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A beaker containing water is placed on the pan of balance which shows a reading of M gms. A lump of sugar of mass m gms and volume Vcc. Is now suspended by a thread in such a way that it is completely immersed in water without touching the beaker and without any overflow of water. What will be the reading of the balance just when the lump of sugar is immersed ? How will the reading change as the time passes on?

Radhika Batra , 10 Years ago
Grade 11
anser 2 Answers
Kevin Nash

Last Activity: 10 Years ago

Hello Student,
Please find the answer to your question
When the lump of suger is just immersed
T = Mg – B
(For equilibrium of lump of sugar )
The reading on the pan balance
= Mg + Vdw g
Where V = Volume of lump of sugar
dw = density of water
When the lump is half dissolved,
The reading on the pan balance = Mg + v / 2 ds g + V / 2 dwg
When ds = density of sugar
Since ds > dw
∴ the reading will increase.
Thus, we can conclude that as the time passes the reading will keep increasing.
Thanks
Kevin Nash
askIITians Faculty

Sher Mohammad

Last Activity: 10 Years ago

Initially , there will be no change, as sugar is mixing in water the weight on balance will increase and hence the reading will increase.
sher mohammad
faculty askiitians
b.tech, iit delhi

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