A weight of 200 , kg is suspended by a vertic | vedclass

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Question

(A) Given: Force $F = mg = 200 \, kg 	imes 9.8 \, m/s^2 \approx 2000 \, N$ (taking $g = 10 \, m/s^2$ for standard calculation),Original length $L = 6

Vedclass · Accepted Answer

$2.35 	imes 10^{12} \, N/m^2$

Vedclass · Answer

(A) Given: Force $F = mg = 200 \, kg 	imes 9.8 \, m/s^2 \approx 2000 \, N$ (taking $g = 10 \, m/s^2$ for standard calculation),Original length $L = 600.5 \, cm - 0.5 \, cm = 600 \, cm = 6 \, m$,Change in length $\Delta L = 0.5 \, cm = 0.5 	imes 10^{-2} \, m$,Area $A = 1 \, mm^2 = 10^{-6} \, m^2$.Using the formula for Young's modulus $Y = \frac{F \cdot L}{A \cdot \Delta L}$:$Y = \frac{2000 	imes 6}{10^{-6} 	imes 0.5 	imes 10^{-2}}$$Y = \frac{12000}{0.5 	imes 10^{-8}} = 24000 	imes 10^8 = 2.4 	imes 10^{12} \, N/m^2$.Considering the provided options,the closest value is $2.35 	imes 10^{12} \, N/m^2$.

$A$ weight of $200 \, kg$ is suspended by a vertical wire of length $600.5 \, cm$. The area of cross-section of the wire is $1 \, mm^2$. When the load is removed,the wire contracts by $0.5 \, cm$. The Young's modulus of the material of the wire will be:

Similar Questions

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The force constant of a wire does not depend on

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