Can you solve this physics problem?

Question

Two identical containers have the same volume of water, and one is placed on each pan of a double-pan balance. Obtain a hollow plastic ball and a steel ball that have the same volume. A string from the ceiling suspends the steel ball so it is completely submerged in the water on the left pan. The hollow ball is held submerged in the water of the right pan by a string fastened to the bottom of the container. Now what will be the condition of the double pan balance?

1. The balance indicator will read zero. The scale is balanced.
2. The left pan (with the steel ball) moves down.
3. The left pan rises.

for what it's worth i would have said 3 also. the answer is actually 2, the plastic ball pan would rise. it is due to the additional upward force the tension of the string puts on the bottom of the container.

Answer 1

the left pan rises

Answer 2

Sorry, but you are wrong. The so called extra buoyancy caused by the plastic ball putting tension on the string and 'pulling' up on the plastic ball pan system is exactly countered by the increased height of the water in the plastic pan systems container. It is a equilibrium system. You have not introduced any new outside force.

The other pan, with the steel ball, does have a increase in the height of the fluid without a commensurate balanced connection to the displacement body. The buoyancy of the steel ball will cause the pan to lower on that side. The increase in forcing is introduce from gravity acting on the steel ball. Part of that force is taken up by the string, the other part from displacement. Put a spring scale on the string when you try the experiment if you want to find out what contribution the ball has. When immersed, the tension will decrease on the string.

So 2 is correct, but not because of the plastic ball.

Answer 3

1. The scale is balanced.

This question is based on the Archimedes Principle. Neither ball is touching the sides of the container. Both balls displace the same amount of water. It's the weight of the displaced water that becomes important.

Edit: Karman V has a point. If you held the hollow ball submerged in the water with a toothpick or similar tool, the scales would be balanced - you would only look at the weight of the water displaced. The fact that the ball is attached to the pan means there can be no net change in force.

(However, I think the additional details provided by the poster intend to say something similar to Karman V's response, if not in as much detail.)

Answer 4

Ans-2
The weight of the steel ball is supported by the tension in the string and the buoyancy of the water. So the weight measured is the weight of the steel ball minus the tension in the string. With the rubber ball it does not matter if there is a string or not. the weight of the rubber ball is transferred to the bottom of the container by the tension in the string and the buoyancy. Because the buoyancy is the same in both cases the left pan would move down because of the additional tension in the string.

Answer 5

The left pan will rise. Each pan holds the same amount of liquid, and although not stated, I am assuming each pan weighs the same. The hollow ball, while it might not weigh anything does weigh something. It's weight would be added to the right pan. If the string were cut, the same would occur, the weight of the hollow ball would be added to the right hand pan as the ball floated on the water.

Answer 6

3. Since the hollow ball is held to the bottom of the container on the right pan, its weight must now be counted as part of the container/water apparatus. This additional weight will tip the scale and raise the left pan.

Answer 7

3

Answer 8

If the pans holding the water have the same weight then the right pan will weigh more, but only by a gram or less.

I hope this is right, good question.

Wait...is the steel ball touching the pan?

Answer 9

Assuming the hollow ball and string have zero mass, (1) - the scale should remain balanced.

Neither ball imparts any forces to either container. The massless hollow ball and string add no weight to the first contianer, and the steel ball is suspended by the string it is not held by the container and the weight does not transfer to the balance.

Hope I'm not all wet.

Answer 10

The left pan would lower... NOT because of the steel ball in it. It would lower because the plastic ball would create more buoyancy with the plastic ball having a direct effect on the container its in...The steel ball is merely suspened in the water, not having any effect on the 'specific gravity' of the container

Answer 11

Left pan rises. String from hollow ball would occupy additional mass.