Question #a28fb

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Answer 1 · 2016-11-28T13:09:35

Given
$"wt of sample in air"=295N$

$"wt of sample in alcohol"=200N$

$"sp.gravity of alcohol"=0.7$

$"density of alcohol"=0.7gcm^-3=(0.7*10^-3kg)/(10^-6m^3)=700kgm^-3$

Now wt of displaced alcohol by the sample

$="Its wt in air - wt in alcohol"$

$=295-200=95N$

a) Now if $vm^3$ be the volume of the sample,then the weight of displaced alcofol by the sample

$=vxx700xx9.8N$

So $vxx700xx9.8=95$

$=>v=95/(700xx9.8)~~0.01385m^3$

b) Density of the material

$="mass"/"volume"$

$=(295/9.8)/(95/(700xx9.8))m^3$

$=(295xx700)/95~~2173.7kgm^-3$