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, as shown in the figure. 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 ____

Calculate the moment of inertia (I) of the system:
The system consists of two point masses attached to the ends of a massless rod. The moment of inertia for a point mass is given by $I = m r^{2}$ , where $m$ is the mass and $r$ is the distance from the axis of rotation.
- Mass $m_{1} = 20$ gm $= 0.020$ kg
- Mass $m_{2} = 30$ gm $= 0.030$ kg
- Length of the rod $L = 10$ cm $= 0.10$ m
Since the rod is massless, the total moment of inertia is the sum of the moments of inertia of the two masses:
$I = m_{1} {(\frac{L}{2})}^{2} + m_{2} {(\frac{L}{2})}^{2}$ $I = (0.020) {(\frac{0.10}{2})}^{2} + (0.030) {(\frac{0.10}{2})}^{2}$ $I = (0.020) {(0.05)}^{2} + (0.030) {(0.05)}^{2}$ $I = (0.020) (0.0025) + (0.030) (0.0025)$ $I = 0.00005 + 0.000075$ $I = 0.000125 kg \cdot m^{2}$
Use the torsional constant (κ) to find the angular frequency (ω):
The angular frequency of a torsional pendulum is given by:
$ω = \sqrt{\frac{κ}{I}}$
where $κ = 1.2 \times 1 0^{- 8} N \cdot m \cdot {rad}^{- 1}$ .
Substituting the values:
$ω = \sqrt{\frac{1.2 \times 1 0^{- 8}}{0.000125}}$ $ω = \sqrt{\frac{1.2 \times 1 0^{- 8}}{1.25 \times 1 0^{- 4}}}$ $ω = \sqrt{9.6 \times 1 0^{- 5}}$ $ω = \sqrt{96 \times 1 0^{- 6}}$ $ω = \sqrt{96} \times 1 0^{- 3}$ $ω \approx 9.80 \times 1 0^{- 3} rad/s$
Determine the value of $n$ :
The angular frequency is given as $n \times 1 0^{- 3} rad/s$ . From the calculation above:
$n \approx 9.80$

Therefore, the value of $n$ is approximately 9.80.

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