A sphere of mass m is tied to end of a string of length l and rotated through the other end along a horizontal circular path with speed v . The work done in full horizontal circle is

Options

A. 0
B. `((mv^(2))/(l)).2pil`
C. `mg.2pi l`
D. `((mv^(2))/(l)).(l)`

Answer

To solve the problem of the work done when a sphere of mass \( m \) is tied to a string of length \( l \) and rotated in a horizontal circular path with speed \( v \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Forces Acting on the Sphere:**
- The sphere experiences two main forces: the tension \( T \) in the string acting upwards and the gravitational force \( mg \) acting downwards.
- In the horizontal circular motion, the net force acting towards the center of the circle is provided by the horizontal component of the tension.

2. **Analyzing the Circular Motion:**
- The sphere moves in a horizontal circle, so the tension in the string must provide the necessary centripetal force required for circular motion.
- The centripetal force \( F_c \) can be expressed as:
\[ F_c = \frac{mv^2}{r} \]
- Here, \( r \) is the radius of the circular path, which is equal to the length of the string \( l \) in this case.

3. **Setting Up the Equation for Tension:**
- The vertical forces must balance since the sphere does not move vertically. Therefore, we can write:
\[ T - mg = 0 \quad \Rightarrow \quad T = mg \]

4. **Finding the Work Done:**
- Work done \( W \) is defined as the dot product of force and displacement:
\[ W = \vec{F} \cdot \vec{d} \]
- In this scenario, the tension \( T \) acts perpendicular to the displacement of the sphere as it rotates. Since the angle \( \theta \) between the tension force and the displacement is \( 90^\circ \):
\[ W = T \cdot d \cdot \cos(90^\circ) \]
- Since \( \cos(90^\circ) = 0 \), the work done is:
\[ W = T \cdot d \cdot 0 = 0 \]

5. **Conclusion:**
- The work done in completing a full horizontal circle is zero because the force (tension) is always perpendicular to the displacement of the sphere.

### Final Answer:
The work done in a full horizontal circle is \( \boxed{0} \).