A block of mass m at the end of a string is whirled round in a vertical circle of radius R. The critical speed of the block at top of its swing below which the string would slacken before the block reaches the bottom is?

Options

A. `sqrt(5Rg)`
B. `sqrt(3Rg)`
C. `sqrt(2Rg)`
D. `sqrt(Rg)`

Explanation

To find the critical speed of a block of mass \( m \) at the top of its circular motion, we need to consider the forces acting on the block at that point. Here’s a step-by-step solution:

### Step 1: Analyze Forces at the Top of the Circle
At the top of the vertical circle, the forces acting on the block are:
- The weight of the block \( mg \) acting downwards.
- The tension \( T \) in the string also acting downwards.

For the block to maintain circular motion, the net force acting towards the center of the circle must equal the centripetal force required to keep the block moving in a circle.

### Step 2: Set Up the Equation for Centripetal Force
At the top of the circle, the centripetal force required is given by:
\[ F_c = \frac{mv^2}{R} \]
where \( v \) is the speed of the block and \( R \) is the radius of the circle.

The net force towards the center (which is the sum of the weight and tension) can be expressed as:
\[ mg + T = \frac{mv^2}{R} \]

### Step 3: Consider the Critical Condition
For the string to just begin to slacken, the tension \( T \) becomes zero. Therefore, we can set \( T = 0 \) in our equation:
\[ mg = \frac{mv^2}{R} \]

### Step 4: Simplify the Equation
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ g = \frac{v^2}{R} \]

### Step 5: Solve for the Critical Speed \( v \)
Rearranging the equation gives us:
\[ v^2 = gR \]
Taking the square root of both sides, we find:
\[ v = \sqrt{gR} \]

### Conclusion
Thus, the critical speed of the block at the top of its swing, below which the string would slacken before the block reaches the bottom, is:
\[ v = \sqrt{gR} \]