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The first real length standard, adopted more than 200 years ago, was a platinum bar with two very fine marks separated by what was defined to be exactly one meter. If this standard bar was to be accurate to within \( \pm 1.0\;{\rm{\mu m}}\), how carefully would the trustees have needed to control the temperature? The coefficient of linear expansion is \(9 \times {10^{ - 6}}\;{\rm{/^\circ C}}\) .

Short Answer

Expert verified

The required temperature is\(\Delta T = \pm 0.11{\rm{^\circ C}}\).

Step by step solution

01

Given data

The standard bar should be within\(\Delta L = \pm 1.0\;{\rm{\mu m}}\).

The coefficient of linear expansion is\(\alpha = 9 \times {10^{ - 6}}\;{\rm{/^\circ C}}\).

The length is \(L = 1\;{\rm{m}}\).

02

Expression for linear expansion

In this problem, use the expression for the linear expansion of a material to determine the temperature required.

03

Calculation of the temperature

The relation to find the requiredtemperature can be written as:

\(\Delta L = \alpha L\Delta T\)

Here,\(\Delta T\)is the required temperature.

On plugging the values in the above relation, you get:

\(\begin{aligned}\left( {1\;{\rm{\mu m}} \times \frac{{{{10}^{ - 6}}\;{\rm{m}}}}{{1\;{\rm{\mu m}}}}} \right) &= \left( {9 \times {{10}^{ - 6}}\;{\rm{/^\circ C}}} \right)\left( {1\;{\rm{m}}} \right)\Delta T\\\Delta T &= \pm 0.11{\rm{^\circ C}}\end{aligned}\)

Thus, \(\Delta T = \pm 0.11{\rm{^\circ C}}\)is the required temperature.

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