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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)) |
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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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