A solid of density 2700 kgm’ and of volume 0.0015 m3 ¡s completely immersed in alcohol of density 800 kgm’.Calculate:
- Weight of solid in SI system.
- Upthrust on solid in SI system.
- Apparent weight of solid in alcohol.
- Will the apparent weight of solid be less or more, if it is immersed completely in brine solution? Give a reason, [g = 10 ms-2]
Density of solid = ρ= 2700 kgm3
Volume of solid = V = 0.0015 m3
Density of alcohol = ρ’ = 800 kgm3
- Mass of solid = m = V x p
m = 0.0015 x 2700 = 4.05 kg
Weight of solid = mg = 4.05 x 10 = 40.5 N
- Volume of alcohol displaced = Volume of solid
V = 0.0015 m3
Mass of alcohol displaced = m’ = V x p’
m’ = 0.0015 x 800 = m’ = 1.2 kg
Upthrust = Weight of alcohol displaced
= m’g= 1.2 x 10= 12N
- Apparent weight of solid in alcohol
= Actual weight of solid – Upthrust
= 40.5 -12 = 28.5 N
- When a solid is immersed completely in brine solution, then upthrust acts on it in an upward direction, as a result, its apparent weight of solid will be less than actual weight of solid will be less than the actual weight of solid.
Concept: Archimedes' Principle
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