Question

A piece of pure gold (p = 19.3 gcm) is suspected to be hollow from inside. It weighs 77.2 g in air and 71.2 g in water. The volume of the hollow portion in gold will be
(1) 1 cm
(2) 2 cm
(3) 3 cm
(4) 4 cm​

Answer 1

Answer:

4 cm

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Answer 2

Solution :-

• Density of pure gold= 1.93 g/cm^3

• Mass of gold in air = 77.2 g

we know that ,

⟹ d = m/v

here ,

• d = density

• m= mass

• v = volume

on rearranging ,

⟹ v = m /d

⟹ v = 77.2 / 19.3

⟹ v = 4 cm ^3

The volume of the pure gold piece = 4 cm^3

When the gold piece is dipped in water ,

As per Archimedes principle,

The volume of water displaced = Apparent loss of weight of the gold piece in water

• Mass of gold piece in water =71.2 g

• Mass of water displaced = 77.2 - 71.2 = 6 g

• Density of water= 1 g /cm^3

we know that ,

⟹ d '= m '/v '

on rearranging ,

⟹ v '= m ' /d '

⟹ v = 6 / 1

⟹ v = 6 cm ^3

The volume of the water displaced = 6 cm^3

Volume of the hollow portion of gold

= Volume of water displaced - Volume of pure gold

= 6 - 4

= 2 cm^3

Volume of the hollow portion of gold is 2 cm^3