(a) A ball, suspended by a thread, swings in a vertical plane so that its acceleration in the extreme and the lowest positions are equal. The angle `θ` of thread deflection in the extreme position will be
( b) The kinetic energy of a particle moving in a circle of radius R depens on the distance covered as `T = as^(2)` , where 'a' is a constant. Find the force acting on the particle as a function of s.

Answer

### Step-by-Step Solution

#### Part (a)

1. **Understanding the Problem**:
- A ball is suspended by a thread and swings in a vertical plane. The acceleration at the extreme position and the lowest position is equal. We need to find the angle \( \theta \) of the thread deflection in the extreme position.

2. **Acceleration at the Extreme Position**:
- At the extreme position, the tangential acceleration \( a_t \) is given by:
\[ a_t = g \sin \theta \]
- At the lowest position, the centripetal acceleration \( a_c \) is given by:
\[ a_c = \frac{v^2}{L} \]
- Since the acceleration in both positions is equal, we can set them equal to each other:
\[ g \sin \theta = \frac{v^2}{L} \]

3. **Using Energy Conservation**:
- The potential energy at the extreme position is converted into kinetic energy at the lowest position. The height \( h \) can be expressed as:
\[ h = L - L \cos \theta = L(1 - \cos \theta) \]
- The potential energy at the extreme position is:
\[ PE = mg h = mg L(1 - \cos \theta) \]
- The kinetic energy at the lowest position is:
\[ KE = \frac{1}{2} mv^2 \]
- By conservation of energy:
\[ mg L(1 - \cos \theta) = \frac{1}{2} mv^2 \]
- Canceling \( m \) from both sides gives:
\[ g L(1 - \cos \theta) = \frac{1}{2} v^2 \]

4. **Substituting for \( v^2 \)**:
- From the earlier equation \( g \sin \theta = \frac{v^2}{L} \), we can express \( v^2 \) as:
\[ v^2 = g L \sin \theta \]
- Substituting this into the energy equation gives:
\[ g L(1 - \cos \theta) = \frac{1}{2} g L \sin \theta \]

5. **Simplifying the Equation**:
- Dividing both sides by \( gL \) (assuming \( gL \neq 0 \)):
\[ 1 - \cos \theta = \frac{1}{2} \sin \theta \]
- Rearranging gives:
\[ 2(1 - \cos \theta) = \sin \theta \]

6. **Using Trigonometric Identities**:
- We can use the identity \( \sin^2 \theta + \cos^2 \theta = 1 \) to express \( \sin \theta \) in terms of \( \cos \theta \):
\[ 2 - 2\cos \theta = \sqrt{1 - \cos^2 \theta} \]
- Squaring both sides and simplifying will lead to the final expression for \( \theta \).

7. **Final Result**:
- After solving the trigonometric equation, we find:
\[ \theta = 2 \tan^{-1}(1/2) \]

#### Part (b)

1. **Understanding the Problem**:
- The kinetic energy \( T \) of a particle moving in a circle of radius \( R \) depends on the distance \( s \) covered as \( T = as^2 \), where \( a \) is a constant. We need to find the force acting on the particle as a function of \( s \).

2. **Relating Kinetic Energy to Velocity**:
- The kinetic energy is given by:
\[ T = \frac{1}{2} mv^2 \]
- Setting this equal to the given expression:
\[ \frac{1}{2} mv^2 = as^2 \]
- Rearranging gives:
\[ mv^2 = 2as^2 \]

3. **Finding Velocity**:
- From the above equation, we can express \( v^2 \):
\[ v^2 = \frac{2as^2}{m} \]

4. **Finding the Centripetal Force**:
- The centripetal force \( F_c \) required to keep the particle moving in a circle is given by:
\[ F_c = \frac{mv^2}{R} \]
- Substituting \( v^2 \):
\[ F_c = \frac{m(2as^2/m)}{R} = \frac{2as^2}{R} \]

5. **Finding the Total Force**:
- The total force acting on the particle is the sum of the centripetal force and any tangential force. Since we are only considering the centripetal force in this case, we have:
\[ F = \frac{2as^2}{R} \]

6. **Final Result**:
- The force acting on the particle as a function of \( s \) is:
\[ F(s) = \frac{2as^2}{R} \]