Assertion For looping a verticla loop of radius, r the minimum velocity at lowest point should be `sqrt(5gr).` Reason In this event the velocityh at the highest point will be zero.

Options

A. Both assertion and reson are true and reason is the correct explanation of assertion
B. Both assetion and reason are true but reason is not the correct explanation of assertion
C. Assertion is true but reason is false
D. Both assetion and reason are flase

Answer

To solve the problem, we need to analyze the assertion and reason provided in the question step by step.

### Step 1: Understanding the Assertion
The assertion states that for looping a vertical loop of radius \( r \), the minimum velocity at the lowest point should be \( \sqrt{5gr} \).

### Step 2: Analyzing the Reason
The reason states that in this event, the velocity at the highest point will be zero. We need to verify this claim.

### Step 3: Applying the Conservation of Energy
We can use the principle of conservation of mechanical energy to analyze the motion of the object in the vertical loop. The total mechanical energy at the highest point must equal the total mechanical energy at the lowest point.

1. **At the Highest Point:**
- Let the velocity at the highest point be \( v_t \).
- The height at the highest point is \( 2r \) (since it is twice the radius).
- The potential energy at the highest point is \( PE_t = mg(2r) = 2mgr \).
- The kinetic energy at the highest point is \( KE_t = \frac{1}{2}mv_t^2 \).

Therefore, the total energy at the highest point is:
\[ E_t = KE_t + PE_t = \frac{1}{2}mv_t^2 + 2mgr \]

2. **At the Lowest Point:**
- Let the velocity at the lowest point be \( v_b \).
- The height at the lowest point is \( 0 \).
- The potential energy at the lowest point is \( PE_b = 0 \).
- The kinetic energy at the lowest point is \( KE_b = \frac{1}{2}mv_b^2 \).

Therefore, the total energy at the lowest point is:
\[ E_b = KE_b + PE_b = \frac{1}{2}mv_b^2 + 0 = \frac{1}{2}mv_b^2 \]

### Step 4: Setting Total Energies Equal
By conservation of energy, we set the total energy at the highest point equal to the total energy at the lowest point:
\[ \frac{1}{2}mv_t^2 + 2mgr = \frac{1}{2}mv_b^2 \]

### Step 5: Finding the Relationship Between Velocities
From the previous step, we can isolate \( v_b \):
\[ \frac{1}{2}mv_b^2 = \frac{1}{2}mv_t^2 + 2mgr \]
Dividing through by \( m \) and multiplying by 2 gives:
\[ v_b^2 = v_t^2 + 4gr \]

### Step 6: Finding the Velocity at the Highest Point
From the analysis of forces at the highest point, for the object to just maintain circular motion, the centripetal force requirement gives:
\[ mg = \frac{mv_t^2}{r} \implies v_t^2 = gr \]

### Step 7: Substitute \( v_t^2 \) into the Energy Equation
Substituting \( v_t^2 = gr \) into our energy equation:
\[ v_b^2 = gr + 4gr = 5gr \]
Thus, we find:
\[ v_b = \sqrt{5gr} \]

### Conclusion
The assertion is true: the minimum velocity at the lowest point should indeed be \( \sqrt{5gr} \). The reason is false because the velocity at the highest point is not zero; it is \( \sqrt{gr} \).

### Final Answer
- **Assertion**: True
- **Reason**: False