A particle of mass m is attached to a string of length L and given velocity `sqrt(10hL)` in the horizontal direction at the lowest point. Find tension in the string when the particle is at (a) (i) lowest position (ii) highest position, (b) when the string makes an angle `60^(@)` with (i) lower vertical and (ii) upper vertical.

Options

A. 11mg , 5 mg , 9.5mg , 6.5 mg
B. 10mg , 5 mg , 9.5mg , 6.5 mg
C. 11mg , 5 mg , 9mg , 6.5 mg
D. 11mg , 5 mg , 9.5mg , 6 mg

Explanation

To solve the problem step by step, we will analyze the motion of the particle attached to the string at different positions and apply the principles of circular motion and energy conservation.

### Given:
- Mass of the particle = \( m \)
- Length of the string = \( L \)
- Initial velocity at the lowest point = \( v = \sqrt{10hL} \)

### (a) Tension in the string at different positions

#### (i) Tension at the Lowest Position

1. **Identify the Forces**: At the lowest position, the forces acting on the particle are:
- Tension \( T \) acting upwards.
- Weight \( mg \) acting downwards.

2. **Centripetal Force Requirement**: The net force providing the centripetal acceleration is given by:
\[ T - mg = \frac{mv^2}{L} \]

3. **Substituting Values**: Substitute \( v^2 = 10hL \) into the equation:
\[ T - mg = \frac{m(10hL)}{L} \]
Simplifying gives:
\[ T - mg = 10mh \]

4. **Solving for Tension**: Rearranging the equation:
\[ T = 10mh + mg = mg(10h + 1) \]
Thus, at the lowest position:
\[ T = 11mg \]

#### (ii) Tension at the Highest Position

1. **Identify the Forces**: At the highest position, the forces acting on the particle are:
- Tension \( T' \) acting downwards.
- Weight \( mg \) acting downwards.

2. **Centripetal Force Requirement**: The net force providing the centripetal acceleration is given by:
\[ T' + mg = \frac{mv'^2}{L} \]

3. **Using Energy Conservation**: We will use conservation of energy to find \( v' \) (velocity at the highest point):
- Initial energy at the lowest point:
\[ KE_i = \frac{1}{2} mv^2 = \frac{1}{2} m(10hL) \]
- Potential energy at the highest point (height = \( 2L \)):
\[ PE_f = mg(2L) \]
- Final kinetic energy at the highest point:
\[ KE_f = \frac{1}{2} mv'^2 \]

Setting initial energy equal to final energy:
\[ \frac{1}{2} m(10hL) = \frac{1}{2} mv'^2 + mg(2L) \]

4. **Solving for \( v' \)**:
\[ 10hL = v'^2 + 4gL \]
\[ v'^2 = 10hL - 4gL = (10h - 4g)L \]

5. **Substituting \( v' \) into the Tension Equation**:
\[ T' + mg = \frac{m(10h - 4g)L}{L} \]
\[ T' + mg = m(10h - 4g) \]
Rearranging gives:
\[ T' = m(10h - 4g) - mg = m(10h - 5g) \]

### (b) Tension when the string makes an angle \( 60^\circ \)

#### (i) When the string makes an angle of \( 60^\circ \) with the lower vertical

1. **Identify the Forces**: At this position, the forces acting are:
- Tension \( T \) acting along the string.
- Weight \( mg \) acting downwards.

2. **Components of Forces**:
- The radial component of tension provides the centripetal force:
\[ T \cos(60^\circ) = \frac{mv^2}{L} \]
- The vertical component of tension balances the weight:
\[ T \sin(60^\circ) = mg \]

3. **Using Energy Conservation**: Similar to previous steps, we can find the velocity \( v \) at this position using energy conservation:
\[ \frac{1}{2} m(10hL) = \frac{1}{2} mv^2 + mgL(1 - \cos(60^\circ)) \]
\[ \frac{1}{2} m(10hL) = \frac{1}{2} mv^2 + mgL(1 - \frac{1}{2}) \]
\[ 10hL = v^2 + gL \]
\[ v^2 = 10hL - gL = (10h - g)L \]

4. **Substituting \( v^2 \)**:
\[ T \cos(60^\circ) = \frac{m(10h - g)L}{L} \]
\[ T \cdot \frac{1}{2} = m(10h - g) \]
\[ T = 2m(10h - g) \]

5. **Using the vertical component**:
\[ T \cdot \frac{\sqrt{3}}{2} = mg \]
\[ T = \frac{2mg}{\sqrt{3}} \]

#### (ii) When the string makes an angle of \( 60^\circ \) with the upper vertical

1. **Repeat the above steps** with the appropriate adjustments for the angle and the energy conservation.

2. **Using energy conservation**:
\[ \frac{1}{2} m(10hL) = \frac{1}{2} mu'^2 + mgL(1 + \cos(60^\circ)) \]
\[ 10hL = u'^2 + mgL \]
\[ u'^2 = 10hL - mgL = (10h - g)L \]

3. **Substituting \( u'^2 \)** into the tension equations:
\[ T' \cos(60^\circ) = \frac{m(10h - g)L}{L} \]
\[ T' \cdot \frac{1}{2} = m(10h - g) \]
\[ T' = 2m(10h - g) \]

4. **Using the vertical component**:
\[ T' \cdot \frac{\sqrt{3}}{2} = mg \]
\[ T' = \frac{2mg}{\sqrt{3}} \]

### Summary of Results:
- Tension at the lowest position: \( T = 11mg \)
- Tension at the highest position: \( T' = m(10h - 5g) \)
- Tension at \( 60^\circ \) with lower vertical: \( T = 2m(10h - g) \)
- Tension at \( 60^\circ \) with upper vertical: \( T' = 2m(10h - g) \)