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A meter scale made of steel, reads accurately at $25^{\circ} \mathrm{C}$. Suppose in an experiment an accuracy of $0.06 \mathrm{~mm}$ in $1 \mathrm{~m}$ is required, the range of temperature in which the experiment can be performed with this meter scale is (Coefficient of linear expansion of steel is $11 \times 10^{-6} /{ }^{\circ} \mathrm{C}$ )
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Verified Answer
The correct answer is:
$19^{\circ} \mathrm{C}$ to $31^{\circ} \mathrm{C}$
Given, Coefficient of linear expansion of steel,
$$
\alpha=11 \times 10^{-6} /{ }^{\circ} \mathrm{C}
$$
We know that,
$$
\Delta l=l \alpha \Delta t \Rightarrow \Delta t=\frac{\Delta l}{l \alpha}
$$
Here, $\Delta l=6 \times 10^{-5} \mathrm{~m} \Rightarrow l=1 \mathrm{~m}$
$$
\begin{aligned}
\Delta t & =\frac{6 \times 10^{-5}}{1 \times 11 \times 10^{-6}} \\
& =5.45^{\circ} \mathrm{C}
\end{aligned}
$$
So, the range of temperature in which the experiment can be hence performed him this metre scale will be $19^{\circ} \mathrm{C}$ to $31^{\circ} \mathrm{C}$
Here, option (a) is correct.
$$
\alpha=11 \times 10^{-6} /{ }^{\circ} \mathrm{C}
$$
We know that,
$$
\Delta l=l \alpha \Delta t \Rightarrow \Delta t=\frac{\Delta l}{l \alpha}
$$
Here, $\Delta l=6 \times 10^{-5} \mathrm{~m} \Rightarrow l=1 \mathrm{~m}$
$$
\begin{aligned}
\Delta t & =\frac{6 \times 10^{-5}}{1 \times 11 \times 10^{-6}} \\
& =5.45^{\circ} \mathrm{C}
\end{aligned}
$$
So, the range of temperature in which the experiment can be hence performed him this metre scale will be $19^{\circ} \mathrm{C}$ to $31^{\circ} \mathrm{C}$
Here, option (a) is correct.
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