The speed of a transverse wave travelling in | vedclass

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(A) Given,length of the wire,$l = 50 	ext{ cm} = 0.5 	ext{ m}$.Cross-sectional area of wire,$A = 1 	ext{ mm}^2 = 1 	imes 10^{-6} 	ext{ m}^2$.Mass

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$8 	imes 10^{-5} 	ext{ m}$

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(A) Given,length of the wire,$l = 50 	ext{ cm} = 0.5 	ext{ m}$.Cross-sectional area of wire,$A = 1 	ext{ mm}^2 = 1 	imes 10^{-6} 	ext{ m}^2$.Mass of the wire,$m = 5 	ext{ g} = 5 	imes 10^{-3} 	ext{ kg}$.Speed of transverse wave,$v = 80 	ext{ ms}^{-1}$.Young's modulus,$Y = 4 	imes 10^{11} 	ext{ Nm}^{-2}$.The speed of a transverse wave in a stretched wire is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu = \frac{m}{l}$ is the linear mass density.So,$v = \sqrt{\frac{T \cdot l}{m}} \implies v^2 = \frac{T \cdot l}{m} \implies T = \frac{v^2 m}{l}$.Young's modulus is defined as $Y = \frac{T/A}{\Delta l/l} \implies \Delta l = \frac{T \cdot l}{A \cdot Y}$.Substituting $T = \frac{v^2 m}{l}$ into the expression for $\Delta l$:$\Delta l = \frac{(v^2 m / l) \cdot l}{A \cdot Y} = \frac{v^2 m}{A Y}$.Substituting the values:$\Delta l = \frac{(80)^2 	imes (5 	imes 10^{-3})}{(1 	imes 10^{-6}) 	imes (4 	imes 10^{11})} = \frac{6400 	imes 5 	imes 10^{-3}}{4 	imes 10^5} = \frac{32000 	imes 10^{-3}}{4 	imes 10^5} = \frac{32}{4 	imes 10^5} = 8 	imes 10^{-5} 	ext{ m}$.Thus,the extension in the length of the wire is $8 	imes 10^{-5} 	ext{ m}$.

The speed of a transverse wave travelling in a wire of length $50 \text{ cm}$,cross-sectional area $1 \text{ mm}^2$ and mass $5 \text{ g}$ is $80 \text{ ms}^{-1}$. The Young's modulus of the material of the wire is $4 \times 10^{11} \text{ Nm}^{-2}$. The extension in the length of the wire is

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