Calculating the Mass of a Piece of Glass Using Archimedes' Principle

AI Thread Summary
The discussion revolves around calculating the mass of a glass piece using Archimedes' Principle, comparing its apparent weight to that of a metal piece in different liquids. The initial problem states that a 20g metal piece has the same apparent weight as the glass when both are submerged in water, and additional weight is needed when the liquid is changed to alcohol. Participants emphasize the importance of correctly applying Archimedes' Principle and understanding the relationship between weight, density, and buoyancy. Clarifications are sought on the equations used and the definitions of variables involved in the calculations. The conversation encourages a step-by-step approach to solving the problem rather than providing direct answers.
soumya_26in
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1.a piece of metal of weight 20gms has equal apparent weight with a piece of glass when both are suspended from the arms of a balance and immersed in water(density 1).if water is replaced by alcohol(density 0.96), 0.84gms must be added to the pan from which the metal is suspended to restore balance.what is the mass of the glass piece?

2.a cube with each side of unit length has the same force F applied normally outwards on all its six faces.what are the expressions of longitudinal & volume strain in terms of the elastic modulii?
 
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Welcome to PF!

Hi soumya! Welcome to PF! :smile:

You must tell us how far you've got, so that we know what's troubling you, and how to help.

Start with the balance … what have you tried?

What equations have you got? :smile:
 
tiny-tim said:
Hi soumya! Welcome to PF! :smile:

You must tell us how far you've got, so that we know what's troubling you, and how to help.

Start with the balance … what have you tried?

What equations have you got? :smile:

actually both prblms were in my class eleven's final Q paper.i have tried 1) as follows:
1)let,
mass of the glass piece=m gms
In 1st case, apparent wt of metal=(20-20/d1),d1=density of the metal
& apparent wt of glass=(m-m/d2),d2=density of the glass
by the prblm,
20-20/d1=m-m/d2.....(i)
In case 2, apparent wt of metal=[20-(20+0.84)*0.96/d1]
& apparent wt of glass=(m-m*0.96/d2)
by the prblm,
[20-(20.84*0.96)/d1]=(m-m*0.96/d2)....(ii)
is this process write?
 
actually both prblms were in my class eleven's final Q paper.i have tried 1) as follows:
1)let,
mass of the glass piece=m gms
In 1st case, apparent wt of metal=(20-20/d1),d1=density of the metal
& apparent wt of glass=(m-m/d2),d2=density of the glass
by the prblm,
20-20/d1=m-m/d2.....(i)
In case 2, apparent wt of metal=[20-(20+0.84)*0.96/d1]
& apparent wt of glass=(m-m*0.96/d2)
by the prblm,
[20-(20.84*0.96)/d1]=(m-m*0.96/d2)....(ii)
is this process write?
 
soumya_26in said:
apparent wt of metal=(20-20/d1),d1=density of the metal

No.

(You understand, I can't give you the answer, I can only help you do it yourself? So …)

20 is the mass of the metal, but d1 is the mass/volume of the water - so 20/d1 isn't anything.

Start again. Do it one step at a time.

First: what is Archimedes principle?

How exactly does it reduce the weight of the metal? What does it depend on? Are there any unknowns?

Same for the glass.

What equation can you write? :smile:

(How did you manage to post twice? :smile: If you click on "edit" under the second post, I think it gives you the option of deleting.)
 
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