- #1

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In the solution, it was given that the thermal volume expansion of mercury = 18 * 10

^{-5}and the thermal volume expansion of glass = 2 x 10

^{-5}

## The Attempt at a Solution

I have attempted to solve the problem, however, I don't really know which volume expansion value should I use in the problem.

I have setup:

[tex]\Delta[/tex] V = V

_{expansion}

(V

_{0}+ [tex]\Delta[/tex]A) * [tex]\Delta[/tex]h = V

_{0}+ V

_{0}* volume expansion value * [tex]\Delta[/tex] Temperature

[tex]\Pi[/tex] *(0.0040 cm/2)

^{2}+ [tex]\Pi[/tex] *(0.0040 cm/2)

^{2}* ( thermal expansion of "what material???" * (30 degree) * ([tex]\Delta[/tex] h =( ([tex]\frac{4}{3}[/tex] * [tex]\Pi[/tex] * (0.25cm/2)

^{3}) + ([tex]\frac{4}{3}[/tex] * [tex]\Pi[/tex] * (0.25cm/2)

^{3}* (volume expansion of "which material" ) * (30 degree)

I have tried to solve with both thermal expansion number, however, I have got 650 cm and 631cm.

The solution of the problem is [tex]\Delta[/tex] h = 3.1cm...

I wonder where did I do wrong? Whcih volume of expansion should I choose?