Question

Two pieces of metals are suspended from the arms of a balance and arc found to be in equilibrium when kept immersed in water. The mass of one piece is $32\;{\rm{gm}}$ and its density is $8\;{\rm{gm}}{\rm{.c}}{{\rm{m}}^{{\rm{ - 3}}}}$, the density of the other is $5\;{\rm{gm}}{\rm{.c}}{{\rm{m}}^{{\rm{ - 3}}}}$, the mass of the other is,
(1) $28\;{\rm{gm}}$
(2) $35\;{\rm{gm}}$
(3) $21\,{\rm{gm}}$
(4) $33.6\;{\rm{gm}}$

Answer 1

Hint: Here, in this question we will use the equilibrium condition for the apparent weight of two metals. We also find the volume of the one metal as it has both the mass as well as density. We divide the mass of the first piece of metal to the density of the metal that gives us the volume of the first metal.

Complete step by step answer:
Given: The mass of the first piece of the metal is ${m_1} = 32\;{\rm{gm}}$, the density of the first piece of the metal is ${\rho _1} = 8\;{\rm{gm}}{\rm{.c}}{{\rm{m}}^{{\rm{ - 3}}}}$, the density of the other piece of the metal is ${\rho _2} = 5\;{\rm{gm}}{\rm{.c}}{{\rm{m}}^{{\rm{ - 3}}}}$.
We first write the formula for volume to find the volume of the first piece of the metal.
${V_1} = \dfrac{{{m_1}}}{{{\rho _1}}}$
We have the relation of volume and just need to substitute the given values in the above relation.
$\begin{array}{l}
{V_1} = \dfrac{{32}}{8}\\
{V_1} = 4\;{\rm{c}}{{\rm{m}}^{\rm{3}}}
\end{array}$
Now, we apply the equilibrium condition which is,
Apparent weight of first piece of metal is equal to apparent weight of second piece of metal
${m_1} - {\rho _w}\dfrac{{{m_1}}}{{{\rho _1}}} = {m_2} - {\rho _w}\dfrac{{{m_2}}}{{{\rho _2}}}$
Here, ${\rho _w}$ is the density of water which is equal to $1\;{\rm{g/c}}{{\rm{m}}^{\rm{3}}}$
We substitute the values in above relation,
$
32 - 1 \times \dfrac{{32}}{8} = {m_2} - 1 \times \dfrac{{{m_2}}}{5}\\
\implies 28 = {m_2} - \dfrac{{{m_2}}}{5}\\
\implies \dfrac{4}{5}{m_2} = 28\\
\therefore {m_2} = 35\,{\rm{g}}
$
The mass of the second piece of the metal is $35\;{\rm{g}}$.

So, the correct answer is “Option 2”.

Note:
An object immersed in water partially or fully always experiences an upward force or upthrust that pushes it in an upward direction. This force is termed a buoyancy force. The weight of the water displaced by the immersed object is equal to the buoyant force acting on the object.