Two point-like objects of masses 20gm and 30gm are fixed at the two...

Given:

${m}_{1} = 30gm$ and ${m}_{2} = 20gm$.

Moment of inertia about the axis of rotation is

$I = {m}_{1}{{r}_{1}}^{2} + {m}_{2}{{r}_{2}}^{2}$

X-coordinate of the centre of mass can be written as,

${x}_{CM} = \dfrac{{m}_{1}{x}_{1} + {m}_{2}{x}_{2}}{{m}_{1} + {m}_{2}} = \dfrac{0 + \left(20 \times {10}^{- 3}\right) \times \left(10 \times {10}^{- 2}\right)}{\left(30 \times {10}^{- 3}\right) + \left(20 \times {10}^{- 3}\right)} = 4 \times {10}^{- 2}m = 4cm = {r}_{1}$

Therefore,

And ${r}_{2} = \left(10 - 4\right)cm = 6cm$

$∴ I = \left(30 \times {10}^{- 3} \times 16 \times {10}^{- 4}\right) + \left(20 \times {10}^{- 3} \times 36 \times {10}^{- 4}\right)$

$⇒ I = 1200 \times {10}^{- 7}kg{m}^{2}$

If the system is rotated by small angle ‘$\theta$’, the restoring torque is ${τ}_{\left(R\right)} = - k\theta$

And $\dfrac{{d}^{2}\theta}{d{t}^{2}} = \dfrac{- k}{I} \cdot \theta = - {\omega}^{2}\theta = \dfrac{- 1 . 2 \times {10}^{- 8}}{1200 \times {10}^{- 7}} \cdot \theta$

$∴ {\omega}^{2} = {10}^{- 4}$

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

Therefore, $n = 10$.