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WAEC Physics 2022 Paper

Question : 40

Total: 50

30^{\circ} \mathrm{C}

1.9 \times 10^{-5} \mathrm{K}^{-1} \mathrm{J}

Solution:

Concept:

The diameter of the ring is a linear dimension, so the change in diameter follows linear expansion.

Explanation:

Given initial diameter

d_1 = 50.0\ \mathrm{cm}

, final diameter

d_2 = 50.29\ \mathrm{cm}

, initial temperature

\theta_1 = 30^\circ\mathrm{C}

, and linear expansivity

\alpha = 1.9 \times 10^{-5}\ \mathrm{K}^{-1}

The linear expansion formula is:

d_2 = d_1(1 + \alpha \Delta T)

So,

\Delta T = \frac{d_2 - d_1}{\alpha d_1}

Substitute the values:

\Delta T = \frac{50.29 - 50.0}{50.0(1.9 \times 10^{-5})}

\Delta T = \frac{0.29}{9.5 \times 10^{-4}} = 305.3\ \mathrm{K}

The required temperature is:

\theta_2 = 30 + 305.3 = 335.3^\circ\mathrm{C}

Answer:

335.3^\circ\mathrm{C}

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