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  • The speed of a transverse wave on a straight wire (mass . length <img alt="80 \mathrm{~cm}"

The speed of a transverse wave on a straight wire (mass 9 \mathrm{~g}. length 80 \mathrm{~cm} and area of cross - section 4 \mathrm{~mm}^{2} ) is 120 \mathrm{~m} / \mathrm{s}. If Young's modulus of wire is 18 \times 1012 \mathrm{~N} / \mathrm{m}^{2}, the extension of wire over its natural length

Option: 1

\Delta l=18 \times 10^{7} \mathrm{~m}


Option: 2

\Delta l=18 \times 10^{8} \mathrm{~m


Option: 3

\Delta l=20 \times 10^{7} \mathrm{~m


Option: 4

\Delta l=18 \times 10^{5} \mathrm{~m


Answers (1)

best_answer

$$ \begin{aligned} & V=\sqrt{\frac{T}{A}} &\quad \frac{T}{A}=y \frac{\Delta l}{l} \\ & v^{2}=\frac{1}{M} \times Y A \frac{\Delta l}{l} \\ && T=Y A+\frac{l}{l} \\ & \Delta l=\frac{V^{2} \mu l}{Y A} \\ && u_{1}=\frac{m}{l} \\ && m=\pi l \\ & \Delta l=\frac{120 \times 120^{30} \times 9 \times 10^{-3}}{18 \times 10^{12} \times 4 \times 10^{-6}} \\ & \Delta l=60 \times 30 \times 10^{-3} \times 10^{6} \times 10^{-12} \\ & \Delta l=18 \times 10^{2} \times 10^{-3} \times 10^{6} \times 10^{-12} \\ & \Delta l=18 \times 10^{7} \mathrm{~m} \end{aligned}

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Divya Prakash Singh

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