A piece of gold (p=19.3 g/cm3) is suspected to be hollow from inside. It weighs 77.2 g in air and 71.2 g in water. What is the volume of hollow portion of the gold?
 
 



Asked by Jerlin George | 23rd Nov, 2014, 12:02: PM

Expert Answer:

Given that the density of gold =19.3 g/cm3
Mass of gold piece in air =77.2 g
So volume of the gold piece will be:
begin mathsize 14px style We space know space that space density space equals Mass over Volume rightwards double arrow Volume space equals Mass over Density Volume space of space gold space piece space equals fraction numerator 77.2 straight g over denominator 19.3 straight g divided by cm cubed end fraction equals 4 space cm cubed end style
Given that the mass of the gold piece in water =71.2 g
The apparent loss in weight of the gold in water will be equal to the weight of the water displaced by the gold piece.
Apparent loss of weight = 77.2g - 71.2 g =6 g
So the weight of water displaced by the gold piece = 6 g
Volume of water displaced =mass of water displaced / density of water
Volume of water displaced = 6g / 1 g/cm3 = 6 cm3
Volume of the gold piece = 4 cm3 and volume of water displaced by the gold piece= 6 cm3
As Volume of the gold piece is less than volume of water displaced by the gold piece we can say that the gold piece is hollow inside
Therefore the volume of hollow portion of the gold piece = volume of water displaced by the gold piece - Volume of the gold piece 
i.e. 6 cm3 - 4 cm3  = 2 cm3
Volume of hollow portion of gold = 2 cm3 .

Answered by Jyothi Nair | 23rd Nov, 2014, 08:35: PM