A) Increases
B) Decreases
C) First increases then decreases
D) No effect
Correct Answer: B
Solution :
Key Idea: Value of acceleration due to gravity decreases on going below the surface of the earth. On going below depth h from the surface of earth, the value of g below is given by \[g'=g\left( 1-\frac{h}{{{R}_{e}}} \right)\]hence, g decreases. Time period \[\text{(T) = }\frac{\text{1}}{\text{frequency(n)}}\] And \[T=2\pi \sqrt{\frac{l}{g}}\] where I is length of pendulum. \[\therefore \] \[\frac{1}{n}2\pi \sqrt{\frac{l}{g\left( 1-\frac{h}{{{R}_{e}}} \right)}}\] Since, \[n\propto g,\]and g decreases therefore frequency also decreases.You need to login to perform this action.
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