A brass rod is 2 m long at a certain temperature. Calculate the linear expansion of the rod for a temperature change of 100 K. (Take linear expansivity of brass as 1.8 × 10\(^{-5}\) K\(^{-1}\))
A
0.0036 m
B
0.360 m
C
0.3600 m
D
0.1800 m
Verified Correct Answer: A.
Physical Principles & Numerical Solution
L = 2 m; \alpha = 1.8 \times 10\(^{-5}\) K\(^{-1}\)); \Delta\theta = 100 K
\alpha = \(\frac{\DeltaL }{ (L\Delta\theta)}\)
\DeltaL = \alpha L \Delta \theta
\DeltaL = 1.8 \times 10\(^{-5}\) x 2 x 100 = 3.6 x 10\(^{-3}\)
\(\therefore\) \(\Delta\) L = 0.0036m.
Source: WAEC / JAMB / NECO Classroom Archive (Record #76351)