As we move above the surface of the Earth,the value of acceleration due to gravity $g$ . . . . . . .

  • A
    increases
  • B
    decreases
  • C
    remains constant
  • D
    becomes zero

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$A$ pendulum is oscillating with frequency $n$ on the surface of the Earth. If it is taken to a depth $d = \frac{R}{2}$ below the surface of the Earth,where $R$ is the radius of the Earth,what is the new frequency of oscillations at this depth?

If the change in the value of $g$ at a height $h$ above the surface of the earth is the same as at a depth $x$ below it,then (where $x$ and $h$ are much smaller than the radius of the earth $R_e$):

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