A pendulum clock which keeps correct time at the bottom of mountain losses $30 s e c / d a y$ when it taken to the top of mountain. If the height of the mountain is $1.1 μ k m$ find the value of $μ (R_{e} = 6400 k m)$ .

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Solution

$T_{e} = 2 π \sqrt{\frac{l}{g}}$
As we know
$g = g_{0} (1 - \frac{2 h}{R})$
$T_{m o u n t a i n} = 2 π \sqrt{\frac{l}{g_{0} (1 - \frac{2 h}{6400})}}$
Consider a simple pendulum which is having time period of $2$ sec
Now, $2 = 2 π \sqrt{\frac{l}{g}} \to (I)$
$2 + \frac{23}{86400} = 2 π \sqrt{\frac{l}{g (1 - \frac{2 h}{6400})}} \to (I I)$
On dividing both equations, we get
$\frac{2}{2 + \frac{30}{86400}} = \sqrt{1 - \frac{2 h}{6400}}$
$0.9 = \sqrt{1 - \frac{2 h}{6400}}$
$0.81 = 1 - \frac{2 h}{6400}$
$2 h = (6400 - 5184)$
$h = 608$ Km.

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A pendulum clock which keeps correct time at the bottom of mountain losses 30 sec/day when it taken to the top of mountain. If the height of the mountain is 1.1μ km find the value of μ(Re=6400 km).

A pendulum clock which keeps correct time at the bottom of mountain losses $30 s e c / d a y$ when it taken to the top of mountain. If the height of the mountain is $1.1 μ k m$ find the value of $μ (R_{e} = 6400 k m)$ .