Thinking Physics Question 6

i'm guessing e, since the level of the water (brim full) already takes into account the "objects" in the icebergs

B

actually if something more dense than ice (or water?) were in the ice then it seems like the water level should decrease after the ice melts, because it would have been displacing more water than it "should have" since it had more mass in the same volume

edit: meant to change to B

ANS: C
explanation: the empty space in in ice A will compensate volume, for ice B, when water is frozen, vol drops unusually, assuming baths are full, the internal water will overcompensate..therefore it will spill..i dont know/never bloody see an railroad spike before..and A and B agrees with answer C.

b)
For floating objects only the total mass of it determines the amount of water in moves aside.
So the phase transition of A and W doesn't make a difference. S is different, because before it melts the amount of water moved aside is proportional to the total mass. Afterwards the spike sinks to the ground where it will move only the volume of the spike itself aside which is less than the equivalent of the mass of the floating so the water sinks.

e) The mass won't change in any of the tubs.

I just read masterofchaos' spoiler though and it seems pretty reasonable, don't have any knowledge of those kind of mechanics though, my instinctive answer would've been e).

D. Ice is less dense than water. In the first two cases, the ice can displace no more than a volume of water equal to its weight, and when it melts, it expands to a volume that was larger than the initial volume of water displaced. In the third case, even if it were to be completely submerged in the water, the volume of the water after melting would exceed the volume of the ice before melting.

My answer is b) The water in S will get lower and the water in A and W will stay exactly brim full

A floating object displaces its weight in water. Any part of the iceberg which was ice turns into water when it melts, displacing the same volume before and after. So we can disregard the ice part of the icebergs.
The floating railroad spike displaces a volume of water equal to its weight, that is, a volume greater than the sinking railroad spike. The water in S will get lower as a result. Air displaces a little bit of water before, and none after, so the water in A will lower a little bit. For the third case, unfrozen water displaces the same amount of water before and afterwards.

D
In order for the ice to melt the temperature around the ice-cold water needs to increase. When the temperature of water increases the volume taken by the water increases, so if each one of the tubs is full to begin with, then melting the ice will overflow the tub.

I think my answer is more chemistry than physics >.>

B
According to Archimede's principle, the weight of the water displaced = however much your object weighs. This means with just ice, the water level will not change since weight of ice = weight of water from iceberg melting. Since air weight is practically negligible, this will mean that iceberg A's water level will not move. The same explanation holds for iceberg W, because the water inside the iceberg can be treated separately from the iceberg via archimede's principle. However, when we look at iceberg S, the water displaced is far more than it was in either situation A or B. However, there is the same amount of ice. The same amount of ice is going to create the exact same amount of water, and because the spike doesn't change, more water is displaced by iceberg S than the amount of water that iceberg S will generate when melted. Thus tub S will grow lower in its water level

I will assume the volume of ice is negligible
Iceberg A will release air which will go into the atmosphere and therefore not increase the volume
=> Brim full
Iceberg W has water in it however the weight of it is equal to the volume it displaces and since water has a relative density of 1
=> Brim full
Iceberg S has something heavier than water so it should displace more water than it would if it was released.
=> Water level lowers
Not sure if I remember the buoyancy (sp?) law correctly butok
That makes my answer:
=> (b)

get glass almost full of water and drop let say 10 coins and see what happens to water level

ANS: C
explanation: the empty space in in ice A will compensate volume, for ice B, when water is frozen, vol drops unusually, assuming baths are full, the internal water will overcompensate..therefore it will spill..i dont know/never seen a railroad spike before..and A and B agrees with answer C.

I certainly couldn't've come up with the correct answer but after reading MasterOfChaos' and others' posts it seems like B is reasonable to me. When inside the iceberg the railroad spike causes additional displacement due to its weight/mass but once it sinks to the bottom it will cause less displacement since only it's volume will matter and not it's mass/weight.

Answer is definitely E.

You can treat each object the same according to Archimedes principle. Since each of lower or higher density objects are inside the icebergs respectively (ice, water, spike), this gets taken into account in how low each iceberg sits in the water. Thus, each iceberg will displace as much volume as proportional to its buoyancy. Therefore, no water level will change.

Gene removed his answer so.. I am assuming micronesia doesn't care about this topic anymore though.

Most of you guys choosing B is kinda amusing namely because unfrozen water is more dense than ice (same as the spike). Thus, if you chose B you SHOULD also being saying that the water level would be lower for iceberg with water inside it too which is obviously not the case as you reasoned out.