Physics, asked by navajyothsraj, 7 months ago

1. The weight of an object in kerosene is 8N and that in air is 10 N. If so what is the buoyant force
exerted by the kerosene?

Answers

Answered by mitraarati59
0

Answer:

The solid apparently weighs less in water when compared to air because of the upward buoyant force acting on it when it is submerged in water. 

According to Archimedes principle, the buoyant force acting on an object is equal to the weight of the fluid displaced by it. 

Therefore, the apparent decrease in the weight of an object is equal to the weight of the fluid displaced by it.

Weight of displaced water = Apparent decrease in weight = 120−105=15 gf

Mass of displaced water = 15g

Volume of solid = volume of water displaced = densitymass=1g/cm315g=15cm3

Actual weight of solid = weight of solid in air =120gf (since buoyant force due to air is negligible)

Mass of solid = 120g

Density of solid = VolumeMass=15120=8g/cm3

Relative density of solid = DensitywaterDensitysolid=1g/cm38g/cm3=8

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