A) Increases
B) Decreases
C) First increases then decreases
D) No effect
Correct Answer: B
Solution :
Ongoing 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 Also, Time period\[(T)=\frac{1}{frequency(n)}\] and \[T=2\pi \sqrt{\frac{l}{g}}\] where \[l\] 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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