$\text{A small electric dipole }{\overrightarrow{p}}_0,\text{having a moment of inertia }I\text{ about its center, is kept at a distance}\mathrm{ r}\text{ from}$

$\text{the center of }\text{a spherical shell of radius}\mathrm{ R}.\text{The surface charge density}\mathrm{ \sigma }\text{ is uniformly distributed on the}$

$\text{spherical shell. }$ $\text{The dipole is initially oriented at a small angle}\mathrm{ \theta }\text{ as shown in the figure.}$

$\text{While staying at distance}r,\text{the dipole is free to rotate about its center.}$

If released from rest, then which of the following statement(s) is(are) correct?

[ε₀ is the permittivity of free space.]

Correct Options:
1. The dipole will undergo small oscillations at any finite value of $r>R$.
2. The dipole will undergo small oscillations with an angular frequency of $\sqrt{\frac{\sigma p_0}{100{\epsilon }_{0}I}}$ at $r=10R$.

Detailed Step-by-Step Explanation:

Step 1: Electric Field due to the Uniformly Charged Spherical Shell
As shown in the figure, a spherical shell of radius $R$ has a uniform surface charge density $\sigma$.
The total charge $Q$ on the shell is given by:

$Q=\sigma ·4\pi R^2$

By Gauss's Law, the electric field $\overrightarrow{E}$ at a distance $r$ from the center of the spherical shell depends on whether the point is inside or outside the shell:
1. Inside the shell ($r<R$): The electric field is zero ($E=0$). Hence, no torque acts on the dipole inside the shell, so oscillations do not occur for $r<R$.
2. Outside the shell ($r>R$): The electric field is radially outwards (assuming positive charge) and has a magnitude of:

$E=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r^2}=\frac{\sigma ·4\pi R^2}{4\pi {\epsilon }_0{r}^2}=\frac{\sigma R^2}{{\epsilon }_0{r}^2}$

Step 2: Restoring Torque on the Dipole
The electric field vector $\overrightarrow{E}$ points along the radial line (dashed line connecting the center of the sphere to the center of the dipole).
The dipole moment vector ${\overrightarrow{p}}_0$ makes a small angle $\theta$ with respect to the radial axis.
The magnitude of the restoring torque $\tau$ acting on the dipole is given by:

$\tau =-p_{0}E\sin \theta$

Since $\theta$ is a small angle, we can use the small-angle approximation $\sin \theta \approx \theta$:

$\tau \approx -p_{0}E\theta$

Because the torque acts to decrease $\theta$ and restore the dipole back towards its radial alignment, the dipole will perform simple harmonic oscillations for any finite distance $r>R$.

Step 3: Calculating Angular Frequency of Oscillations
Using Newton's second law for rotation, $\tau =I\alpha =I\frac{d^2\theta }{d{t}^2}$:

$I\frac{d^2\theta }{d{t}^2}=-p_{0}E\theta$

$\frac{d^2\theta }{d{t}^2}+\left(\frac{p_{0}E}{I}\right)\theta =0$

Comparing this with the standard Simple Harmonic Motion (SHM) equation $\frac{d^2\theta }{d{t}^2}+{\omega }^2\theta =0$, the angular frequency $\omega$ is:

$\omega =\sqrt{\frac{p_{0}E}{I}}=\sqrt{\frac{p_0\sigma R^2}{{\epsilon }_{0}I{r}^2}}$

Step 4: Evaluating Angular Frequency at Specific Positions
Now, let's substitute $r=10R$ into the expression for $\omega$:

$\omega =\sqrt{\frac{p_0\sigma R^2}{{\epsilon }_{0}I{\left(10R\right)}^2}}=\sqrt{\frac{\sigma p_0{R}^2}{100{\epsilon }_{0}I{R}^2}}=\sqrt{\frac{\sigma p_0}{100{\epsilon }_{0}I}}$

Thus, at $r=10R$, the dipole undergoes small oscillations with an angular frequency of $\sqrt{\frac{\sigma p_0}{100{\epsilon }_{0}I}}$.

Hence, the correct options are:
- The dipole will undergo small oscillations at any finite value of $r>R$.
- The dipole will undergo small oscillations with an angular frequency of $\sqrt{\frac{\sigma p_0}{100{\epsilon }_{0}I}}$ at $r=10R$.