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Standard XII

Physics

Thermal Expansion

A steel tape ...

Question

A steel tape is calibrated at $20^{o}$ C. On a cold day when the temperature is $- 15^{o}$ C, percentage error in the tape will be [ $α_{s t e e l}$ = 1.2 x $10^{- 5^{o}} C^{- 1}$ ].

-0.035%

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-0.042%

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-0.012%

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-0.018%

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Solution

The correct option is B -0.042%
$Δ l = l α Δ t W h e r e Δ l = c h a n g e i n l e n g t h Δ t = C h a n g e i n t e m p e r a t u r e H e r e Δ t = - 15 - 20 = - 35^{0} C S o Δ l = l \times 1.2 \times 10^{- 5} \times - 35 a n d p e r c e n t a g e e r r o r = \frac{Δ l}{l} \times 100 = - 0.042$

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Similar questions

Q. A steel tape is calibrated at

20^{\circ} C

. On a cold day when the temperature is

- 15^{\circ} C

percentage error in the tape will be

[α_{steel} = 1.2 \times 10^{- 5 \cdot 0} C^{- 1}]

Q. A steel tape is calibrated at

20^{\circ} C

. On a cold day when the temperature is

- 15^{\circ} C

, the percentage error in the tape is

- \frac{x}{1000} %

. Find

x

Coefficient of linear thermal expansion for steel

α = 12 \times 10^{- 6}

K^{- 1}

Q. A steel tape is calibrated at

20^{\circ} C

. when the temperature of the day is

- 10^{\circ} C

, the percentage error in the measurement with the tape is (

α = 12 \times 10^{- 6} /^{\circ} C

)

Q. A surveyor's 30m steel tape is correct at a temperature of

20^{\circ} C

. The distance between two points, as measured by this tape on a day when the temperature is

35^{\circ} C

, is 26m. What is the true distance between the points? (

α_{s t e e l}

1.2 \times 10^{- 5} /^{\circ} C

A steel tape 1m long is correctly calibrated for a temperature of 27.0 °C. The length of a steel rod measured by this tape is found to be 63.0 cm on a hot day when the temperature is 45.0 °C. What is the actual length of the steel rod on that day? What is the length of the same steel rod on a day when the temperature is 27.0 °C? Coefficient of linear expansion of steel = 1.20 × 10^–5 K^–1.

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A steel tape is calibrated at 20oC. On a cold day when the temperature is −15oC, percentage error in the tape will be [αsteel = 1.2 x 10−5oC−1].

The correct option is B -0.042%Δl=lαΔtWhereΔl=changeinlengthΔt=ChangeintemperatureHereΔt=−15−20=−350CSoΔl=l×1.2×10−5×−35andpercentageerror=Δll×100=−0.042

A steel tape is calibrated at $20^{o}$ C. On a cold day when the temperature is $- 15^{o}$ C, percentage error in the tape will be [ $α_{s t e e l}$ = 1.2 x $10^{- 5^{o}} C^{- 1}$ ].