Refer to the following figure for question,

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As shown in figure a monkey of 20 kg mass is holding a light rope that passes over a frictionless pulley.

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C) Remains unchanged.

Explanation:

It is given that the monkey and the bananas have an equal mass of 20 kg, and they are on the two ends of a rope.

The same forces are acting on both objects (the force of gravity and the tension force in the string; since the monkey and bananas have same mass of 20 kg; and F = mg). Therefore, the net force on both the monkey and the bananas is the same, which means the acceleration (direction and magnitude) is also equal. the tension in the string is equal on the monkey's end and the banana's end.

There are 2 cases for the motion of the monkey, as follows:

1. When the monkey climbs up the rope, the bananas move up as well. This is because the acceleration is the same in both direction and magnitude on both ends of the rope.

The distance between the monkey and the bananas stays constant because they have the same velocity (in magnitude and direction) so they are always moving up or down at the same rate.

2. If the monkey lets go of the rope, the distance between the monkey and the bananas again stays the same because both him and the bananas will be in free fall and they have the same mass and the same initial velocity, they will move at the same rate and the distance between them will not change.

Therefore, in all cases, the distance between them will be constant.

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How will bananas experience free fall ,aren't they already on ground?