A) ? 2 ´ 10?3
B) ? 1 ´ 10?3
C) 2 ´ 10?3
D) 1 ´ 10?3
Correct Answer: D
Solution :
Fractional change in period \[\frac{\Delta T}{T}=\frac{1}{2}\alpha \Delta \theta =\frac{1}{2}\times 2\times {{10}^{-6}}\times 10={{10}^{-5}}\] % change\[=\frac{\Delta T}{T}\times 100={{10}^{-5}}\times 100={{10}^{-3}}\,%\]
You need to login to perform this action.
You will be redirected in
3 sec
