Question

A piece of pure gold (density = 19.3 gcm^-3 ) 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

• A. 3 cm³
• B. 2 cm³
• C. 4 cm³
• D. 1 cm³

Answer 1

Answer: (B)

Step 1: Given that:

The density of pure gold = $19.3gc{m}^{-3}$

Weight of gold in air($w_g$) = $77.2g$

Weight of gold in water($w_w$) = $71.2g$

Step 2: Calculation of volume of gold in air and water:

Volume(V) of a body = $\frac{Massofthebody}{Densityofthebody}$

Now,

The volume of the gold in the air will be;

$V_{air}=\frac{77.2g}{19.3gc{m}^{-3}}$

$V_{air}=4c{m}^3$

The apparent loss in weight of the gold = Weight of gold in the air - Weight of gold in the water

= $77.2g-71.2g$

= $6g$

Apparent loss in weight = Volume of the body in fluid $\times$ density of fluid

Here, density of fluid(water) = $1gc{m}^{-3}$

$6g=V_{water}\times 1gc{m}^{-3}$

$V_{water}=\frac{6}{1}$

$V_{water}=6c{m}^3$

Thus,

The volume of hollow portion in gold = Difference in the volume of gold in air and in water

= $6c{m}^3-4c{m}^3$

= $2c{m}^3$

Thus,

The volume of the hollow portion of gold will be $2c{m}^3$

Hence,

Option A) $2c{m}^3$ is the correct option.