Question

An iron anchor of density$\mathbf{7870}\mathbf{}\raisebox{1ex}{$\mathbf{kg}$}\!\left/ \!\raisebox{-1ex}{${\mathbf{m}}^{\mathbf{3}}$}\right.$appears $\mathbf{200}\mathbf{}\mathbf{N}$ lighter in water than in air. (a) What is the volume of the anchor? (b) How much does it weigh in air?

Short answer

a) The volume of the anchor is$0.0204{\mathrm{m}}^3$

b) Its weight in air is$1573\mathrm{N}$

Step-by-step solution

The given data

a) Density of iron anchor, ${\rho }_{Fe}=7870\mathrm{kg}/{\mathrm{m}}^3$

b) Buoyant force,$F_b=200\mathrm{N}$

Understanding the concept of Archimedes Principle

We can use the Archimedes principle to find the force of buoyancy in the water. Using this force, we can find the volume. Also, apparent weight is the difference between actual weight and buoyant force. Using this concept, it is possible to find the weight of the anchor.

Formula:

Buoyant force exerted by fluid on body, ${\mathrm{F}}_{\dot{\mathrm{i}}}={\mathrm{\rho }}_{\text{water}}\times \mathrm{g}\times \mathrm{V}$ (i)

The apparent weight of a body, ${\mathrm{W}}_{\text{app}}=\mathrm{W}-{\mathrm{F}}_{\mathrm{b}}$(ii)

The weight of a body, $\mathrm{W}=\mathrm{mg}$(iii)

a) Calculation of volume of the anchor

Comparing and substituting values of equations (i), (ii) & (iii) and using the given values, we get

$\mathrm{W}-{\mathrm{W}}_{\text{app}}={\mathrm{\rho }}_{\text{woter}}\times \mathrm{g}\times \mathrm{V}$

$\mathrm{mg}-{\mathrm{W}}_{\mathrm{app}}={\mathrm{\rho }}_{\text{water}}\times \mathrm{g}\times \mathrm{V}$

$\frac{200}{{\mathrm{\rho }}_{\text{water}}\times \mathrm{g}}=\mathrm{V}$

$\frac{200}{\left(1000\right)(9.8)}=\mathrm{V}$

$\mathrm{V}=0.0204{\mathrm{m}}^3$

Hence, the volume of iron anchor is found to be$0.0204{\mathrm{m}}^3$

b) Calculation of weight of anchor in air

Density can be written as

$\mathrm{Density}=\frac{\mathrm{Mass}}{\mathrm{Volume}}$

So, the mass of the anchor is$\mathrm{m}={\mathrm{\rho }}_{60}\times \mathrm{V}$

From equation (iii) and (1), we get

$\mathrm{W}={\mathrm{\rho }}_{\mathrm{g}}\times \mathrm{V}\times \mathrm{g}$

$=7870\times 0.0204\times 9.8$

$=1573.4\mathrm{N}$

Hence, the weight of anchor in air is$1573.4\mathrm{N}$