A steel scale is to be prepared such that the millimeter intervals are to be accurate within 6-10-5 mm- The maximum temperature variation from the temperature of calibration during the reading of the millimeter marks is -12-10-6-C

The correct option is C 5.0^∘C The change in length allowed for 1mm interval is 1mm×αΔθ=6×10^ 5mm #160;So, Δθ=6×10^ 5/1.2×10^ 6=5^oC ...

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Standard XII

Physics

Current Density Vector

A steel scale...

Question

A steel scale is to be prepared such that the millimeter intervals are to be accurate within $6 \times 10^{- 5} m m$ . The maximum temperature variation from the temperature of calibration during the reading of the millimeter marks is $(α = 12 \times 10^{- 6} /^{\circ} C)$

{4.0}^{\circ}

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{4.5}^{\circ} C

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{5.0}^{\circ} C

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{5.5}^{\circ} C

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Solution

The correct option is C ${5.0}^{\circ} C$
The change in length allowed for $1 m m$ interval is $1 m m \times α Δ θ = 6 \times 10^{- 5} m m$
So, $Δ θ = 6 \times 10^{- 5} / 1.2 \times 10^{- 6} = 5^{o} C$

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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 (α=12×10−6/∘C)

The correct option is C 5.0∘CThe change in length allowed for 1mm interval is 1mm×αΔθ=6×10−5mm So, Δθ=6×10−5/1.2×10−6=5oC

A steel scale is to be prepared such that the millimeter intervals are to be accurate within $6 \times 10^{- 5} m m$ . The maximum temperature variation from the temperature of calibration during the reading of the millimeter marks is $(α = 12 \times 10^{- 6} /^{\circ} C)$

The correct option is C ${5.0}^{\circ} C$
The change in length allowed for $1 m m$ interval is $1 m m \times α Δ θ = 6 \times 10^{- 5} m m$
So, $Δ θ = 6 \times 10^{- 5} / 1.2 \times 10^{- 6} = 5^{o} C$