Answer is 98.78

Classical Mechanics Level 2

A piece of pure gold of relative density 9.3 9.3 weighs 38.25 g 38.25\text{ g} in the air and 33.865 g 33.865\text{ g} in water. It is suspected to have a cavity inside.

Calculate the volume of the cavity in the gold in cm 3 , \text{cm}^3, to 3 decimal places.


The answer is 0.272.

Sumukh Bansal by Sumukh Bansal , 16, India

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1 solution

Munem Shahriar
Nov 9, 2017

Given that,

m = 38.250 g m = 38.250g

ρ = 9.3 ρ= 9.3

Now,

V = m ρ \text{V} = \dfrac{\text{m}}{\text{ρ}}

⇒ 38.250 9.3 ≈ 4.1129 \Rightarrow \dfrac{38.250}{9.3} \approx 4.1129

Loss of mass = 38.250 − 33.865 = 4.385 g = 38.250- 33.865 = 4.385g

According to laws of flotation:

Mass of water = volume of object \text{Mass of water = volume of object}

So volume of gold in water is 4.385

Hence,

Volume of hollow portion = 4.385 − 4.1129 ≈ 0.272 c m 3 \text{Volume of hollow portion} = 4.385 - 4.1129 \approx \boxed{0.272} cm^3

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