A small disc is on the top of a hemisphere of radius `R`. What is the smallest horizontal velocity `v` that should fbe given to the disc for it to leave the hemisphere and not slide down it?[There is no friction]

Options

A. `v=sqrt(2gR)`
B. `v=sqrt(gR)`
C. `v=(g)/(R )`
D. `v=sqrt(g^(2)R)`

Explanation

To solve the problem of finding the smallest horizontal velocity \( v \) that should be given to the disc for it to leave the hemisphere and not slide down it, we can follow these steps:

### Step 1: Understand the Forces Acting on the Disc
When the disc is at the top of the hemisphere, two forces act on it:
1. The gravitational force \( mg \) acting downwards.
2. The normal force \( N \) acting perpendicular to the surface of the hemisphere.

### Step 2: Apply the Condition for Circular Motion
For the disc to maintain contact with the hemisphere while moving in a circular path, the net force acting towards the center of the hemisphere must provide the necessary centripetal force. The equation for centripetal force is given by:
\[ N + mg \cos(\theta) = \frac{mv^2}{R} \]
where \( \theta \) is the angle the radius makes with the vertical, \( R \) is the radius of the hemisphere, and \( v \) is the velocity of the disc.

### Step 3: Determine the Condition for Losing Contact
The disc will lose contact with the hemisphere when the normal force \( N \) becomes zero. Therefore, we set \( N = 0 \) in the centripetal force equation:
\[ mg \cos(\theta) = \frac{mv^2}{R} \]

### Step 4: Simplify the Equation
Since \( N = 0 \), we can simplify the equation:
\[ mg \cos(\theta) = \frac{mv^2}{R} \]
Dividing both sides by \( m \) (mass of the disc):
\[ g \cos(\theta) = \frac{v^2}{R} \]
Rearranging gives us:
\[ v^2 = gR \cos(\theta) \]

### Step 5: Find the Minimum Velocity
The minimum velocity occurs at the point where the disc is just about to lose contact with the hemisphere. This happens when \( \theta = 0 \) (at the top of the hemisphere). Thus:
\[ v^2 = gR \cos(0) = gR \]
Taking the square root gives us:
\[ v = \sqrt{gR} \]

### Conclusion
The smallest horizontal velocity \( v \) that should be given to the disc for it to leave the hemisphere and not slide down it is:
\[ \boxed{v = \sqrt{gR}} \]