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  4. A steel scale is to be prepared such that the millimeter intervals are to be accurate within 6× 10- 5mm . The maximum temperature variation from the temperature of calibration during the reading of the millimeter marks is (linear coefficient of expansion for steel is given as .α=12 × 10-6 ° C -1)

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Q. A steel scale is to be prepared such that the millimeter intervals are to be accurate within $6\times 10^{- 5}mm$ . The maximum temperature variation from the temperature of calibration during the reading of the millimeter marks is (linear coefficient of expansion for steel is given as $\left.\alpha=12 \times 10^{-6} \quad{ }^{\circ} C ^{-1}\right)$

NTA AbhyasNTA Abhyas 2022

Solution:

We know that thermal extension is given by,
$\Delta l=l_{0}\alpha \Delta T$
$6\times 10^{- 5}=1\times 12\times 10^{- 6}\times \Delta \top$
$\Delta T=\frac{6 \times 10^{- 5}}{12 \times 10^{- 6}}$ $\Delta T=5^\circ C$