A point mass m is moved in vertical circle of radius r with help of string. Velocity of mass is `root(7gr)` at lowest point. Tension in string at lowest point is

Options

A. 6 mg
B. 7 mg
C. 8 mg
D. 1 mg

Answer

To find the tension in the string at the lowest point of the vertical circle, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Mass**:
At the lowest point of the vertical circle, two forces act on the mass \( m \):
- The gravitational force acting downward, which is \( mg \).
- The tension \( T \) in the string acting upward.

2. **Determine the Centripetal Force Requirement**:
The mass moving in a circular path requires a centripetal force to keep it in that path. This centripetal force is provided by the net force acting towards the center of the circle. At the lowest point, the net force is the difference between the tension and the weight of the mass:
\[ T - mg = \frac{mv^2}{r} \]
where \( v \) is the velocity of the mass and \( r \) is the radius of the circle.

3. **Substitute the Given Velocity**:
The problem states that the velocity at the lowest point is \( v = \sqrt{7gr} \). We can substitute this into the centripetal force equation:
\[ T - mg = \frac{m(\sqrt{7gr})^2}{r} \]

4. **Simplify the Equation**:
Squaring the velocity gives:
\[ (\sqrt{7gr})^2 = 7gr \]
Thus, the equation becomes:
\[ T - mg = \frac{m(7gr)}{r} \]
Simplifying further:
\[ T - mg = 7mg \]

5. **Solve for Tension \( T \)**:
Rearranging the equation to find \( T \):
\[ T = mg + 7mg \]
\[ T = 8mg \]

### Final Answer:
The tension in the string at the lowest point is \( T = 8mg \).