1.

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?

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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