$A$ steel tape $1 \; m$ long is correctly calibrated for a temperature of $27.0 \; ^{\circ}C$. 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 $45.0 \; ^{\circ}C$. 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 $27.0 \; ^{\circ}C$? (Coefficient of linear expansion of steel $\alpha = 1.20 \times 10^{-5} \; K^{-1}$)

(N/A) The length of the steel tape at $T = 27.0 \; ^{\circ}C$ is $l = 100 \; cm$.
At $T_1 = 45.0 \; ^{\circ}C$,the length of the tape $l'$ increases due to thermal expansion:
$l' = l(1 + \alpha \Delta T) = 100 \times [1 + 1.20 \times 10^{-5} \times (45.0 - 27.0)]$
$l' = 100 \times [1 + 1.20 \times 10^{-5} \times 18] = 100 \times [1 + 0.000216] = 100.0216 \; cm$.
The measured length of the rod is $63.0 \; cm$ using the expanded tape. The actual length $l_2$ at $45.0 \; ^{\circ}C$ is:
$l_2 = \frac{l'}{l} \times 63.0 = \frac{100.0216}{100} \times 63.0 = 63.0136 \; cm$.
To find the length at $27.0 \; ^{\circ}C$,we use $l_2 = l_0(1 + \alpha \Delta T)$:
$63.0136 = l_0(1 + 1.20 \times 10^{-5} \times 18) = l_0(1.000216)$
$l_0 = \frac{63.0136}{1.000216} \approx 63.0 \; cm$.