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 at the Lowest Point:**
At the lowest point of the vertical circle, two forces act on the mass \( m \):
- The gravitational force \( mg \) acting downward.
- The tension \( T \) in the string acting upward.

2. **Apply Newton's Second Law:**
According to Newton's second law, the net force acting on the mass is equal to the mass times its acceleration. At the lowest point, the net force can be expressed as:
\[ T - mg = \frac{mv^2}{r} \]
where \( v \) is the velocity of the mass at the lowest point, and \( r \) is the radius of the vertical circle.

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

4. **Simplify the Equation:**
Calculate \( (\sqrt{7gr})^2 \):
\[ (\sqrt{7gr})^2 = 7gr \]
Substitute this back into the equation:
\[ T - mg = \frac{m(7gr)}{r} \]
Simplifying further gives:
\[ T - mg = 7mg \]

5. **Solve for Tension \( T \):**
Now, we can isolate \( T \):
\[ T = mg + 7mg \]
\[ T = 8mg \]

Thus, the tension in the string at the lowest point is:
\[ \boxed{8mg} \]