A metre scale made of steel reads accurately at $25^{\circ} C$ . Suppose in an experiment an accuracy of $0.06\, mm$ in $1\,m$ is required, the range of temperature in which the experiment can be performed with this metre scale is (coefficient of linear expansion of steel is $ 11 \times 10^{-6} /^{\circ} \, C$ )
- $19^{\circ} C $ to $31^{\circ} C $
- $25^{\circ} C $ to $32^{\circ} C $
- $18^{\circ} C $ to $25^{\circ} C $
- $18^{\circ} C $ to $32^{\circ} C $
The Correct Option is A
Solution and Explanation
Given, Coefficient of linear expansion of steel,
\(\alpha=11 \times 10^{-6} /{ }^{\circ} C\)
We know that,
\(\Delta l=l \alpha \Delta t\)
\(\Rightarrow \Delta t=\frac{\Delta l}{l \alpha}\)
Here, \(\Delta l=6 \times 10^{-5} m \Rightarrow l=1 m\)
\(\Delta t=\frac{6 \times 10^{-5}}{1 \times 11 \times 10^{-6}}\)
\(=5.45^{\circ} C\)
So, the range of temperature in which the experiment can be hence performed him this metre scale will be \(19^{\circ} C\) to \(31^{\circ} C\)
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Concepts Used:
Thermal Expansion
Thermal expansion is the tendency of matter to change its shape, area, and volume in response to a change in temperature. Temperature is a monotonic function of the average molecular kinetic energy of a substance.
The expansion of the solid material is taken to be the linear expansion coefficient, as the expansion takes place in terms of height, thickness and length. The gaseous and liquid expansion takes the volume expansion coefficient. Normally, if the material is fluid, we can explain the changes in terms of volume change.
The bonding force among the molecules and atoms differs from material to material. These characteristics of the compounds and elements are known as the expansion coefficient.
