Physics JAMB

A metal of volume 40 cm³ is heated from 30°C to 90°C; the increase in volume is?

A. 0.12 cm³
B. 4.00 cm³
C. 1.20 cm³
D. 0.40 cm³
E. no correct option

Correct Answer: Option E

Explanation

increase in volume, DV = RV1 ∆θ = 3α V1
= 3 x 2.0 X 10-5 x 40 x (90 – 30)
= 6 x 10-5 x 40 x 60
0.144 cm³
∴ No correct option

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