Physics
JAMB 2011
A metal of volume 40 cm
3 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 = RV
1 ∆ θ = 3α Vsub>1 Dθ
= 3 x 2.0 X 10
-5 x 40 x (90 - 30)
= 6 x 10
-5 x 40 x 60
0.144 cm
3
∴ No correct option
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