A

Step 1: Analyze the System
We have a conical pendulum where a sphere of mass 200 gm (0.2 kg) is swinging in a horizontal circle with a radius of 50 cm (0.5 m). The length of the string is 130 cm (1.3 m).

Step 2: Setup the Geometry
From the setup, let's denote:
- L = length of the string = 1.3 m
- r = radius of the horizontal circle = 0.5 m
- h = vertical height of the mass from the ceiling (the vertical component of the string)

Using the right triangle formed by the string, the height, and the radius, we have:
$$ h = \sqrt{L^2 - r^2} = \sqrt{(1.3)^2 - (0.5)^2} = \sqrt{1.69 - 0.25} = \sqrt{1.44} = 1.2 m $$

Step 3: Calculate the angle
We can find the angle θ that the string makes with the vertical using:
$$ \cos(\theta) = \frac{h}{L} = \frac{1.2}{1.3} = 0.923 $$
Hence,
$$ \theta = \cos^{-1}(0.923) \approx 22.6^{\circ} $$

Step 4: Forces Acting on Sphere
The forces acting on the sphere are:
- Weight (mg) acting downwards
- Tension (T) acting along the string

Resolving the tension into components:
- Vertical component: \( T \cos(\theta) = mg \)
- Horizontal component: \( T \sin(\theta) = \frac{mv^2}{r} \)

Step 5: Establish the relationship between tension and speed
From the vertical component, we have:
$$ T \cos(\theta) = mg \Rightarrow T = \frac{mg}{\cos(\theta)} $$
Substituting the values:
$$ m = 0.2 kg, \; g = 9.8 m/s^2 $$
$$ T = \frac{0.2 \cdot 9.8}{\cos(22.6^{\circ})} \approx \frac{1.96}{0.923} \approx 2.12 N $$

Step 6: Relation of speed and tension in horizontal circle
From the horizontal component:
$$ T \sin(\theta) = \frac{mv^2}{r} $$
Substitute the expression for T:
$$ \frac{mg \sin(\theta)}{\cos(\theta)} = \frac{mv^2}{r} $$
Cancel m and rearrange:
$$ v^2 = \frac{gr \tan(\theta)}{1} = g r \tan(\theta) $$
Calculating \( g \tan(\theta) \):
Substituting the values, we have:
$$ \tan(\theta) = \frac{\sin(22.6^{\circ})}{\cos(22.6^{\circ})} \approx 0.414 $$
$$ v^2 = (9.8)(0.5)(0.414) \approx 2.03 $$
Thus, $$ v \approx \sqrt{2.03} \approx 1.42 m/s $$

Step 7: Calculate the period
The period (T) of revolution for the conical pendulum is given by:
$$ T = \frac{2\pi r}{v} \approx \frac{2\pi (0.5)}{1.42} \approx \frac{3.14}{1.42} \approx 2.21 s $$

Final Answers
Thus, the period time of the conical pendulum is approximately 2.21 seconds and the tension in the string is approximately 2.12 N.