Question

A solid body with R.D. 9 weighs 3N in air. Find the weight of the solid inside water

and inside a liquid (R.D. 1.5).​

Answer 1

Answer:

A solid body with R.D. 9 weighs 3N in air. Find the weight of the solid inside water

and inside a liquid (R.D. 1.5).

answer is 3N

Answer 2

Given: weight of a body = 3N

          Relative density of liquid = 1.5

          Relative density of body = 9

To find: weight of solid body inside the liquid

Solution:

• The relative density of a solid is the ratio of the density of a substance to the density of a standard, usual water for a liquid or solid, and air for gas.

• The relative density of a solid is the ratio of the weight of solids in the air to the weight of water displaced by the body.

According to the question, the relative density of the solid is 9

Therefore, 9 = weight of solid in air/weight of water displaced by the body.

9 = 3/x

x = 1/3 = 0.33N

Therefore, the weight of water displaced by the body = 0.33N

As the weight of the body in water is equal to the  = weight of the body in the air - the weight of water displaced by the body

= 3 - 0.33

=2.67 N

Therefore, the weight of the body in the water of relative density 1.5 is 2.67N.