A simple pendulum is hollow from within and its time period is T. How is the time period of this pendulum affected when one quarter of the bob is filled with mercury? When three quarters of the bob are filled with mercury? And when the bob is completely filled with mercury?  

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The period of a simple pendulum can be approximated for small displacements by the following famous equation:`T = 2*pi*sqrt(l/g)`  Here, g is gravity's acceleration at the location and l denotes the length of the pendulum. This length is the distance from the center of mass of our pendulum to...

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The period of a simple pendulum can be approximated for small displacements by the following famous equation:

`T = 2*pi*sqrt(l/g)` 

Here, g is gravity's acceleration at the location and l denotes the length of the pendulum. This length is the distance from the center of mass of our pendulum to the fixed point (where the line is attached to the ceiling). Since this is a simple pendulum, we can neglect the mass of the line. So our length is just the distance from the fixed point to the bob's center of mass.

Adding mercury to the bob will change the distribution and amount of mass in it, moving the center of mass downwards. To calculate this amount, we would need more information about the size of the bob. At the end, this will lead to an increase in l, resulting in a higher period.

But, if the bob is small enough, then we can also neglect this change in the position of the center of mass and the period will remain T for all cases given. That is to say that the simple pendulum does not depend on its mass (approximately) .

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