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Question
A solid body weighs 2.10 N. in air. Its relative density is 8.4. How much will the body weigh if placed
(1) in water,
(2) in liquid of relative density 1.2?
Solution
Weight of solid body in air = 2.10 N
R.D. of solid = 8.4
R.D. of solid = `("Density of solid"(rho_"solid"))/"Density of water"`
`8.4 = rho_"solid"/rho_"water"`
`rho_"solid" = 8.4 xx rho_"water" = 8.4 xx 1000 = 8400` kgm-3
(1) R.D. of solid = `"Weight of solid in air"/"Weight of water displaced by body"`
`8.4 = 2.10/"Weight of water displaced by body"`
Weight of water displaced by body = `2.10/8.4 = 0.25` N
Weight of body in water = Weight of body in air – Weight of water displaced by body
= 2.10 - 0.25 = 1.85 N
(2) Upthrust due to water = Weight of water displaced by body
= 0.25 N
Upthrust due to liquid = Upthrust due to water ×R.D. of liquid
= 0.25 x 1.2 = 0.30 N
Weight of body in liquid = Weight of body in air – Upthrust due to liquid
= 2.10-0.30= 1.80 N
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