A meter scale made of steel reads accurately | vedclass

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Question

$(A)$ Given: Coefficient of linear expansion of steel, $\alpha = 11 	imes 10^{-6} /{ }^{\circ} C$. Required accuracy $\Delta l = 0.06 \,mm = 6 	imes

Vedclass · Accepted Answer

$19^{\circ} C$ to $31^{\circ} C$

Vedclass · Answer

$(A)$ Given: Coefficient of linear expansion of steel, $\alpha = 11 	imes 10^{-6} /{ }^{\circ} C$. Required accuracy $\Delta l = 0.06 \,mm = 6 	imes 10^{-5} \,m$ for a length $l = 1 \,m$. We know the formula for linear expansion: $\Delta l = l \alpha \Delta t$. Rearranging for the temperature change: $\Delta t = \frac{\Delta l}{l \alpha}$. Substituting the values: $\Delta t = \frac{6 	imes 10^{-5}}{1 	imes 11 	imes 10^{-6}} = \frac{60}{11} \approx 5.45^{\circ} C$. Since the scale is accurate at $25^{\circ} C$, the allowed temperature range is $25^{\circ} C \pm 5.45^{\circ} C$. This gives a range from $19.55^{\circ} C$ to $30.45^{\circ} C$. Rounding to the nearest integer values provided in the options, the range is approximately $19^{\circ} C$ to $31^{\circ} C$. Thus, option $A$ is correct.

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