11.6 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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Payal Gupta | Contributor-Level 10

5 months ago

11.6 Length of the steel tape, l = 1 m = 100 cm, At temperature T = 27 $°$ C

Coefficient of linear expansion of steel $α$ = 1.2 $\times 10^{- 5}$ / K

Let $l_{1} = 63 c m$ be the length of the steel rod at temperature $T_{1}$ = 45.0 °C and

$l_{2}$ be the length of the steel rod and l’ be the length of the steel tape at 45.0 °C

We have l’ = l + $α l (T_{1} - T)$ = 100 + 1.2 $\times 10^{- 5} \times 100 \times$ (45-27) = 100.0216 cm

$l_{2}$ can be calculated as $l_{2}$ = $\frac{l'}{l} \times 63$ = 63.0136 cm

11.6 Length of the steel tape, l = 1 m = 100 cm, At temperature T = 27 <math><mo>°</mo></math> CCoefficient of linear expansion of steel <math><mi>α</mi></math> = 1.2 <math><mo>×</mo><mi></mi><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>5</mn></mrow></mrow></msup></math> / KLet <math><msub><mrow><mrow><mi>l</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub><mo>=</mo><mn>63</mn><mi></mi><mi>c</mi><mi>m</mi><mi></mi></math> be the length of the steel rod at temperature <math><msub><mrow><mrow><mi>T</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></math> = 45.0 °C and<math><msub><mrow><mrow><mi>l</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub><mi></mi></math> be the length of the steel rod and l’ be the length of the steel tape at 45.0 °CWe have l’ = l + <math><mi>α</mi><mi>l</mi><mo> (</mo><msub><mrow><mrow><mi>T</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub><mo>-</mo><mi></mi><mi>T</mi><mo>)</mo></math> = 100 + 1.2 <math><mo>×</mo><mi></mi><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>5</mn></mrow></mrow></msup><mo>×</mo><mn>100</mn><mi></mi><mo>×</mo></math>  (45-27) = 100.0216 cm<math><msub><mrow><mrow><mi>l</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></math> can be calculated as <math><msub><mrow><mrow><mi>l</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></math> = <math><mfrac><mrow><mrow><mi>l</mi><mi>'</mi></mrow></mrow><mrow><mrow><mi>l</mi></mrow></mrow></mfrac><mi></mi><mo>×</mo><mn>63</mn></math> = 63.0136 cm

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