1.

A piece of iron of density 7.8xx10^(3)kgm^(-3) and volume 100cm^(3) is completely immersed in water (rho=1000kgm^(-3)). Calculate (i) the weight of iron piece in air (ii) the upthrust and (iii) its apparent weight in water . (g=10ms^(-2))

Answer»

Solution :Given : VOLUME of iron piece `=100cm^(3)`
`=100xx10^(-6)m^(3)=10^(-4)m^(3)`
(i) Weight of iron piece in AIR
`=` Volume `xx` density of iron `xxg`
`=10^(-4)xx(7.8xx10^(3))xx10=7.8N`
(ii) Upthrust `=` (Volume of water DISPLACED) `xx` density of water `xxg`
But volume of water displaced = volume of iron piece when it is completely immersed `=10^(-4)m^(3)`
`:.` Upthrust `=10^(-4)xx1000xx10=1N`
(III) Apparent weight `=` True weight - Upthrust
`=7.8-1=6.8N`


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