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The coefficients of thermal expansion of steel and a metal $X$ are respectively $12 × 10^{-6}$ and $2 × 10^{-6} per^o C$. At $40^o C$, the side of a cube of metal $X$ was measured using a steel vernier callipers. The reading was $100 \,\,mm$.Assuming that the calibration of the vernier was done at $0^o C$, then the actual length of the side of the cube at $0^o C$ will be
$> 100 mm$
$< 100 mm$
$= 100 mm$
data insufficient to conclude
Solution
$\cdot$ Coefficient of thermal expansion of steel $=12 \times 10^{-6} /^{\circ} \mathrm{C}$
$\cdot$ Coefficient of thermal expansion of metal $X=2 \times 10^{-6} /^{\circ} \mathrm{C}$
$\cdot$ side of a cube at $40^{\circ} \mathrm{C}=100 \mathrm{mm}$
Calibration is done at $0^{\circ} \mathrm{C},$ suppose actual length is $=\mathrm{L}$
We know that,
easure value $=$ Calibrated value $\times[1+(a 0-a s) \Delta T]$
where ao is the thermal expansion of metal and as is the thermal expansion of steel.
By substituting the values we get
$100=\mathrm{Lo}\left[1+(2-12) 10^{-6} \Delta \mathrm{T}\right]$
$100=\mathrm{Lo}\left[1-10 \times 10^{-6} \Delta T\right]$
So, Lo > $100 \mathrm{mm}$
