Question

A solid weighs 120 GF in air and 105 GF when it is completely immersed in water calculate the relative density of solid.​

Answer 1

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Given, weight of solid in air W1 = 50 gf and weight of solid in water w2 = 44 gf

(i) Upthurst = loss in weight when immersed in water = W1 - W2 = 50 - 44 = 6 gf

(ii) Weight of water displaced = upthurst = 6 gf

Since density of water is 1 g/m³, therefore volume of water displaced = 6cm³

But the solid displaces water equal to its own volume, therefore volume of solid = 6cm³.

(iii) R.D. of solid = Weight of solid in air / Weight in air - Weight in water

= W1 / W1 - W2 = 50 / 50 - 44 = 50 / 6 = 8.33

Answer 2

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

Explanation:

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