Question

A steel tape gives correct measurement at $20^\circ \;{\rm{C}}$. A piece of wood is being measured with the steel tape at $0^\circ \;{\rm{C}}$. The reading is 25cm on the tape, the real length of the given piece of wood must be?
A. 25 cm
B. <25 cm
C. >25 cm
D. Cannot say

Answer 1

We know that an increase in length during linear thermal expansion $L = {L_0}(1 + \alpha \Delta T)$, where $L$ and ${L_0}$ are final and initial lengths, $\Delta T$ is temperature change and $\alpha$ is the coefficient of linear expansion. Here, we will try to find whether the tape is expanding or contracting.

Complete step by step solution:
As given in the question:
The initial temperature of the tape is $T = 20^\circ \;{\rm{C}}$.
The final temperature of the tape is $t = 0^\circ \;{\rm{C}}$.
The final reading of the tape at $0^\circ \;{\rm{C}}$ is $L = 25\;{\rm{cm}}$.
Let the initial reading of the tape at $20^\circ \;{\rm{C}}$ is ${L_0}$.
We know that the formula for linear thermal expansion is:
$L = {L_0}(1 + \alpha \Delta T)$
$\Rightarrow L = {L_0}(1 + \alpha \left( {t - T} \right))$
Here $\alpha$ is the coefficient of thermal expansion.
We will now substitute $T = 20^\circ \;{\rm{C}}$ $t = 0^\circ \;{\rm{C}}$ $L = 25\;{\rm{cm}}$ to simplify the equation.
$\Rightarrow 25\;{\rm{cm}} = {L_0}(1 + \alpha \left( {0^\circ \;{\rm{C}} - 20^\circ \;{\rm{C}}} \right))$
$\Rightarrow 25 = {L_0}(1 - 20\alpha )$
$\Rightarrow \dfrac{{25 - {L_0}}}{{{L_0}}} = - 20\alpha$
Since we know that $\alpha$ the coefficient of linear expansion for steel is positive, we can say that the negative sign with 20 $\alpha$ is because ${L_0}$ is greater than 25 cm.

Therefore, the final length is greater than 25 cm, and the correct option is (C).

Additional information: It is quite evident that for metals, when the temperature is increased, the coefficient of thermal expansion $\alpha$ is positive. When the temperature is decreased, the value of the coefficient of thermal contraction ${\alpha _c}$ is negative.

Note: We must be cautious while calculating the temperature difference. We must always subtract the initial temperature with the final temperature to avoid mistakes. If the temperature is increased and the length increased, then the value $\alpha$ is positive.