A steel rod at 25^{circ} C is observed to be | vedclass

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(B) Let $L_s$ be the length of the steel rod at $25^{\circ} C$ as measured by the metal scale. The scale is calibrated at $0^{\circ} C$, so at $25^{\c

Vedclass · Accepted Answer

$1.0002$

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(B) Let $L_s$ be the length of the steel rod at $25^{\circ} C$ as measured by the metal scale. The scale is calibrated at $0^{\circ} C$, so at $25^{\circ} C$, its length expands. The length of $1 \ m$ on the scale at $25^{\circ} C$ corresponds to an actual length $L_{actual} = 1(1 + \alpha_{	ext{metal}} \Delta T) = 1(1 + 20 	imes 10^{-6} 	imes 25) = 1 + 0.0005 = 1.0005 \ m$.Now, this is the actual length of the steel rod at $25^{\circ} C$. Let $L_0$ be the length of the steel rod at $0^{\circ} C$.Using the thermal expansion formula for the steel rod: $L_{actual} = L_0(1 + \alpha_{	ext{steel}} \Delta T)$.$1.0005 = L_0(1 + 12 	imes 10^{-6} 	imes 25)$.$1.0005 = L_0(1 + 0.0003) = L_0(1.0003)$.$L_0 = \frac{1.0005}{1.0003} \approx 1.0001999 \ m \approx 1.0002 \ m$.

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