JAMB - Physics (2011 - No. 8)
A metal of volume 40 cm3 is heated from 30 °c to 90 °c, the increase in volume is
0.12 cm3
4.00 cm3
1.20 cm3
0.40 cm3
no correct option
설명
increase in volume, DV = RV1 ∆ 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 cm3
∴ No correct option
= 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