Question

An aluminium wire is ${{m}_{cork}}$. The densities ${{\rho }_{cork}},{{\rho }_{al}} and {{\rho }_{w}}$ of cork, aluminium and water are $0.5\times {{10}^{3}}kg/{{m}^{3}}$,$2.7\times {{10}^{3}}kg/{{m}^{3}}$ and $1\times {{10}^{8}}kg/{{m}^{3}}$ respectively. Determine the minimum mass ${{m}_{al}}$ of the wire that should be wound on the cork so that the cork with wire is completely submerged in water.

Answer 1

Hint: This problem can be solved by the basic principles of buoyancy by checking the density of substance with respect to water. In this section we will discuss the conditions for wire to be submerged in water and calculate the mass of aluminium.

Formula used:
Density
$\Rightarrow \rho =\dfrac{m}{V}$

Complete Step by step solution:
Buoyancy is a type of force which causes objects to float on the surface of liquid. It is the force exerted on an object that is partly or fully submerged in liquid and can also be termed as buoyant force, SI unit is Newton (N). Force of buoyancy is a vertical force, so the centre of buoyancy is majorly at the position of centre of gravity .

We know the condition of complete submergence of the body in liquid (here water) can be given as,
$\Rightarrow {{\rho }_{W}}\le {{\rho }_{object}}$
But,
 $\Rightarrow \rho =\dfrac{m}{V}$

Here, m (mass of object) , V(volume of object) so,
$\Rightarrow {{\rho }_{W}}\le \dfrac{M}{V}$
$\Rightarrow {{\rho }_{W}}\le \dfrac{M}{V}\Rightarrow {{\rho }_{W}}V\le M$…………(1)
Here M (mass of the body), V (total volume of body), so according to the problem,
$\Rightarrow M={{m}_{al}}+{{m}_{cork}}$
$\Rightarrow V=\dfrac{{{m}_{cork}}}{{{\rho }_{cork}}}+\dfrac{{{m}_{al}}}{{{\rho }_{al}}}$

Now, putting these value in eq (1), we have
$\Rightarrow {{\rho }_{W}}\left( \dfrac{{{m}_{cork}}}{{{\rho }_{cork}}}+\dfrac{{{m}_{al}}}{{{\rho }_{al}}} \right)\le {{m}_{al}}+{{m}_{cork}}$
$\Rightarrow {{m}_{al}}=\dfrac{{{\rho }_{al}}({{\rho }_{W}}-{{\rho }_{cork}})}{({{\rho }_{al}}-{{\rho }_{W}}){{\rho }_{cork}}}{{m}_{cork}}$ , for boundary condition,
$\Rightarrow {{m}_{al}}=\dfrac{2.7\times {{10}^{3}}(1\times {{10}^{8}}-0.5\times {{10}^{3}})}{(2.7\times {{10}^{3}}-1\times {{10}^{8}})0.5\times {{10}^{3}}}{{m}_{cork}}\simeq 1.6{{m}_{cork}}$
$\Rightarrow {{m}_{al}}=\dfrac{2.7\times {{10}^{3}}(1\times {{10}^{8}}-0.5\times {{10}^{3}})}{(2.7\times {{10}^{3}}-1\times {{10}^{8}})0.5\times {{10}^{3}}}{{m}_{cork}}\simeq 1.6{{m}_{cork}}$

$\therefore $the minimum amount of mass of aluminium required to completely submerged wire into the water is 1.6 times the mass of cork used.

Note:
Density of material is defined as mass per unit volume, it’s a measurement of how tightly the matter is packed together. In fluid, relative density is defined as the ratio of the density of the substance to the density of the water and also known as specific gravity of the substance. When an object is fully or partially submerged in water it experiences an upward force which is known as Upthrust.