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Q10.6: NCERT Physics - Part 2 THERMAL PROPERTIES OF MATTER 10.6 Solution: A steel tape 1m long is correctly calibrated for a temperature…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students usually treat the 63.0 cm reading as the rod’s true length at 45 °C, forgetting that the steel tape itself has expanded; the correct correction is $L_{actual}=L_{measured}\times[1+\alpha( T_{meas}-T_{cal})]$, and the reverse division when converting back to the calibration temperature.

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__ K__.

Step-by-Step Verified Solution

**Answer:** Actual length at 45�C = 63.0136 cm. Length at 27�C = 62.9864 cm. **Step-by-Step Explanation:** At 45�C, the steel tape has expanded. The actual length of 1 cm on the tape is:
Lactual=1×[1+α(45−27)]=1+1.20×10−5×18=1.000216 cmL_{actual} = 1 \times [1 + \alpha(45 - 27)] = 1 + 1.20 \times 10^{-5} \times 18 = 1.000216 \text{ cm}
The measured length is 63.0 cm, so the actual length of the rod at 45�C is:
L45=63.0×1.000216=63.0136 cmL_{45} = 63.0 \times 1.000216 = 63.0136 \text{ cm}
At 27�C, the actual length of the rod (which is the calibrated length of the tape) is:
L27=63.01361+1.20×10−5×18=62.9864 cmL_{27} = \frac{63.0136}{1 + 1.20 \times 10^{-5} \times 18} = 62.9864 \text{ cm}
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All Questions in this Chapter

Q10.1 Q10.2 Q10.3 Q10.4 Q10.5 Q10.6 Q10.7 Q10.8 Q10.9 Q10.10 Q10.11 Q10.12 Q10.13 Q10.14 Q10.15 Q10.16 Q10.17 Q10.18 Q10.19 Q10.20

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