A solid sphere of density `sigma` with one -third of tis volume submerged . When the sphere is pressed down slightly and released , it executes S.H.M of time period T. Then (viscous effect is ignored)

Options

A. `p=(sigma)/(3)`
B. `p=3sigma`
C. `(4pi)/(3)sqrt((R )/(g))`
D. `(3pi)/(2)sqrt((R )/(g))`

Explanation

To solve the problem of a solid sphere of density \( \sigma \) with one-third of its volume submerged, executing simple harmonic motion (SHM) when pressed down and released, we will follow these steps:

### Step 1: Determine the mass of the sphere
The volume \( V \) of a solid sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
The mass \( m \) of the sphere can be calculated using the formula:
\[ m = V \cdot \sigma = \frac{4}{3} \pi r^3 \cdot \sigma \]

### Step 2: Calculate the volume submerged in the liquid
Since one-third of the sphere's volume is submerged, the volume \( V_s \) submerged is:
\[ V_s = \frac{1}{3} V = \frac{1}{3} \cdot \frac{4}{3} \pi r^3 = \frac{4}{9} \pi r^3 \]

### Step 3: Calculate the buoyant force acting on the sphere
The buoyant force \( F_b \) acting on the sphere is equal to the weight of the liquid displaced, which can be calculated as:
\[ F_b = V_s \cdot \rho \cdot g = \frac{4}{9} \pi r^3 \cdot \rho \cdot g \]
where \( \rho \) is the density of the liquid.

### Step 4: Establish the relationship between densities
For the sphere to float, the buoyant force must equal the weight of the sphere:
\[ F_b = m \cdot g \]
Substituting the expressions we have:
\[ \frac{4}{9} \pi r^3 \cdot \rho \cdot g = \frac{4}{3} \pi r^3 \cdot \sigma \cdot g \]
Cancelling common terms gives:
\[ \frac{4}{9} \rho = \frac{4}{3} \sigma \]
From this, we can derive:
\[ \rho = 3 \sigma \]

### Step 5: Determine the restoring force when the sphere is displaced
When the sphere is pressed down by a distance \( x \), the volume of liquid displaced is:
\[ V_d = A \cdot x \]
where \( A \) is the cross-sectional area of the sphere at the water surface, given by:
\[ A = \pi r^2 \]
Thus, the volume displaced is:
\[ V_d = \pi r^2 x \]
The buoyant force when displaced is:
\[ F_b' = V_d \cdot \rho \cdot g = \pi r^2 x \cdot (3\sigma) \cdot g \]
The restoring force \( F \) acting on the sphere is:
\[ F = -F_b' = -\pi r^2 x \cdot (3\sigma) \cdot g \]

### Step 6: Relate the restoring force to acceleration
Using Newton's second law, \( F = m a \):
\[ -\pi r^2 x \cdot (3\sigma) \cdot g = m \cdot a \]
Substituting \( m = \frac{4}{3} \pi r^3 \sigma \):
\[ -\pi r^2 x \cdot (3\sigma) \cdot g = \left(\frac{4}{3} \pi r^3 \sigma\right) a \]
Cancelling \( \pi \sigma \) from both sides:
\[ -3 r^2 g x = \frac{4}{3} r^3 a \]
Thus,
\[ a = -\frac{9g}{4r} x \]

### Step 7: Determine the angular frequency and time period
The equation for SHM is \( a = -\omega^2 x \), where:
\[ \omega^2 = \frac{9g}{4r} \]
Taking the square root gives:
\[ \omega = \frac{3}{2} \sqrt{\frac{g}{r}} \]
The time period \( T \) of SHM is given by:
\[ T = \frac{2\pi}{\omega} = \frac{2\pi}{\frac{3}{2} \sqrt{\frac{g}{r}}} = \frac{4\pi}{3} \sqrt{\frac{r}{g}} \]

### Final Result
The time period \( T \) of the simple harmonic motion of the sphere is:
\[ T = \frac{4\pi}{3} \sqrt{\frac{r}{g}} \]