Question

A bar measured with a Vernier caliper is found to be $180mm$ long. The temperature during the measurement is ${10}^{\circ }C$. The measurement error will be if the scale of the Vernier caliper has been graduated at a temperature of ${20}^{\circ }C:(\alpha =1.2\times {10}^{-5}{}^{\circ }{C}^{-1}$. Assume that the length of the bar does not change).

• A. $1.98\times {10}^{-1}mm$
• B. $1.98\times {10}^{-2}mm$
• C. $1.98\times {10}^{-3}mm$
• D. $1.98\times {10}^{-4}mm$

Answer 1

Answer: (B)

$1.98\times {10}^{-2}mm$

Due to thermal expansion, the reading is different from the true reading.

True reading $=$ scale reading $(1+\alpha \Delta \theta )$

And the error is $=$ scale reading $-$ True reading $=-$ scale reading $\left(\alpha \Delta \theta \right)$

$\alpha =1.2\times {10}^{-5}/{}^{o}C$ is the coefficient of thermal expansion.

$\Delta \theta =10-20=-{10}^{o}C$

Scale reading is $180mm$.

Thus, the error is $180mm\times 1.2\times {10}^{-5}\times 10=1.98\times {10}^{-2}mm$