A thin 1 , m long rod has a radius of 5 , mm. | vedclass

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Question

(A) The formula for Young's modulus is $Y = \frac{F \ell}{A \Delta \ell} = \frac{F \ell}{\pi r^2 \Delta \ell}$.Given: $\ell = 1 \, m = 1000 \, mm$,$r

Vedclass · Accepted Answer

The maximum value of $Y$ that can be determined is $10^{14} \, N/m^2$.

Vedclass · Answer

(A) The formula for Young's modulus is $Y = \frac{F \ell}{A \Delta \ell} = \frac{F \ell}{\pi r^2 \Delta \ell}$.Given: $\ell = 1 \, m = 1000 \, mm$,$r = 5 \, mm$,$F = 50 \pi 	imes 10^3 \, N$,and $\Delta \ell = \Delta r = 0.01 \, mm$.The relative error is given by $\frac{\Delta Y}{Y} = \frac{\Delta \ell}{\ell} + 2 \frac{\Delta r}{r} + \frac{\Delta(\Delta \ell)}{\Delta \ell}$.Since $\Delta \ell$ is the smallest measurable length,$\Delta(\Delta \ell) = 0.01 \, mm$.Calculating contributions: $\frac{\Delta \ell}{\ell} = \frac{0.01}{1000} = 10^{-5}$,$2 \frac{\Delta r}{r} = 2 	imes \frac{0.01}{5} = 4 	imes 10^{-3}$,and $\frac{\Delta(\Delta \ell)}{\Delta \ell} = \frac{0.01}{\Delta \ell}$.Since $\Delta \ell$ is typically very small,the term $\frac{\Delta(\Delta \ell)}{\Delta \ell}$ dominates the error.Statement $A$ is false because the maximum $Y$ depends on the minimum measurable $\Delta \ell$. If $\Delta \ell = 0.01 \, mm$,$Y = \frac{50 \pi 	imes 10^3 	imes 1000}{\pi 	imes 5^2 	imes 0.01} = 2 	imes 10^{11} \, N/m^2$. The value $10^{14}$ is incorrect.

$A$ thin $1 \, m$ long rod has a radius of $5 \, mm$. $A$ force of $50 \, \pi \, kN$ is applied at one end to determine its Young's modulus. Assume that the force is exactly known. If the least count in the measurement of all lengths is $0.01 \, mm$,which of the following statements is false?

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