A load of $2 \,kg$ produces an extension of $1 \,mm$ in a wire of $3 \,m$ in length and $1 \,mm$ in diameter. The Young's modulus of wire will be $.......... Nm ^{-2}$
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We know
$\frac{\text { Force } \times \text { Length }}{\text { Area of cross-section } \times \text { elongation }}=$ Young's Modulus
$\frac{F \times L}{A \times \Delta L}=Y$ $\left\{\begin{array}{l}F=2 \times 10 N , A=\pi \times(1 / 2)^2 \times 10^{-6} \,m ^2 \\ L=3 \,m \quad, \Delta L=1 \times 10^{-3} \,m \end{array}\right\}$
Substituting values
$\frac{20 \times 3}{\pi \times \frac{1}{4} \times 10^{-6} \times 1 \times 10^{-3}}=Y$
$\frac{20 \times 3 \times 4}{3.14 \times 10^{-9}}=Y$
$7.48 \times 10^{10} \,Nm ^{-2}=Y$
$\frac{\text { Force } \times \text { Length }}{\text { Area of cross-section } \times \text { elongation }}=$ Young's Modulus
$\frac{F \times L}{A \times \Delta L}=Y$ $\left\{\begin{array}{l}F=2 \times 10 N , A=\pi \times(1 / 2)^2 \times 10^{-6} \,m ^2 \\ L=3 \,m \quad, \Delta L=1 \times 10^{-3} \,m \end{array}\right\}$
Substituting values
$\frac{20 \times 3}{\pi \times \frac{1}{4} \times 10^{-6} \times 1 \times 10^{-3}}=Y$
$\frac{20 \times 3 \times 4}{3.14 \times 10^{-9}}=Y$
$7.48 \times 10^{10} \,Nm ^{-2}=Y$
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