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The temperature of glass vessel containing 100cm3 of mercury is raised from 10oC to 100oC....

Physics
WAEC 1995

The temperature of glass vessel containing 100cm3 of mercury is raised from 10oC to 100oC. Calculate the apparent cubic expansion of the mercury. (Real cubic expansivity of mercury = 1.82 x 10-4K-1) (Cubic expansivity of glass = 2.4 x 10-5K-1)

  • A. 0.52cm3
  • B. 1.42cm3
  • C. 1.87cm3
  • D. 5.22cm3
  • E. 14.22cm3
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Correct Answer: Option B
Explanation

To calculate the apparent cubic expansion of the mercury, we use the formula for apparent expansion:

\(\text{Apparent Expansion} = V_0 \cdot \left( \beta_m - \beta_g \right) \cdot \Delta T\)

Where:
- \( V_0 = 100 \, \text{cm}^3 \) (initial volume of mercury)
- \( \beta_m = 1.82 \times 10^{-4} \, \text{K}^{-1} \) (real cubic expansivity of mercury)
- \( \beta_g = 2.4 \times 10^{-5} \, \text{K}^{-1} \) (cubic expansivity of glass)
- \( \Delta T = 100^\circ C - 10^\circ C = 90 \, \text{K} \) (change in temperature)

First, we calculate \( \beta_m - \beta_g \):

\(\beta_m - \beta_g = 1.82 \times 10^{-4} - 2.4 \times 10^{-5} = 1.58 \times 10^{-4} \, \text{K}^{-1}\))

Now substitute the values into the apparent expansion formula:

\(\text{Apparent Expansion} = 100 \, \text{cm}^3 \cdot (1.58 \times 10^{-4}) \cdot 90\)

Calculating this gives:

\(= 100 \cdot 1.58 \times 10^{-4} \cdot 90\)

\(= 100 \cdot 1.422 \times 10^{-2}\)

\(= 1.422 \, \text{cm}^3\)

Thus, the apparent cubic expansion of the mercury is approximately:

\(\text{Answer: } 1.42 \, \text{cm}^3\)


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