Question

a piece of silver weight 66 gf in air 55 gf in water . find relative density of silver​

Answer 1

Answer:

Let us consider that D

w

and D be the density of water and solid respectively.

The actual weight of the solid is 32gf.

Apparent weight of the body = Actual weight - Buoyant force (V×D

w

×g).

Let V be the volume of the solid.

Density of water D

w

=1gcm

3

.

In this case, 28.8=mg−V×D

w

×g

That is, 28.8=32−V×1

Therefore, V=3.2cm

3

.

The density of the body is given as mass/Volume, that is 32/3.2=10gcm

3

.

The relative density of a solid is given as the ratio of the density of a substance to the density of a standard substance under specified conditions.

That is, Relativedensity=

Densityofwater

Densityofsolid

=

1gcm

3

10gcm

3

=10

Hence, X-4 = 10-4 = 6.