A small block is freely sliding down from top of a smooth inclined plane. The block reaches bottom of inclined plane then the block describes a vertical circle of radius `0.5 m` along smooth track. The minimum vertical height of inclined plane should be

Options

A. `1m`
B. `1.25 m`
C. `3m`
D. `2.5 m`

Explanation

To solve the problem of finding the minimum vertical height of the inclined plane from which a block must slide to complete a vertical circle of radius 0.5 m, we can follow these steps:

### Step 1: Understand the Energy Conservation Principle
When the block slides down the inclined plane, it converts its potential energy at the top into kinetic energy at the bottom. The total mechanical energy is conserved since there is no friction.

### Step 2: Set Up the Energy Conservation Equation
Let the height of the inclined plane be \( h \). The potential energy at the top (point A) is given by:
\[ PE_A = mgh \]
At the bottom (point B), the potential energy is zero, and the kinetic energy is:
\[ KE_B = \frac{1}{2} mv^2 \]
By conservation of energy:
\[ PE_A = KE_B \]
Thus,
\[ mgh = \frac{1}{2} mv^2 \]

### Step 3: Simplify the Equation
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ gh = \frac{1}{2} v^2 \]
Rearranging gives:
\[ v^2 = 2gh \]
(1)

### Step 4: Determine the Minimum Velocity for Vertical Circular Motion
For the block to complete a vertical circle, it must have a minimum velocity at the bottom of the circle. The minimum velocity \( v_{min} \) required at the bottom of the circle is given by:
\[ v_{min} = \sqrt{g r} \]
where \( r \) is the radius of the circle (0.5 m in this case).

### Step 5: Substitute the Radius
Substituting \( r = 0.5 \) m into the equation for minimum velocity:
\[ v_{min} = \sqrt{g \cdot 0.5} = \sqrt{0.5g} \]

### Step 6: Relate the Two Velocities
From equation (1), we have:
\[ v^2 = 2gh \]
Now substituting \( v_{min} \) into this equation:
\[ 0.5g = 2gh \]
Thus,
\[ (0.5g)^2 = 2gh \]
Substituting \( v^2 \) gives:
\[ 0.5g = 2gh \]

### Step 7: Solve for Height \( h \)
Now we can equate:
\[ 2gh = 0.5g \]
Cancelling \( g \) (assuming \( g \neq 0 \)):
\[ 2h = 0.5 \]
Thus,
\[ h = \frac{0.5}{2} = 0.25 \text{ m} \]

### Step 8: Final Calculation for Minimum Height
Now, we need to find the height in terms of \( r \):
\[ h = \frac{5r}{2} \]
Substituting \( r = 0.5 \):
\[ h = \frac{5 \cdot 0.5}{2} = \frac{2.5}{2} = 1.25 \text{ m} \]

### Conclusion
The minimum vertical height of the inclined plane should be **1.25 m**.

---