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:

MCQ, simple harmonic motion mcq / August 18, 2022 August 20, 2022

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 km
2.2 km
3.3 km
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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