A) cannot be compared.
B) equal to T
C) less than T
D) more than T
Correct Answer: D
Solution :
Time period of a simple pendulum at the surface of earth, \[T=2\pi \sqrt{\frac{l}{g}}\] \[T\propto \frac{1}{\sqrt{g}}\] In a mine (below the surface of earth) value of 'g' will decrease, so time period 'T' will increase.You need to login to perform this action.
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