Consider a case of fixed smooth sphere of radius R . A block of mass m is placed at the most poit of the sphere . A sharp impulse is applied on the block to impart it a speed v .
What si the maximum possible value of v so that block does not lose contact with the sphere at the top point ?

Options

A. `sqrt(5gR)`
B. `sqrt(3gR)`
C. `sqrt(gR)`
D. 0

Explanation

To solve the problem, we need to analyze the forces acting on the block at the topmost point of the sphere and apply the principles of circular motion.

### Step-by-step Solution:

1. **Identify the Forces Acting on the Block:**
At the topmost point of the sphere, the forces acting on the block of mass \( m \) are:
- The weight of the block acting downwards: \( mg \)
- The normal reaction force \( N \) from the sphere acting upwards.

2. **Apply the Condition for Circular Motion:**
For the block to maintain circular motion, the net force acting towards the center of the sphere must provide the necessary centripetal force. This can be expressed as:
\[ mg - N = \frac{mv^2}{R} \]
where \( v \) is the speed of the block at the topmost point and \( R \) is the radius of the sphere.

3. **Condition for Maintaining Contact:**
The block will lose contact with the sphere when the normal force \( N \) becomes zero. Therefore, for the block to just maintain contact, we set \( N = 0 \):
\[ mg - 0 = \frac{mv^2}{R} \]
Simplifying this gives:
\[ mg = \frac{mv^2}{R} \]

4. **Cancel Mass \( m \):**
Since \( m \) is non-zero, we can divide both sides by \( m \):
\[ g = \frac{v^2}{R} \]

5. **Rearranging for \( v \):**
Rearranging the equation to solve for \( v \) gives:
\[ v^2 = gR \]
Taking the square root of both sides, we find:
\[ v = \sqrt{gR} \]

6. **Conclusion:**
The maximum possible value of \( v \) so that the block does not lose contact with the sphere at the top point is:
\[ v_{\text{max}} = \sqrt{gR} \]