A particle is projected with velocity `u` horizontally from the top of a smooth sphere of radius a so that it slides down the outside of the sphere. If the particle leaves the sphere when it has fallen a height `(a)/(4)` , the value of `u` is

Options

A. `sqrt(ag)`
B. `(sqrt(ag))/(4)`
C. `(sqrt(ag))/(2)`
D. `(sqrt(ag))/(3)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the scenario
A particle is projected horizontally from the top of a smooth sphere of radius \( a \). It slides down and leaves the sphere after falling a height of \( \frac{a}{4} \). We need to find the initial horizontal velocity \( u \).

### Step 2: Determine the height fallen
The total height of the sphere is \( a \). If the particle falls a height of \( \frac{a}{4} \), then the height above the ground when it leaves the sphere is:
\[ h = a - \frac{a}{4} = \frac{3a}{4} \]

### Step 3: Apply energy conservation
When the particle falls a height of \( \frac{a}{4} \), we can use the conservation of energy principle. The potential energy lost will convert into kinetic energy.

The potential energy lost when falling from height \( a \) to \( \frac{3a}{4} \) is:
\[ \Delta PE = mg \left( \frac{a}{4} \right) \]

The kinetic energy gained is:
\[ KE = \frac{1}{2} mv^2 \]

At the top, the particle has kinetic energy due to its initial velocity \( u \):
\[ KE_{initial} = \frac{1}{2} mu^2 \]

At the point just before leaving the sphere, the total kinetic energy will be:
\[ KE_{final} = \frac{1}{2} mu^2 + mg \left( \frac{a}{4} \right) \]

Setting the potential energy lost equal to the kinetic energy gained:
\[ mg \left( \frac{a}{4} \right) = \frac{1}{2} mv^2 - \frac{1}{2} mu^2 \]

### Step 4: Find the velocity at the point of leaving
Using the kinematic equation for vertical motion:
\[ v^2 = u^2 + 2gh \]
where \( h = \frac{a}{4} \) and \( g \) is the acceleration due to gravity. Thus:
\[ v^2 = u^2 + 2g \left( \frac{a}{4} \right) = u^2 + \frac{ga}{2} \]

### Step 5: Apply centripetal force condition
At the point of leaving the sphere, the centripetal force required for circular motion is provided by the component of the weight acting towards the center of the sphere. The normal force \( N \) becomes zero when the particle leaves the sphere.

Using the angle \( \theta \) at which the particle leaves, we have:
\[ mg \cos \theta = \frac{mv^2}{a} \]

### Step 6: Determine \( \cos \theta \)
From the geometry of the problem, when the particle has fallen \( \frac{a}{4} \), the height from the center of the sphere is:
\[ OP = \frac{3a}{4} \]
Thus, using the right triangle formed:
\[ \cos \theta = \frac{3a/4}{a} = \frac{3}{4} \]

### Step 7: Substitute \( \cos \theta \) into the force equation
Substituting \( \cos \theta \) into the centripetal force equation:
\[ mg \left( \frac{3}{4} \right) = \frac{m}{a} \left( u^2 + \frac{ga}{2} \right) \]
Canceling \( m \) and rearranging gives:
\[ g \left( \frac{3}{4} \right) = \frac{u^2 + \frac{ga}{2}}{a} \]
Multiplying through by \( a \):
\[ \frac{3ga}{4} = u^2 + \frac{ga}{2} \]

### Step 8: Solve for \( u^2 \)
Rearranging gives:
\[ u^2 = \frac{3ga}{4} - \frac{ga}{2} = \frac{3ga}{4} - \frac{2ga}{4} = \frac{ga}{4} \]

### Step 9: Calculate \( u \)
Taking the square root:
\[ u = \sqrt{\frac{ga}{4}} = \frac{1}{2} \sqrt{ga} \]

### Final Answer
The value of \( u \) is:
\[ u = \frac{1}{2} \sqrt{ga} \]