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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?
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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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