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Question: A pendulum clock which shows the correct time on earth is taken to the moon. Then the time shown wil...

A pendulum clock which shows the correct time on earth is taken to the moon. Then the time shown will be
A) the correct time.
B) six times faster.
C) 6\sqrt 6 times slower.
D) none of these.

Explanation

Solution

The period of the pendulum of the clock depends on the acceleration due to gravity on the surface. The acceleration due to gravity on the surface of the moon is found to be 16\dfrac{1}{6} times the acceleration due to gravity on the surface of the earth.

Formula used:
-The period of the pendulum is given by, T=2πlgT = 2\pi \sqrt {\dfrac{l}{g}} where ll is the length of the pendulum and gg is the acceleration due to gravity on the surface of the earth.

Complete step by step answer.
Step 1: Describe the problem at hand.
It is given that a pendulum clock is taken to the moon. This pendulum clock kept the correct time on the earth. The time shown on the moon is to be determined. This can be obtained using the relation for the period of the pendulum.
Step 2: Express the relation for the period of the pendulum on the surface of the earth and correspondingly obtain an expression for the period on the surface of the moon.
The period of the pendulum on the surface of the earth is given by, T=2πlgT = 2\pi \sqrt {\dfrac{l}{g}} -------- (1)
where ll is the length of the pendulum and gg is the acceleration due to gravity on the surface of the earth.
The above relation suggests that the period of the pendulum only depends on the acceleration due to gravity on the surface.
The acceleration due to gravity on the moon’s surface is less than that on the earth’s surface.
Let gm{g_m} be the acceleration due to gravity on the moon’s surface. It is given by, gm=g6{g_m} = \dfrac{g}{6}
Now using equation (1), the period of the pendulum on the moon’s surface can be obtained as Tm=2πlgm=2π6lg{T_m} = 2\pi \sqrt {\dfrac{l}{{{g_m}}}} = 2\pi \sqrt {\dfrac{{6l}}{g}} ------------ (2).
Step 3: Using equations (1) and (2) determine if the clock runs slower or faster on the moon’s surface.
Equation (1) gives the period on the earth’s surface as T=2πlgT = 2\pi \sqrt {\dfrac{l}{g}} and equation (2) gives the period on the moon’s surface as Tm=2π6lg{T_m} = 2\pi \sqrt {\dfrac{{6l}}{g}} .
Dividing equation (1) by (2) we get, TTm=2πlg2π6lg\dfrac{T}{{{T_m}}} = \dfrac{{2\pi \sqrt {\dfrac{l}{g}} }}{{2\pi \sqrt {\dfrac{{6l}}{g}} }}
Cancelling out the similar terms in the numerator and denominator we get, TTm=16\dfrac{T}{{{T_m}}} = \dfrac{1}{{\sqrt 6 }} or Tm=6T{T_m} = \sqrt 6 T
This implies that the period of oscillation is 6\sqrt 6 times longer on the moon’s surface and thus the clock will run slower on the moon’s surface.
Thus the time shown on the clock on the moon will be 6\sqrt 6 times slower.

Hence the correct option is C.

Note: Here, we assume that the length of the pendulum ll remains the same when the clock is taken to the moon i.e., we neglected the change in length that can occur due to a rise or drop in temperature. The gravitational force on the moon’s surface is lesser than the force on the earth because the mass of the moon is much smaller when compared to the mass of the earth.