A) increases
B) decreases
C) remains unchanged
D) becomes infinite
Correct Answer: A
Solution :
Time period of a simple pendulum of length I is given by \[T=2n\sqrt{\frac{l}{g}}\] where g is acceleration due to gravity. On moon \[{{g}_{m}}=\frac{g}{6}\] \[\therefore \] \[T'=2\pi \sqrt{\frac{l}{g/6}}\] \[=2\pi \sqrt{\frac{6l}{g}}\] \[=\sqrt{6}.2\pi \sqrt{\frac{l}{g}}\] \[\Rightarrow \] \[T'=\sqrt{6}T\] Hence, time period increases on the surface of moon.
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