A) 5.7 ´ 10?5/°C
B) 0.63 ´ 10?5/°C
C) 1.9 ´ 10?5/°C
D) 16.1 ´ 10?5/°C
Correct Answer: A
Solution :
\[\alpha =\frac{\Delta L}{{{L}_{0}}(\Delta \theta )}\]\[=\frac{0.19}{100\,(100-0)}=1.9\times {{10}^{-5}}/{}^\circ C\] Now g = 3a = 3 ´ 1.9 ´ 10?5/°C = 5.7 ´ 10?5/°CYou need to login to perform this action.
You will be redirected in
3 sec