A body is revolving in a vertical circle with constant mechanical energy. The speed of the body at the highest point is `sqrt(2rg)`. The speed of the body at the lowest point is

Options

A. `sqrt(7gr)`
B. `sqrt(6gr)`
C. `sqrt(8gr)`
D. `sqrt(9gr)`

Answer

To solve the problem, we will use the principle of conservation of mechanical energy. The total mechanical energy (kinetic energy + potential energy) remains constant for the body revolving in a vertical circle.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Speed at the highest point (V_top) = \( \sqrt{2rg} \)
- Height at the highest point (h_top) = 2r (since it is at the top of the vertical circle)
- Height at the lowest point (h_bottom) = 0 (taking the lowest point as the reference level)

2. **Write the Expression for Mechanical Energy:**
- Mechanical energy at the highest point (E_top):
\[ E_{top} = \text{K.E. at top} + \text{P.E. at top} \]
\[ E_{top} = \frac{1}{2} m V_{top}^2 + mgh_{top} \]
Substituting the values:
\[ E_{top} = \frac{1}{2} m (\sqrt{2rg})^2 + mg(2r) \]
\[ E_{top} = \frac{1}{2} m (2rg) + mg(2r) \]
\[ E_{top} = mrg + 2mgr = 3mgr \]

3. **Write the Expression for Mechanical Energy at the Lowest Point:**
- Mechanical energy at the lowest point (E_bottom):
\[ E_{bottom} = \text{K.E. at bottom} + \text{P.E. at bottom} \]
\[ E_{bottom} = \frac{1}{2} m V_{bottom}^2 + mg(0) \]
\[ E_{bottom} = \frac{1}{2} m V_{bottom}^2 \]

4. **Set the Mechanical Energies Equal:**
Since mechanical energy is conserved:
\[ E_{top} = E_{bottom} \]
\[ 3mgr = \frac{1}{2} m V_{bottom}^2 \]

5. **Solve for V_bottom:**
- Cancel out the mass (m) from both sides:
\[ 3gr = \frac{1}{2} V_{bottom}^2 \]
- Multiply both sides by 2:
\[ 6gr = V_{bottom}^2 \]
- Take the square root:
\[ V_{bottom} = \sqrt{6rg} \]

### Final Answer:
The speed of the body at the lowest point is \( \sqrt{6rg} \).

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