Question

A rod is measured at ${10}^{\circ }\text{C}$ using a scale calibrated at that temperature. Find the percentage error in length, if thermal coefficients of linear expansion of the rod and scale are $3\times {10}^{-5}{/}^{\circ }\text{C}$ and $2\times {10}^{-5}{/}^{\circ }\text{C}$ respectively, and the rod is measured at ${50}^{\circ }\text{C}$.

• A. $0.01\%$
• B. $0.02\%$
• C. $0.03\%$
• D. $0.04\%$

Answer 1

The correct option is D $0.04\%$

Let the length of the rod at temperature $T_0$ [actual length] be $L_0$
${\alpha }_0=3\times {10}^{-5}{/}^{\circ }\text{C}$
${\alpha }_s=2\times {10}^{-5}{/}^{\circ }\text{C}$
$T_0={10}^{\circ }\text{C}$
$T={50}^{\circ }\text{C}$
Measured length at temperature $T$
$L=L_0[1+({\alpha }_0-{\alpha }_s\left)\right(T-T_0\left)\right]$
Percentage error $=\left(\frac{L-L_0}{L_0}\right)\times 100$
$=\left[\frac{L_0[1+({\alpha }_0-{\alpha }_s\left)\right(T-T_0\left)\right]-L_0}{L_0}\right]\times 100$
$=\left[\right({\alpha }_0-{\alpha }_s\left)\right(T-T_0\left)\right]\times 100$
$=\left[\right({10}^{-5}\left)\right(40\left)\right]100$
$=4\times {10}^{-2}$
$=0.04\%$