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Physics
Density of river water is less than density of sea water. When a body is immersed completely, the loss in weight will be?
a) equal in both
b) more in sea water
c) less in sea water
d) depended on the mass
Which one of the following have longer wavelength?
a) radiowaves
b) soundwaves
c)X-rays
d) infrared rays
a) equal in both
b) more in sea water
c) less in sea water
d) depended on the mass
Which one of the following have longer wavelength?
a) radiowaves
b) soundwaves
c)X-rays
d) infrared rays
Answers
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For more on marking an answer as the "Best Answer", please visit our FAQ.1. There is no loss of weight. The body's weight (i.e mass on earth) obviously remains the same. But because seawater is denser than freshwater, the body displaces less of the seawater than of the freshwater.
2. Audible sound waves have much longer wavelengths than those electromagnetic waves, except in the case of very very low-frequency radio waves. So the question is unanswerable.
2. Audible sound waves have much longer wavelengths than those electromagnetic waves, except in the case of very very low-frequency radio waves. So the question is unanswerable.
chakka - the question specifically asks about weight, not mass.
I questioned the term myself, but since one of the answer options [ie, "d)"], mentions mass, I assume weight to be used in the correct context.
The body's mass will remain the same whether immersed, on Earth, or anywhere else in the solar system.
However, the body's weight (measured in Newtons or a force equivalent) in water will be equal to the force due to gravity less the upthrust from the water, which will be proportional to the volume displaced and its density.
"..the body displaces less of the seawater than of the freshwater.
Err..., what is "less"? Volume? Mass? Weight? Moles?
Assuming the body to be solid and rigid, it will displace equal volumes of water, irrespective of being saline or fresh.
I questioned the term myself, but since one of the answer options [ie, "d)"], mentions mass, I assume weight to be used in the correct context.
The body's mass will remain the same whether immersed, on Earth, or anywhere else in the solar system.
However, the body's weight (measured in Newtons or a force equivalent) in water will be equal to the force due to gravity less the upthrust from the water, which will be proportional to the volume displaced and its density.
"..the body displaces less of the seawater than of the freshwater.
Err..., what is "less"? Volume? Mass? Weight? Moles?
Assuming the body to be solid and rigid, it will displace equal volumes of water, irrespective of being saline or fresh.
Brachiopod, I completely agree with your explanation, but the answer to the first question should be b) more in sea water. You accidentally said C.
As for comparing how much water will be displaced depends on the situation. If the object is fully submersed, then the same volume of water will be displaced whether it is fresh or sea water. However, if the object is floating like the ship that you use in your example, less water will be displaced in sea water. Thus, as you pointed out the boat will sit higher in sea water.
As for comparing how much water will be displaced depends on the situation. If the object is fully submersed, then the same volume of water will be displaced whether it is fresh or sea water. However, if the object is floating like the ship that you use in your example, less water will be displaced in sea water. Thus, as you pointed out the boat will sit higher in sea water.
brachiopod, the weight of an object (to us) is its mass on earth i.e 1lb mass = 1lb weight. On the moon and on Jupiter the weight of that amount of mass would be different. On earth it cannot change just by being supported or buoyed up. If I am standing on bathroom scales which are showing my weight and then someone supports me under the armpits so as to reduce the reading on the scales , are you suggesting that my weight has changed?
I apologise for not having noticed that the questioner stipulated that the body is immersed completely, in which case the amount of water displaced (whether volume , mass or weight) is the same in both cases. And, of course, since buoying up does not occur, the density of the water is irrelevant.
I apologise for not having noticed that the questioner stipulated that the body is immersed completely, in which case the amount of water displaced (whether volume , mass or weight) is the same in both cases. And, of course, since buoying up does not occur, the density of the water is irrelevant.
brachiopod I don't know what altitude has got to do with it, but upthrust does not affect my weight; it merely supports it; my weight is unchanged.
If there are gravitational anomalies on this planet then, yes, my weight would change accordingly and the scales would show that change, just as they would if I experienced the 'gravitational anomaly' of going to another planet with a different mass from that of earth.
To get back to the original question, an object's weight does not change purely because the object is immersed in a liquid.
