Physics, asked by vsusmitha122004, 7 months ago

A solid of mass 76g and displaces 20cm of water at 4°C. The RD of solid in​

Answers

Answered by farhaanaarif84
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 =

density

mass

=

1g/cm

3

15g

=15cm

3

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 =

Volume

Mass

=

15

120

=8g/cm

3

Relative density of solid =

Density

water

Density

solid

=

1g/cm

3

8g/cm

3

=8

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