In question `3` of the same exercise, the radius of the sphere is `a`. The length of the string is also `a`. Find tension in the string.

Explanation

To find the tension in the string when the radius of the sphere is \( a \) and the length of the string is also \( a \), we can follow these steps:

### Step 1: Understand the Setup
We have a sphere with radius \( a \) and a string of length \( a \) attached to it. The sphere is in a position where the string makes an angle of \( 30^\circ \) with the vertical.

### Step 2: Identify Forces Acting on the Sphere
The forces acting on the sphere are:
- The gravitational force \( mg \) acting downward.
- The tension \( T \) in the string acting at an angle of \( 30^\circ \) to the vertical.

### Step 3: Resolve the Tension into Components
The tension \( T \) can be resolved into two components:
- The vertical component \( T \cos(30^\circ) \), which acts upward.
- The horizontal component \( T \sin(30^\circ) \), which acts towards the wall.

### Step 4: Apply Equilibrium Conditions
Since the sphere is at rest, the forces must balance:
1. In the vertical direction:
\[ T \cos(30^\circ) = mg \]
2. In the horizontal direction:
\[ T \sin(30^\circ) = N \]
where \( N \) is the normal force exerted by the wall on the sphere.

### Step 5: Solve for Tension
From the vertical force balance, we can express \( T \):
\[ T = \frac{mg}{\cos(30^\circ)} \]
We know that \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \), so:
\[ T = \frac{mg}{\frac{\sqrt{3}}{2}} = \frac{2mg}{\sqrt{3}} \]

### Final Answer
Thus, the tension in the string is:
\[ T = \frac{2mg}{\sqrt{3}} \]