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WAEC Physics 2022 Paper

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Question : 40

Total: 50

The diameter of a brass ring at

30^{\circ} \mathrm{C}

is 50.0 cm . To what temperature must this ring be heated to increase its diameter to 50.29 cm ? [ linear expansivity of brass =

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

152.6 °C
182.6 °C
306.1 °C
335.3 °C

Solution:

\text{Area expansivity} (B)=\frac{A_2-A_1}{A_1(\theta_2-\theta_1.)}

\text{Area} = \frac{\pi d^2}{4}

d_1=50.0 \mathrm{cm},\ d_2=50.29 \mathrm{cm},\ \theta_1=30^{\circ} \mathrm{C},\ \theta_2=\text{ ? and } B=2 a

2 \times 1.9 \times 10^{-5} = \frac{50.29^2-50^2}{50^2(\theta_2-30)}

\theta_2-30 = \frac{50.29^2-50^2}{50^2(2 \times 1.9 \times 10^{-5})}

\theta_2 = 30 + \frac{50.29^2-50^2}{50^2(2 \times 1.9 \times 10^{-5})} = 306.1^{\circ} \mathrm{C}.

note:

\frac{\pi}{4}

cancels out.

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