Q.

A piece of gold weighs 10 g in air and 9 g in water. What is the volume of cavity? (Density of gold = 19.3 g cm-3)

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a

0.282 cc

b

0.482 cc

c

0.682 cc

d

None

answer is B.

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Detailed Solution

Actual volume of gold = 1019.3 = 0.518 cm3Now, if V be the apparent volume of the piece of gold, then loss in weight of gold in water = weight of V volume of water⇒ 10-9 =V×1 = 1 ccVolume of cavity = 1-0.518 = 0.482 cc
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