Question

A block of wood weighs $12\,{\text{kg}}$ and has a relative density 0.6. It is to be in water with 0.9 of its volume immersed. The weight of a metal which is needed (a) if the metal is on the top of wood. (b) if the metal is attached below the wood [Relative density of metal=14]
A. $6\,{\text{kg}}$, $7.5\,{\text{kg}}$
B. $6\,{\text{kg}}$, $6.5\,{\text{kg}}$
C. $6.5\,{\text{kg}}$, $7.5\,{\text{kg}}$
D. $7\,{\text{kg}}$, $8.5\,{\text{kg}}$

Answer 1

Use the equation for the density of the object. Determine the volume of the wood and metal needed using the equation of density. Using Archimedes’ principle, determine the weight of the metal in both the cases mentioned in the question.

Formulae used:
The density of an object is given by
$\rho = \dfrac{m}{V}$ …… (1)
Here, $\rho$ is the density of the object, $m$ is the mass of the object and $V$ is the volume of the object.

Complete step by step answer:
It is given that the block of wood weighs $12\,{\text{kg}}$ and has a relative density 0.6.
${m_w} = 12\,{\text{kg}}$
${\rho _w} = 0.6$
We can determine the 0.9 of the volume of the liquid using equation (1).
Rewrite equation (1) for the relative density of the wood.
${\rho _w} = \dfrac{{{m_w}}}{{{V_w}}}$
Rearrange the above equation for ${V_w}$.
${V_w} = \dfrac{{{m_w}}}{{{\rho _w}}}$
Substitute $12\,{\text{kg}}$ for ${m_w}$ and $0.6$ for ${\rho _w}$ in the above equation.
${V_w} = \dfrac{{12\,{\text{kg}}}}{{0.6}}$
$\Rightarrow {V_w} = 20\,{\text{kg/}}{{\text{m}}^3}$
Hence, the total volume of the wood is $20\,{\text{kg/}}{{\text{m}}^3}$.

But only 0.9 of the total volume of the wood is immersed in water. Hence, the volume $V$ of the wood immersed in the water is $\left( {0.9 \times 20} \right)\,{\text{kg/}}{{\text{m}}^3}$.
$V = \left( {0.9 \times 20} \right)\,{\text{kg/}}{{\text{m}}^3}$

Let us now determine the volume of the metal weighing ${m_m}$.
Rewrite equation (1) for the relative density of the wood.
${\rho _m} = \dfrac{{{m_m}}}{{{V_m}}}$
Rearrange the above equation for ${V_m}$.
${V_m} = \dfrac{{{m_m}}}{{{\rho _m}}}$
Substitute $14$ for ${\rho _w}$ in the above equation.
${V_m} = \dfrac{{{m_m}}}{{14}}$

Hence, the total volume of the metal is $\dfrac{{{m_m}}}{{14}}$.

(a)The buoyant force is given by
${\text{Buoyant force}} = \rho Vg$
When the metal is placed on the top of the wood, the volume of the water displaced is equal to the volume of the wood immersed in the water.

According to the Archimedes’ principle, the weight of the metal and wood is equal to the buoyant force on the top of the wood immersed in the water.
$\left( {{m_w} + {m_m}} \right)g = \rho Vg$
$\Rightarrow {m_w} + {m_m} = \rho V$
Rearrange the above equation for the mass ${m_m}$ of the metal.
$\Rightarrow {m_m} = \rho V - {m_w}$

Substitute $1\,{\text{kg/}}{{\text{m}}^3}$ for $\rho$, $\left( {0.9 \times 20} \right)\,{\text{kg/}}{{\text{m}}^3}$ for $V$ and $12\,{\text{kg}}$ for ${m_w}$ in the above equation.
$\Rightarrow {m_m} = \left[ {1\,{\text{kg/}}{{\text{m}}^3}} \right]\left[ {\left( {0.9 \times 20} \right)\,{\text{kg/}}{{\text{m}}^3}} \right] - \left( {12\,{\text{kg}}} \right)$
$\Rightarrow {m_m} = 6\,{\text{kg}}$

Hence, the weight of the metal when metal is placed on top of wood is $6\,{\text{kg}}$.

(b)When the metal is attached below the wood, the volume of the water displaced is equal to the volume of the wood immersed in the water and the volume of the metal.

According to the Archimedes’ principle, the weight of the metal and wood is equal to the buoyant force on the top of the wood immersed in the water.
$\left( {{m_w} + {m_m}} \right)g = \rho \left( {V + {V_m}} \right)g$
$\Rightarrow {m_w} + {m_m} = \rho \left( {V + {V_m}} \right)$

Rearrange the above equation for the mass ${m_m}$ of the metal.
$\Rightarrow {m_m} = \rho \left( {V + {V_m}} \right) - {m_w}$

Substitute $1\,{\text{kg/}}{{\text{m}}^3}$ for $\rho$, $\dfrac{{{m_m}}}{{14}}$ for ${V_m}$, $\left( {0.9 \times 20} \right)\,{\text{kg/}}{{\text{m}}^3}$ for $V$ and $12\,{\text{kg}}$ for ${m_w}$ in the above equation.
$\Rightarrow {m_m} = \left[ {1\,{\text{kg/}}{{\text{m}}^3}} \right]\left[ {\left( {0.9 \times 20} \right)\,{\text{kg/}}{{\text{m}}^3} + \dfrac{{{m_m}}}{{14}}} \right] - \left( {12\,{\text{kg}}} \right)$
$\Rightarrow {m_m} - \dfrac{{{m_m}}}{{14}} = \left( {0.9 \times 20} \right)\,{\text{kg/}}{{\text{m}}^3} - \left( {12\,{\text{kg}}} \right)$
$\Rightarrow \dfrac{{13{m_m}}}{{14}} = \left( {0.9 \times 20} \right)\,{\text{kg/}}{{\text{m}}^3} - \left( {12\,{\text{kg}}} \right)$
$\Rightarrow {m_m} = 6.5\,{\text{kg}}$
Hence, the mass of metal when it is attached below the wood is $6.5\,{\text{kg}}$.

Therefore, the weight of the metal should be $6\,{\text{kg}}$ and $6.5\,{\text{kg}}$.

So, the correct answer is “Option B”.

Note:
The students may assume that the mass of the metal obtained by using Archimedes’ principle should be multiplied by the acceleration due to gravity as the weight of the metal needed is to be determined. But the weight mentioned in the question is the mass of the metal as the options given has the unit of mass.