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 conditions for an object to successfully complete a vertical loop of radius \( r \). We will derive the minimum velocity required at the lowest point of the loop and check the validity of the assertion and reason provided.

### Step 1: Understanding the Forces at the Highest Point
At the highest point of the loop, the object must have enough velocity to maintain circular motion. The forces acting on the object are:
- The gravitational force \( mg \) acting downwards.
- The centripetal force required for circular motion, which is provided by the gravitational force and any tension in the string (if applicable).

For the object to just maintain contact at the highest point, we set the centripetal force equal to the gravitational force:
\[ mg = \frac{mv^2}{r} \]
Where \( v \) is the velocity at the highest point. Simplifying this, we find:
\[ v^2 = rg \quad \Rightarrow \quad v = \sqrt{rg} \]

### Step 2: Energy Conservation from Lowest to Highest Point
Now, we will use the principle of conservation of mechanical energy. The total mechanical energy at the lowest point (point A) must equal the total mechanical energy at the highest point (point B).

Let \( v_1 \) be the velocity at the lowest point and \( v_2 \) be the velocity at the highest point. The height difference between the lowest and highest points is \( 2r \).

Using conservation of energy:
\[ \frac{1}{2}mv_1^2 = \frac{1}{2}mv_2^2 + mg(2r) \]
Substituting \( v_2 = \sqrt{rg} \):
\[ \frac{1}{2}mv_1^2 = \frac{1}{2}m(\sqrt{rg})^2 + mg(2r) \]
\[ \frac{1}{2}mv_1^2 = \frac{1}{2}mgr + 2mgr \]
\[ \frac{1}{2}mv_1^2 = \frac{1}{2}mgr + 4mgr = \frac{9}{2}mgr \]
Cancelling \( m \) and multiplying by 2:
\[ v_1^2 = 9gr \quad \Rightarrow \quad v_1 = 3\sqrt{gr} \]

### Step 3: Finding the Minimum Velocity
To ensure that the object just makes it around the loop, we need to consider the minimum velocity at the lowest point. However, if we want to ensure that the object has enough energy to reach the top of the loop with zero velocity, we need to set the velocity at the highest point to zero.

Using the energy conservation principle again, if we want the velocity at the highest point to be zero:
\[ \frac{1}{2}mv_1^2 = mg(2r) \]
\[ \frac{1}{2}mv_1^2 = 2mgr \]
Cancelling \( m \) and multiplying by 2:
\[ v_1^2 = 4gr \quad \Rightarrow \quad v_1 = 2\sqrt{gr} \]

### Step 4: Correcting the Assertion
The assertion states that the minimum velocity at the lowest point should be \( \sqrt{5gr} \). However, based on our calculations, the minimum velocity required to ensure that the object can complete the loop with zero velocity at the highest point is actually \( 2\sqrt{gr} \).

### Conclusion
- **Assertion**: The assertion that the minimum velocity at the lowest point should be \( \sqrt{5gr} \) is **incorrect**.
- **Reason**: The reason that the velocity at the highest point will be zero is also **incorrect** because the velocity at the highest point should be \( \sqrt{gr} \) when the object just makes it around the loop.

Thus, both the assertion and the reason are false.