In an experiment with a beam balance on unknown mass $m$ is balanced by two known mass $m$ is balanced by two known masses of $16\;{\rm{kg}}$ and $4\;{\rm{kg}}$ as shown in figure. Find the value of unknown mass $m$ in $kg$.

Hint: Here, we will use the principle of moments. We will apply this principle for the balance in both the figures and solve the obtained equations.

Complete step by step answer:The moment of force about a point of rotation can be written as
$\begin{array}{c}
\tau = \overrightarrow r \times \overrightarrow F \\
 = rF\sin \theta
\end{array}$

where $r$ is the perpendicular distance from the line of application of force $F$ to the point of rotation and $\theta $ is the angle between $\overrightarrow r $ and $\overrightarrow F $.

In the given spring balance, two masses are hanged from the ends of the bar. Moment of force or torque acts about the point of rotation due to each mass. In the case of the spring balance, we see that $\theta = 90^\circ $. Hence, the torque due to a particular mass $M$ can be written as
$
\begin{align*}
\tau &= rF\sin 90^\circ \\
 \Rightarrow &rF
\end{align*}
$
Since $F = Mg$,

$\tau = Mgr$

In the first figure, let ${m_1} = 16\;{\rm{kg}}$. Then, we say that the spring balance is balanced by two masses ${m_1}$ and $m$. The moment due to one mass acts in the clockwise direction and the moment due to the other mass acts in the anticlockwise direction.

From the principle of moments, the moment of force acting in the clockwise direction is equal to the moment acting in the anticlockwise direction for a balanced body. Hence, we write

${\tau _m} = {\tau _1}$

Here ${\tau _m}$ is the moment of force due to mass $m$ and ${\tau _1}$ is the moment of force due to mass ${m_1}$. So, similar to the torque equation $\tau = Mgr$, we write

$mg{l_2} = {m_1}g{l_1}$
$m{l_2} = {m_1}{l_1}$

Since ${m_1} = 16\;{\rm{kg}}$, we get
$m{l_2} = 16{l_1}$ ……(1)
Similarly, for the second figure, let ${m_2} = 4\;{\rm{kg}}$. Applying the principle of moments, we get
$m{l_1} = {m_2}{l_2}$
Since ${m_2} = 4\;{\rm{kg}}$,
$m{l_1} = 4{l_2}$
$\dfrac{{{l_1}}}{{{l_2}}} = \dfrac{4}{m}$ ……(2)
Now, we will divide equation (1) by (2). Hence, we get
m=8

Therefore, the value of mass $m$ is $8\;{\rm{kg}}$.

Note:If ${m_1}$ and ${m_2}$ are two masses hanging from two points of a balanced body and ${r_1}$ and ${r_2}$ are the distances of the masses from the axis of rotation of the body, then ${m_1}{r_1} = {m_2}{r_2}$. We can directly apply this equation in problems which mention that the body is balanced.