A pendulum clock, which keeps correct time at sea level, loses 15 s per day when taken to the top of mountain. If the radius of the earth is 6400 km, the height of the mountain is:

Q. A pendulum clock, which keeps correct time at sea level, loses 15 s per day when taken to the top of mountain. If the radius of the earth is 6400 km, the height of the mountain is:

  1. 1.1 km
  2. 2.2 km
  3. 3.3 km
  4. 4.4 km

Solution:

We know, the time period of a simple pendulum can be calculated by a formula:

    \[T = 2\pi \sqrt{\frac{l}{g}}\]

considering length doesn’t change when taken to the top of a mountain. We can write,

    \[ T \propto \frac{1}{\sqrt{g}} \]

(1)   \begin{equation*}  \frac{T_1}{T_2} = \sqrt{\frac{g_2}{g_1}}\end{equation*}

Where, a) T_1 = time period on surface

b) T_2 = time period on mountain

c) g_1 = acceleration due to gravity on surface

d) g_2 = acceleration due to gravity on mountain

As acceleration due to gravity is inversely proportional to the square of the distance i.e a \propto \frac{1}{R^2}

we can write:

    \[ \frac{g_2}{g_1} \propto \frac{R^2}{\left( R+h \right)^2}\]

(2)   \begin{equation*} \sqrt{\frac{g_2}{g_1}} = \frac{R}{R+h}\end{equation*}

From equation 1 and 2, we can write:

    \[\frac{T_1}{T_2} = \frac{R}{R+h}\]

    \[ \frac{T_2}{T_1} = \frac{R+h}{R} \]

    \[\frac{T_2}{T_1} -1 = \frac{R+h}{R} -1 \]

    \[ \frac{T_2 - T_1}{T_1} = \frac{R+h-R}{R} \]

    \[ \frac{T_2 - T_1}{T_1} = \frac{h}{R} \]

    \[ T_2 - T_1 = \frac{h}{R}.T_1 \]

Finally,

    \[\Delta T = \frac{h}{R}. T\]

According to question, pendulum loses 15 s per day:

    \[ 15 = \frac{h}{6400 \text{km}}. 1 day  \]

    \[ 15 = \frac{h}{6400 \text{km}}. 24 \times 3600 \text{s} \]

    \[h = 1.1 \text{km}\]

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