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Q.

Figure shows a graph of the extension (A) 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 $10^{- 6} m^{2}$ , the Young's modulus of the material of the wire is $x \times 10^{11} N / m^{2}$ . The value of $x$ is

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Detailed Solution

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$Stress = \frac{force}{area} = \frac{W}{A} : Strain = \frac{Δl}{l} .$ Young's modulus is given by

$Y = \frac{stress}{strain} = \frac{W / A}{Δl / l} = \frac{W}{Δl} \times \frac{l}{A}$ ……..(i)

Now, slope of graph is $\frac{Δl}{W} = \frac{4 \times 10^{- 4}}{80} m / N$ . Using this in i), we get $(given l = 1 m and A = 10^{- 6} m^{2})$

$Y = \frac{80}{4 \times 10^{- 4}} \times \frac{1}{10^{- 6}} = 2 \times 10^{11} N / m^{2}$

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Figure shows a graph of the extension (A) 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 10-6m2 , the Young's modulus of the material of the wire is x×1011N/m2. The value of x is