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A solid body weighs 2.10 N in air. Its relative density is 8.4. How much will the body weigh if placed (i) in water (ii) in a liquid of relative density 1.2? |
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Answer» Solution :(i) Given: Weight of the BODY in air `W_(1)=2.10N`. R.D. of body `=8.4`, weight of body in WATER `W_(2)=?` `R.D.=(W_(1))/(W_(1)-W_(2))"":.""8.4=2.1/(2.1-W_(2))` or `8.4(2.1-W_(2))=2.1` or `W_(2)=(2.1xx7.4)/8.4=1.85N` Thus weight of body in water `=1.85N` (ii) Upthrust due to water `=W_(1)-W_(2)=2.10-1.85` `0.25N` Upthrust due to liquid `=` Upthrust due to water `xx` R.D. of liquid `=0.25xx1.2=0.30N` `:.` Weight of body in liquid `=` Weight of body in air - Upthrust due to liquid `=2.10-0.30=1.8N` ALTERNATIVE method: Let weight of body in liquid be x N. Then R.D. `=("Weight of body in air")/("Weight of body in air - Weight of body in liquid")xx` R.D. of liquid or `8.4=2.1/(2.1-x)xx1.2` or `4(2.1-x)=1.2` or `x=7.2/4=1.8N` |
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