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The increase in temperature is needed to increase the length of an aluminum meter-stick by 1.0 mm is 42 K. The option 1 is the correct option.
Thermal expansion is the property of the matter to change its shape, size, length, volume, density etc with change in the temperature.
The thermal expansion can be calculate as,
[tex]\Delta L=aL\Delta T[/tex]
Here, [tex]L[/tex] is the original length, [tex]\Delta T[/tex] is the change in the temperature and [tex]a[/tex] is the coefficient of the expansion.
Given information-
The change in the length of the aluminum mater-stick is 1.00 mm.
Let the initial size of the aluminium mater-stick is 1.00 m or 1000 mm . The thermal expansion coefficient of aluminium is [tex](2.4\times10^{-5})[/tex] per unit Kelvin.
Put the values in the above formula as,
[tex]1=(2.4\times10^{-5})\times 1000\times\Delta T[/tex]
Solve for temperature as,
[tex]\Delta T=\dfrac{1}{(2.4\times10^{-5})\times 1000} \\\Delta T=42K[/tex]
Hence the increase in temperature is needed to increase the length of an aluminum meter-stick by 1.0 mm is 42 K. The option 1 is the correct option.
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