A small ball is kept on the top of a sphere of radius `R`. The sphere start accelerating with constant acceleration of `10m//s^(2)` horizontally. The angle of radial line with the vertical at which small ball leaves the sphere is `1/2 sin^(-1)(K//9)`. Find the value of `K`. [take `g=10m//s^(2)`]

Explanation

To solve the problem, we need to analyze the situation where a small ball is placed on top of a sphere that is accelerating horizontally. We are tasked with finding the value of \( K \) given that the angle of the radial line with the vertical at which the ball leaves the sphere is \( \frac{1}{2} \sin^{-1}\left(\frac{K}{9}\right) \).

### Step-by-Step Solution:

1. **Identify Forces Acting on the Ball**:
- The ball experiences two main forces: the gravitational force acting downwards \( (mg) \) and the normal force \( (N) \) exerted by the sphere.
- As the sphere accelerates horizontally, the ball will also experience an inertial force due to this acceleration.

2. **Set Up the Free Body Diagram**:
- The gravitational force \( mg \) acts vertically downwards.
- The normal force \( N \) acts perpendicular to the surface of the sphere.
- The inertial force due to horizontal acceleration \( a \) acts horizontally to the left (if we assume the sphere accelerates to the right).

3. **Components of Forces**:
- The gravitational force can be resolved into two components when the ball is at an angle \( \theta \):
- \( mg \cos(\theta) \) acts perpendicular to the surface of the sphere.
- \( mg \sin(\theta) \) acts parallel to the surface of the sphere.
- The inertial force can be expressed as \( ma \), where \( a = 10 \, \text{m/s}^2 \).

4. **Equilibrium Condition**:
- For the ball to remain on the sphere, the net force acting towards the center of the sphere must equal the centripetal force required to keep the ball in circular motion.
- The centripetal force needed is \( \frac{mv^2}{R} \), where \( v \) is the speed of the ball at the moment it leaves the sphere.

5. **Force Balance**:
- The radial component of the forces gives us:
\[ N - mg \cos(\theta) = \frac{mv^2}{R} \]
- The tangential component gives us:
\[ mg \sin(\theta) = ma \]
- From the second equation, we can express \( \sin(\theta) \):
\[ \sin(\theta) = \frac{a}{g} = \frac{10}{10} = 1 \]
- This indicates that the angle \( \theta \) at which the ball leaves the sphere is \( 90^\circ \).

6. **Finding \( K \)**:
- We know from the problem statement that:
\[ \theta = \frac{1}{2} \sin^{-1}\left(\frac{K}{9}\right) \]
- Setting \( \theta = 90^\circ \) gives:
\[ 90^\circ = \frac{1}{2} \sin^{-1}\left(\frac{K}{9}\right) \]
- Therefore, \( \sin^{-1}\left(\frac{K}{9}\right) = 180^\circ \), which implies:
\[ \frac{K}{9} = 1 \implies K = 9 \]

### Final Answer:
The value of \( K \) is \( 9 \).