Question

Question #c9111

Answer 1

`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`