A glass container holds a solid silica cylinder at its bottom, with mercury filling the rest up to the top of the silica. Silica volume is constant. Mercury has γ = 1.8 × 1 0 − 4 ∘ C − 1 and the glass

Question

A glass container holds a solid silica cylinder at its bottom, with mercury filling the rest up to the top of the silica. Silica volume is constant. Mercury has $γ = 1.8 \times 1 0^{- 4}^{\circ} C^{- 1}$ and the glass has $α =$ $9.0 \times 1 0^{- 6}^{\circ} C^{- 1}$ . The mercury top stays level with the silica top at all temperatures. Take the glass height unchanged relative to the silica (only its cross-section expansion matters, as in the book). Find $V_{silica} / V_{mercury}$ .

Answer:

9

Solution:

$A_{1} / A_{2} = \frac{γ}{2 α} - 1 = \frac{1.8 \times 1 0 ^{- 4}}{1.8 \times 1 0 ^{- 5}} - 1 = 10 - 1 = 9$ ; equal heights give the same ratio for volumes.

TestHub · Accepted Answer

A 1 ​ / A 2 ​ = 2 α γ ​ − 1 = 1.8 × 1 0 − 5 1.8 × 1 0 − 4 ​ − 1 = 10 − 1 = 9 ; equal heights give the same ratio for volumes.

Physics - Heat & Thermo Question with Solution | TestHub

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