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Question

A piece of gold with a cavity present in it weighs 10 g in air and 9 g in water. What is the volume of the cavity?
[Density of gold =19.3 g/cm3]

A
0.182 cm3
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B
0.482 cm3
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C
1.382 cm3
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D
2.482 cm3
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Solution

The correct option is B 0.482 cm3
Loss in weight of the piece of gold is due to the buoyant force acting on it inside water.


Wapp=WairFB
Wapp=Wairρg(Vgold+Vcavity)
9×103×g=10×103×g103×g×(Vgold+Vcavity)
Vgold+Vcavity=1×106
mg(air)ρg+Vcavity=1×106
10×10319.3×103+Vcavity=1×106
Vcavity=0.482×106 m3=0.482 cm3

Why this question?Caution: Please note that while obtaining FBon gold piece, the total volume of liquiddisplaced by it is inclusive of the volume ofcavity.

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