A) 200 \[\pi \]sec
B) 2\[\pi \]/10 sec
C) 2\[\pi \] sec
D) 4\[\pi \] sec
Correct Answer: B
Solution :
Time period of oscillation \[T=2\pi \sqrt{\frac{\Delta l}{g}}\] \[=2\pi \sqrt{\frac{9.8\times {{10}^{-2}}}{9.8}}=\frac{2\pi }{10}\sec \]You need to login to perform this action.
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