Question #c9111

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Question #c9111

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Answer 1 · 2016-11-01T08:10:25

$M_{a} \to Weight of alloy in air = 87 g w t$ $M_a->"Weight of alloy in air"=87gwt$

$M_{w} \to Weight of alloy in watwr = 71 g w t$ $M_w->"Weight of alloy in watwr"=71gwt$

$M_{a} - M_{w} \to Apparent loss in weight$ $M_a-M_w->"Apparent loss in weight"$

$= 87 - 71 = 16 g w t$ $=87-71=16gwt$

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

So mass of displaced water $16 g$ $16g$

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

Hence density of the alloy is

$d_{alloy} = \frac{Its mass}{Its volume} = \frac{87 g}{16 c m^{3}} = 5.437 g c m^{- 3}$ $d_"alloy"="Its mass"/"Its volume"=(87g)/(16cm^3)=5.437gcm^-3$

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