Sum

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}]

Advertisement Remove all ads

#### Solution

Density of solid = ρ= 2700 kgm^{3}Volume of solid = V = 0.0015 m^{3}Density of alcohol = ρ’ = 800 kgm^{3}

- 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 m^{3}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

Is there an error in this question or solution?

Advertisement Remove all ads

#### APPEARS IN

Advertisement Remove all ads

Advertisement Remove all ads