Question

A steel tape $1m$ 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 $63.0cm$ on a hot day when the temperature is ${45}^{\circ }C$. Coefficient of linear expansion of steel $=1.20\times {10}^{–5}/K$. What is the actual length of the steel rod on that day?

• A. $63.1526cm$
• B. $63.0136cm$
• C. $63.2134cm$
• D. $63.3136cm$

Answer 1

The correct option is B $63.0136cm$

Given: $T_i={27}^{\circ }C,T_f={45}^{\circ }C$,
$\alpha =1.20\times {10}^{–5}/K$, $L=63cm$
The steel tape gives correct reading only at the temperature ${27}^{\circ }C$ at which it has been calibrated.
At any other temperature ${45}^{\circ }C$ the scale will expand and give less reading than the true value. Hence, length of the steel rod at ${27}^{\circ }C$,
i.e., $L=63cm$
Let $\Delta L$ be the increase in the length of the steel tape when temperature
rises from ${27}^{\circ }C$ to ${45}^{\circ }C$,
𝑖.𝑒. $\Delta T={45}^{\circ }C-{27}^{\circ }C={18}^{\circ }C=18K$
We know that: $\Delta L={\alpha }_{S}L\Delta T$
Clearly, $\Delta L={\alpha }_{S}L\Delta T$
$\Delta L=(1.20\times {10}^{-5})\times \left(63cm\right)\times 18K=0.0136cm$
Actual length of the rod at ${45}^{\circ }C$
$L_{Actual}=63+0.013=63.0136cm$
Final answer (a)