Question #c9111

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Answer 1 · 2016-11-01T08:10:25

$M_a->"Weight of alloy in air"=87gwt$

$M_w->"Weight of alloy in watwr"=71gwt$

$M_a-M_w->"Apparent loss in weight"$

$=87-71=16gwt$

As per Archmedes principle the apoarent loss in weight here is equal to the weight of displaced water $=16gwt$ .

So mass of displaced water $16g$

Taking density of water $=1gcm^-3$ we can say that the volume of displaced water or the volume of the alloy-mass is
$=(16g)/(1gcm^-3)=16cm^3$

Hence density of the alloy is

$d_"alloy"="Its mass"/"Its volume"=(87g)/(16cm^3)=5.437gcm^-3$