}$ .

Slope = $(\frac{l}{Y A})$

$Y = (\frac{l}{A}) (\frac{1}{slope})$

$= (\frac{1.0}{10^{- 6}}) \frac{(80 - 20)}{(4 - 1) \times 10^{- 4}}$

$= 2.0 \times 10^{11} N / m^{2}$

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