A piece of gold weighs $10 g$ in air and $9 g$ in water. What is the volume of cavity?
(Density of gold $= 19.3 g c m^{- 3}$ )

0.182 c c

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.282 c c

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.382 c c

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.482 c c

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is B $0.482 c c$
By using Archimedes principle which states that the buoyant force is equal to the weight of the body.

$m - ρ (V_{g} + V_{c}) = 9 g$

By substituting the value of $V_{g}$ from $m = ρ V_{g}$ in above equation we get,

$m - ρ_{w} \frac{m}{ρ_{g}} - ρ_{w} V_{c} = 9 g$

On solving the above equation we get the value of

$V_{c} = 0.48 c m^{3}$

Suggest Corrections

Similar questions

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 {c m}^{- 3}

)

A piece of gold (p=19.3 g cm^-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 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

Q. A piece of gold