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Question
A simple pendulum of length 40 cm is taken inside a deep mine. Assume for the time being that the mine is 1600 km deep. Calculate the time period of the pendulum there. Radius of the earth = 6400 km.
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Solution
It is given that:
Length of the pendulum, l = 40 cm = 0.4 m
Radius of the earth, R = 6400 km
Acceleration due to gravity on the earth's surface, `g = 9.8 "ms"^(- 2)`
Let
\[g'\] be the acceleration due to gravity at a depth of 1600 km from the surface of the earth.
Its value is given by,
\[g' = g\left( 1 - \frac{d}{R} \right)\] \[\text{where d is the depth from the earth surfce, }\] \[\text { R is the radius of earth, and }\]
\[\text {g is acceleration due to gravity .}\]
\[\text{ On substituting the respective values, we get: }\] \[ g' = 9 . 8\left( 1 - \frac{1600}{6400} \right)\]
\[= 9 . 8\left( 1 - \frac{1}{4} \right)\]
\[= 9 . 8 \times \left( \frac{3}{4} \right) = 7 . 35 {\text{ms}}^{- 2}\]
Time period is given as,
\[\Rightarrow T = 2\pi\sqrt{\left( \frac{0 . 4}{7 . 35} \right)}\]
\[ \Rightarrow T = 2 \times 3 . 14 \times 0 . 23\]
\[ = 1 . 465 \approx 1 . 47 s\]
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