A bob is suspended from an ideal string of length 'l'. Now it is pulled to a side through `60^(@)` to vertical and whirled along a horizontal circle. Then its period of revolution is

Options

A. A) `pi sqrt((l)/(g))`
B. B) `pi sqrt((l)/(2g))`
C. C) `pi sqrt((2l)/(g))`
D. D) `2pi sqrt((l)/(g))`

Answer

To solve the problem, we need to find the period of revolution of a bob that is suspended from a string and is whirled in a horizontal circle at an angle of \(60^\circ\) to the vertical.

### Step-by-step Solution:

1. **Identify the Forces Acting on the Bob:**
- The forces acting on the bob are:
- Tension \(T\) in the string, acting along the string.
- Weight \(mg\) acting downward.
- Centrifugal force \(F_c\) acting horizontally due to circular motion.

2. **Resolve the Tension into Components:**
- The tension \(T\) can be resolved into two components:
- Vertical component: \(T \cos \theta\)
- Horizontal component: \(T \sin \theta\)
- Here, \(\theta = 60^\circ\).

3. **Set Up the Vertical Force Equation:**
- In the vertical direction, the forces must balance:
\[ T \cos \theta = mg \]
- Substituting \(\theta = 60^\circ\):
\[ T \cos 60^\circ = mg \]
- Since \(\cos 60^\circ = \frac{1}{2}\):
\[ T \cdot \frac{1}{2} = mg \implies T = 2mg \]

4. **Set Up the Horizontal Force Equation:**
- In the horizontal direction, the tension's horizontal component provides the necessary centripetal force:
\[ T \sin \theta = m \omega^2 r \]
- The radius \(r\) of the circular path can be expressed in terms of the length of the string \(L\):
\[ r = L \sin \theta \]
- Substituting \(\theta = 60^\circ\):
\[ r = L \sin 60^\circ = L \cdot \frac{\sqrt{3}}{2} \]
- Now substituting \(r\) back into the horizontal force equation:
\[ T \sin 60^\circ = m \omega^2 \left(L \cdot \frac{\sqrt{3}}{2}\right) \]

5. **Substituting the Value of Tension:**
- We already found \(T = 2mg\):
\[ 2mg \cdot \frac{\sqrt{3}}{2} = m \omega^2 \left(L \cdot \frac{\sqrt{3}}{2}\right) \]
- Simplifying gives:
\[ mg\sqrt{3} = m \omega^2 \left(L \cdot \frac{\sqrt{3}}{2}\right) \]
- Cancel \(m\) and \(\sqrt{3}\) from both sides:
\[ g = \frac{\omega^2 L}{2} \]

6. **Solving for Angular Velocity \(\omega\):**
- Rearranging gives:
\[ \omega^2 = \frac{2g}{L} \implies \omega = \sqrt{\frac{2g}{L}} \]

7. **Finding the Period of Revolution \(T\):**
- The period \(T\) is related to angular velocity by:
\[ T = \frac{2\pi}{\omega} \]
- Substituting \(\omega\):
\[ T = \frac{2\pi}{\sqrt{\frac{2g}{L}}} = 2\pi \cdot \sqrt{\frac{L}{2g}} = \pi \sqrt{\frac{2L}{g}} \]

### Final Answer:
The period of revolution is:
\[ T = \pi \sqrt{\frac{2L}{g}} \]