determine the period of the oscillation of mercury of mass m =200gm poured into a bent tube whose right forms angle =30 degrees with the vertical. the cross sectional area is 0.50 cm2.the viscosity of mercury to be neglected

To determine the period of oscillation of mercury in the bent tube, we treat the system as a simple harmonic oscillator governed by hydrostatic principles.

### Given data:
- Mass of mercury: \( m = 200 \) g = \( 0.200 \) kg
- Angle of the inclined tube with the vertical: \( \theta = 30^\circ \)
- Cross-sectional area of the tube: \( A = 0.50 \) cm² = \( 5.0 \times 10^{-8} \) m²
- Acceleration due to gravity: \( g = 9.81 \) m/s²

### Step 1: Determine the effective length of oscillating mercury
The mercury oscillates in the bent U-tube, moving up and down in the inclined arm. The length of the oscillating mercury column is related to the vertical displacement.

Let \( x \) be the displacement of the mercury column along the inclined arm. The corresponding vertical displacement is:
\[
h = x \sin\theta
\]
The restoring force arises due to the difference in hydrostatic pressure and is given by:

\[
F = -\rho A g h
\]

Since mass \( m \) of mercury in the column is:

\[
m = \rho A L
\]

where \( L \) is the length of the oscillating column.

Using Newton’s second law:

\[
m \frac{d^2 x}{dt^2} = -\rho A g x \sin\theta
\]

Substituting \( m = \rho A L \):

\[
\rho A L \frac{d^2 x}{dt^2} + \rho A g x \sin\theta = 0
\]

Canceling \( \rho A \):

\[
L \frac{d^2 x}{dt^2} + g x \sin\theta = 0
\]

Comparing with the standard simple harmonic motion equation:

\[
\frac{d^2 x}{dt^2} + \omega^2 x = 0
\]

where angular frequency \( \omega \) is:

\[
\omega^2 = \frac{g \sin\theta}{L}
\]

The period of oscillation is given by:

\[
T = 2\pi \sqrt{\frac{L}{g \sin\theta}}
\]

### Step 2: Determine the effective length \( L \)
The mass of mercury is related to its density \( \rho \):

\[
m = \rho A L
\]

For mercury, \( \rho = 13600 \) kg/m³, so:

\[
0.200 = (13600) (5.0 \times 10^{-8}) L
\]

Solving for \( L \):

\[
L = \frac{0.200}{(13600 \times 5.0 \times 10^{-8})}
\]

\[
L = \frac{0.200}{6.8 \times 10^{-4}}
\]

\[
L = 294.1 \text{ cm} = 2.941 \text{ m}
\]

### Step 3: Compute the period \( T \)
\[
T = 2\pi \sqrt{\frac{2.941}{(9.81 \times \sin 30^\circ)}}
\]

Since \( \sin 30^\circ = 0.5 \):

\[
T = 2\pi \sqrt{\frac{2.941}{(9.81 \times 0.5)}}
\]

\[
T = 2\pi \sqrt{\frac{2.941}{4.905}}
\]

\[
T = 2\pi \sqrt{0.6}
\]

Approximating \( \sqrt{0.6} \approx 0.775 \):

\[
T \approx 2\pi \times 0.775
\]

\[
T \approx 4.87 \text{ s}
\]

### Final Answer:
The period of oscillation of the mercury column is **4.87 seconds**.