From the two arms of a beam balance, a metal body weighing 20 g and a piece of glass are suspended. The apparent weights of these two bodies when measured in water are found to be the same. If immersed in alcohol instead of water, the mass of the metal needs to be increased by 0.84 g for balancing the beam. Calculate the mass of the glass piece. [Given, density of water = `1g*cm^(-3)`, density of alcohol = `0.96g*cm^(-3)`]

Explanation

Let us assume that the weight of the glass piece is mg
Let the density of the metal body = `d_(1)g*cm^(-3)`, and the density of glass = `d_(2)g*cm^(-3)`
According to the question, when the bodies are immersed in water,
apparent weight of the metal body = apparent weight of the glass piece
or, real weight of the metal body - weight of the water displaced by the metal body = real weight of the glass piece - weight of the water displaced by the glass piece
or, `(20-(20xx1)/d_(1))xxg=(m-(mxx1)/d_(2))xxg`
or, `m/d_(2)-20/d_(1)=m-20" "...(1)`
When immersed in alcohol,
`(20-20/d_(1)xx0.96)xxg+0.84xxg=(m-m/d_(2)xx0.96)xxg`
or, `m(1-0.96/d_(2))-20(1-0.96/d_(1))=0.84`
or, `m-(mxx0.96)/d_(2)-20+(20xx0.96)/d_(1)=0.84`
or, `m-0.96(m/d_(2)-20/d_(1))=20.84`
or, m - 0.96(m - 20) = 20.84 [From equation (1)]
or, 0.04m = 20.84 - 19.2
or, `m=1.64/0.04=41`
`therefore` The mass of the glass piece is 41 g.