# Thread: Quiz: 2 simple physics questions

1. concept of bouyancy for question 1...
when the rock was in the boat, a mass of water equal to the mass of the rock had to be be displaced in order for the rock to stay afloat. lets ignore the mass of man and boat because these 2 are constant.

but because water is less dense than the rock, more water (volumetric) has to be displaced than the volume of the rock. thus when the rock is dropped into the water, the only volume it takes up is its own...

thus, the answer is the water level will drop.

2. Originally Posted by szekiat
However, this is irrelevent to the water level on the boat since we assume that the boat is still on the surface and is only affected by the water around it. However, the water level relative to the side of the pool will have increased since the new vol is greater with the introduction of a rock.
But I would think that the volume of water displaced by the boat is now less, since it has less weight in it. For a heavy rock, the difference in the amount of water displaced by the boat with rock and boat without rock would be greater than by the amount of water displaced by the rock - so in general the water level in the pool will go down.

3. Originally Posted by showtime
concept of bouyancy for question 1...
when the rock was in the boat, a mass of water equal to the mass of the rock had to be be displaced in order for the rock to stay afloat. lets ignore the mass of man and boat because these 2 are constant.

but because water is less dense than the rock, more water (volumetric) has to be displaced than the volume of the rock. thus when the rock is dropped into the water, the only volume it takes up is its own...

thus, the answer is the water level will drop.
Yep, that is right; that's why I said it is dependant on the density of the rock, if the rock is more dense than water, water level will drop, but it could be one of those light weight volcanic rocks or something that are less dense than water.

4. 1. Water level will drop.
2. Balloon without air will tip up. (Same composition means same N2, O2, CO2 etc, nothing said about pressure, so I assume that the elastic nature of the balloon will cancel of the higher pressure of the air inside. Therefore, higher pressure equal more dense.)

5. Originally Posted by showtime
concept of bouyancy for question 1...
when the rock was in the boat, a mass of water equal to the mass of the rock had to be be displaced in order for the rock to stay afloat. lets ignore the mass of man and boat because these 2 are constant.

but because water is less dense than the rock, more water (volumetric) has to be displaced than the volume of the rock. thus when the rock is dropped into the water, the only volume it takes up is its own...

thus, the answer is the water level will drop.
This is exactly the explanation I would give for qn 1.

So the answer is, the water level will fall, both with respect to the poll side, as well as with respect to the boat.

6. Originally Posted by roygoh
Two simple but interesting questions to refresh your O Level physics knowledge:

1. There is a man and a piece of rock the size of a basket ball in a little boat that is floating in a swimming pool. The man throws the rock into the water. Does the water level in the swimming pool rise or fall?

2. A deflated balloon is place on one side of a balance and a balloon filled with air (same composition as the ambient atmosphere) is placed on the other side. If both the balloons have identicle mass before one of them is inflated, which side does the balance tip towards?

- Roy
1. no change in water level
2. tip towards deflated baloon

7. Originally Posted by zodnm
2)if ambient atmosphere is used to inflate the balloon normally, then the air inside the balloon is "compressed" and balance will tip towards the balloon filled with air.
This would be my solution for qn 2.

Originally Posted by zodnm
if the air inside the balloon is exact same composition as ambient atmosphere(ie same density and pressure) then balance will stay still.
This is true also, but we are not talking about a rigid container but an elastic ballon, so the first explanation above is more accurate.

Originally Posted by zodnm
***note that in 2nd case, if the opening of the balloon is opened, the balloon will not deflate, as pressure and density of air outside and inside the balloon is same. this also means that 2nd case is not practical
Right...it is not possible for the air inside the filled balloon to have the same density as the ambient atmosphere because the balloon is elastic.

8. Originally Posted by scud
1. no change in water level
2. tip towards deflated baloon

9. Originally Posted by roygoh

10. more physics qns Roy.... more ! this is rather good mind workouts

11. Originally Posted by SniperD
more physics qns Roy.... more ! this is rather good mind workouts
Try the geometry question first?

12. Originally Posted by eddietkm
Air is weightless.
I think Air is not weightless. Every mole or 24 dm3 of air will consist of about Nitrogen-78.1%
Oxygen- 20.9%
and 1% other gases (with Carbon Dioxide- 0.035%)..

Assuming 79% of Nitrogen and 21% of oxygen are present, then each mole of air will weigh 0.79*14 + and 0.21*16= 14.42 g.

Its just that air's density is much lighter than most matter, so its weight is not so easily measured, plus you can hardly "capture" and measure air as air molecules are in constant motion.

My 2cents worth..

13. if air is weightless, there won't be such a thing called atmospheric pressure

14. ## Primary school Question for number two....

Air has weight.....so balance tip to the baloon with air....

rgds,
sulhan

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