WAEC - Physics (1994 - No. 35)
A metal rod of length 40cm at 20°C is heated to a temperature of 45°C. If the new length of the rod is 40.05cm. Calculate its linear expansitivity
1.2 x \(10^{-5}K{-1}\)
2.5x \(10^{-5}K{-1}\)
3.5x \(10^{-5}K{-1}\)
4.5x \(10^{-5}K{-1}\)
5x \(10^{-5}K{-1}\)
Giải thích
\(\alpha = \frac{I_2-I_1}{I_1(\theta_2-\theta_1)} \\ \Rightarrow \frac{40.05 - 40}{40 \times (45-20)} = 5.0 \times10^{-5}\)