If I have a bucket of water weighing a total of X pounds and a bucket of treacle weighing a total of Y pounds and I put a lump of iron weighing one pound into each bucket, then the first bucket will weigh X+1 pounds and the second Y+1 pounds.
This is a non-question and I am surprised that it has given rise to so much chat. The answer, as I said before, is 'there will be no loss of weight".
If there are gravitational anomalies on this planet then, yes, my weight would change accordingly and the scales would show that change, just as they would if I experienced the 'gravitational anomaly' of going to another planet with a different mass from that of earth.
To get back to the original question, an object's weight does not change purely because the object is immersed in a liquid.
If I have a bucket of water weighing a total of X pounds and a bucket of treacle weighing a total of Y pounds and I put a lump of iron weighing one pound into each bucket, then the first bucket will weigh X+1 pounds and the second Y+1 pounds.
This is a non-question and I am surprised that it has given rise to so much chat. The answer, as I said before, is 'there will be no loss of weight".
Buckets? Treacle? Iron?
I think you are confusing mass with weight again, aren't you, chaka
I suggest the following.
Buy yourself a "weight measuring device" like a spring-balance (the type you'd weigh a fish with, like this).
Now take your lump of iron and weigh it using the spring balance.
Now immerse the lump of iron in water and weigh it again.
Then come back on here and tell us all what you observed.
I think you are confusing mass with weight again, aren't you, chaka
I suggest the following.
Buy yourself a "weight measuring device" like a spring-balance (the type you'd weigh a fish with, like this).
Now take your lump of iron and weigh it using the spring balance.
Now immerse the lump of iron in water and weigh it again.
Then come back on here and tell us all what you observed.
I don't need to do that experiment because all it will show is that the water is capable of supporting some of the weight of the iron. Instead of the water I could use my hand to take some of the weight. Better still, I could lower the iron to the floor so that the spring balance reads zero. Would this mean that the iron has lost all its weight? Does my weight reduce to zero when my bed supports me at night? If instead of lowering the iron to the floor I lower it onto bathroom scales, those scales will show the full weight while the spring balance shows zero!
The weight of a body on earth is equal to its mass. You can change that weight only by altering the value of gravity by (a) going to another planet, or (b) flying a parabola in an aircraft so that gravity is neutralised.
Write out a hundred times: Weight is not reduced just because it is supported.
The weight of a body on earth is equal to its mass. You can change that weight only by altering the value of gravity by (a) going to another planet, or (b) flying a parabola in an aircraft so that gravity is neutralised.
Write out a hundred times: Weight is not reduced just because it is supported.
I refer to my point "You are confusing weight with mass"
So, flying around a bit in an aircraft can change my weight,
but,
immersing myself in water can not possibly, in any circumstances whatsoever, affect that physical, absolute constant that is weight?
Is not weight just a force?
Use the force, chaka, use the force....
So, flying around a bit in an aircraft can change my weight,
but,
immersing myself in water can not possibly, in any circumstances whatsoever, affect that physical, absolute constant that is weight?
Is not weight just a force?
Use the force, chaka, use the force....
I have never, since doing physics at school, confused weight with mass, and I don't intend to start now. And yes, weight is force - the force that gravity exerts on mass. I repeat: weight is not reduced just because it is supported. Let's use that spring-balance of yours again:
I hang A 10lb lump of iron on the balance; the dial shows 10lb. I place my hand under the iron, taking some of the weight so that the dial now shows 7lb. Has the iron suddenly lost 3lb in weight? Of course not, and I can now prove it. I replace my hand with bathroom scales, again allowing them to take some of the weight so that the spring-balance dial again shows 7lbs. And what do the bathroom scales show? 3lbs. The 10lb weight has not been reduced; it has merely been shared between the spring and on the balance and the spring under the scales platform.
I hang A 10lb lump of iron on the balance; the dial shows 10lb. I place my hand under the iron, taking some of the weight so that the dial now shows 7lb. Has the iron suddenly lost 3lb in weight? Of course not, and I can now prove it. I replace my hand with bathroom scales, again allowing them to take some of the weight so that the spring-balance dial again shows 7lbs. And what do the bathroom scales show? 3lbs. The 10lb weight has not been reduced; it has merely been shared between the spring and on the balance and the spring under the scales platform.