A metre scale made of steel reads accurately at $25^{o} C$ . Suppose in an experiment an accuracy of $0.06 m m$ in $1 m$ is required, the range of temperature in which the experiment can be performed with this metre scale is (coefficient of linear expansion of steel is $11 \times 10^{- 6} /^{o} C$ )

19^{o} C

31^{o} C

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25^{o} C

32^{o} C

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18^{o} C

25^{o} C

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18^{o} C

32^{o} C

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Solution

The correct option is B $19^{o} C$ to $31^{o} C$
$Δ l = l α Δ t$
$Δ t = \frac{Δ l}{l α} = \frac{6 \times 10^{- 5}}{1 \times 11 \times 10^{- 6}} = {5.45}^{o} C$
$19^{o} C$ to $31^{o} C$

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A metre scale made of steel reads accurately at $20^{\circ} C$ . In a sensitive experiment, distances accurate up to 0.055 mm in 1 m are required. Find the range of temperature in which the experiment can be performed with this metre scale. Coefficient of linear expansion of steel $= 11 \times 10^{- 6}^{\circ} C^{- 1}$

Q. A metre scale is made up of steel and measures correct length at

16^{o} C

. What will be the percentage error if this scale is used
(a) on a summer day when the temperature is

46^{o} C

and
(b) on a winter day when the temperature is

6^{o} C ?

Coefficient of linear expansion of steel

= 11 \times 10^{- 6}

^{o} C^{- 1}

A metre scale made of steel is calibrated at $20^{\circ} C$ to give correct reading. Find the distance between 50 cm mark and 51 cm mark if the scale is used at $10^{\circ} C$ . Coefficient of linear expansion of steel is $1.1 \times 10^{- 5}^{\circ} C^{- 1}$

A metre scale is made up of steel and measures correct length at $16^{\circ} C$ . What will be the percentage error if this scale is used (a) on a summer day when the temperature is $46^{\circ} C$ and (b) on a winter day when the temperature is $6^{\circ} C$ ? Coefficient of linear expansion of steel $= 11 \times 10^{- 6}^{\circ} C^{- 1}$

Q. A metre scale made of steel is calibrated at 20°C to give correct reading. Find the distance between the 50 cm mark and the 51 cm mark if the scale is used at 10°C. Coefficient of linear expansion of steel is 1.1 × 10^–5 °C^–1.

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A metre scale made of steel reads accurately at 25oC. Suppose in an experiment an accuracy of 0.06mm in 1m is required, the range of temperature in which the experiment can be performed with this metre scale is (coefficient of linear expansion of steel is 11×10−6/oC)

The correct option is B 19oC to 31oCΔl=lαΔtΔt=Δllα=6×10−51×11×10−6=5.45oC19oC to 31oC

The correct option is B $19^{o} C$ to $31^{o} C$
$Δ l = l α Δ t$
$Δ t = \frac{Δ l}{l α} = \frac{6 \times 10^{- 5}}{1 \times 11 \times 10^{- 6}} = {5.45}^{o} C$
$19^{o} C$ to $31^{o} C$