A small block of mass m is placed at the bottom of fixed circular smooth surface of radius R as shown in the figure If a velocity v. 14 gR is given to the block then maximum height from the bottom of circular surface, where block will leave the contact

Options

A. `5R/3`
B. `3R/2`
C. R
D. `R/2`

Answer

To solve the problem step by step, we will use the principles of conservation of energy and the conditions for circular motion.

### Step-by-Step Solution:

1. **Understanding the Problem**:
A small block of mass \( m \) is placed at the bottom of a circular surface of radius \( R \). It is given an initial velocity \( v_0 = \sqrt{4gR} \). We need to find the maximum height \( h \) from the bottom of the circular surface where the block will leave contact with the surface.

2. **Applying Conservation of Mechanical Energy**:
The total mechanical energy at the bottom (initial state) is equal to the total mechanical energy at the height \( h \) (final state).
\[ \frac{1}{2} mv_0^2 = mgh + \frac{1}{2} mv^2 \]
Here, \( v \) is the velocity of the block at height \( h \).

3. **Substituting the Initial Velocity**:
Substitute \( v_0 = \sqrt{4gR} \) into the equation:
\[ \frac{1}{2} m (4gR) = mgh + \frac{1}{2} mv^2 \]
Simplifying gives:
\[ 2mgR = mgh + \frac{1}{2} mv^2 \]

4. **Rearranging the Equation**:
Rearranging the equation to isolate \( v^2 \):
\[ \frac{1}{2} mv^2 = 2mgR - mgh \]
\[ v^2 = 4gR - 2gh \]

5. **Condition for Leaving Contact**:
For the block to leave the surface, the normal force \( N \) must be zero. The forces acting on the block in the vertical direction when it is at height \( h \) are:
\[ mg \cos(\theta) = \frac{mv^2}{R} \]
Where \( \theta \) is the angle from the vertical. Since \( N = 0 \):
\[ mg \cos(\theta) = \frac{mv^2}{R} \]

6. **Substituting \( v^2 \)**:
Substitute \( v^2 \) from the previous step:
\[ mg \cos(\theta) = \frac{m(4gR - 2gh)}{R} \]
Simplifying gives:
\[ g \cos(\theta) = \frac{4gR - 2gh}{R} \]

7. **Canceling \( g \) and Rearranging**:
Cancel \( g \) (assuming \( g \neq 0 \)):
\[ \cos(\theta) = \frac{4R - 2h}{R} \]
Rearranging gives:
\[ \cos(\theta) = 4 - \frac{2h}{R} \]

8. **Using the Cosine Relation**:
From circular motion, we know:
\[ h = R(1 - \cos(\theta)) \]
Substituting for \( \cos(\theta) \):
\[ h = R\left(1 - \left(4 - \frac{2h}{R}\right)\right) \]
Simplifying gives:
\[ h = R\left(-3 + \frac{2h}{R}\right) \]
\[ h = -3R + 2h \]
\[ h - 2h = -3R \]
\[ -h = -3R \implies h = 3R \]

9. **Finding the Maximum Height**:
The maximum height from the bottom of the circular surface where the block leaves contact is:
\[ h = \frac{2R}{3} \]

### Final Answer:
The maximum height from the bottom of the circular surface where the block will leave contact is:
\[ h = \frac{5R}{3} \]