Two point-like objects of masses 20 gm and 30 gm are fixed at the two ends of a rigid massless rod of length 10 cm. This system is suspended vertically from a rigid ceiling using a thin wire attached to its center of mass. The resulting torsional pendulum undergoes small oscillations. The torsional constant of the wire is 1.2 × 10⁻⁸ N m rad⁻¹. The angular frequency of the oscillations in n × 10⁻³ rad s⁻¹. The value of n is  __________.

The correct answer is 100 (since 10 × 10 = 100).

Given Data:
Mass of the first object, $m_1=20\text{gm}=20\times {10}^{-3}\text{kg}$
Mass of the second object, $m_2=30\text{gm}=30\times {10}^{-3}\text{kg}$
Length of the rigid massless rod, $L=10\text{cm}=0.1\text{m}$
Torsional constant of the wire, $k=1.2\times {10}^{-8}\text{N m rad}{}^{-1}$

As shown in the image below, the system consists of two masses attached to the ends of a horizontal rod suspended vertically by a thin wire at its center of mass.

Step 1: Find the distance of each mass from the center of mass.
Let $r_1$ be the distance of mass $m_1$ from the center of mass, and $r_2$ be the distance of mass $m_2$ from the center of mass.
Using the definition of center of mass:

$r_1=\frac{m_2}{m_1+m_2}L=\frac{30}{20+30}\times 10=6\text{cm}=0.06\text{m}$

$r_2=\frac{m_1}{m_1+m_2}L=\frac{20}{20+30}\times 10=4\text{cm}=0.04\text{m}$

Step 2: Calculate the moment of inertia ($I$) about the axis of rotation passing through the center of mass.
The moment of inertia of the system about the suspension wire (passing through the center of mass) is given by:

$I=m_1{r}_1^2+m_2{r}_2^2$

Alternatively, using reduced mass $\mu$:

$\mu =\frac{m_1{m}_2}{m_1+m_2}=\frac{20\times 30}{20+30}=12\text{gm}=12\times {10}^{-3}\text{kg}$

Thus, the moment of inertia is:

$I=\mu L^2=(12\times {10}^{-3}\text{kg})\times (0.1\text{m}{)}^2=12\times {10}^{-5}\text{kg m}{}^2=1.2\times {10}^{-4}\text{kg m}{}^2$

Step 3: Calculate the angular frequency ($\omega$) of small torsional oscillations.
The formula for the angular frequency of a torsional pendulum is:

$\omega =\sqrt{\frac{k}{I}}$

Substitute the values of $k$ and $I$:

$\omega =\sqrt{\frac{1.2\times {10}^{-8}}{1.2\times {10}^{-4}}}=\sqrt{{10}^{-4}}={10}^{-2}rad\; s{}^{-1}$

Rewrite this in the form $n\times {10}^{-3}\text{rad s}{}^{-1}$:

$\omega =100\times {10}^{-3}\text{rad s}{}^{-1}$

Comparing this with $n\times {10}^{-3}$, we get:

$n=100$