Two stones of masses `m` and `2m` are whirled in horizontal circles, the heavier one in a radius `(r)/(2)` and the lighter one in radius `r` . The tangential speed of light stone is `n` times that of the value heavier stone when they experience same centripetal forces. The value of `n` is:

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the problem, we need to analyze the centripetal forces acting on both stones and relate their speeds based on the given conditions.

### Step-by-Step Solution:

1. **Identify the masses and radii**:
- Let the mass of the lighter stone be \( m \).
- The mass of the heavier stone is \( 2m \).
- The radius of the lighter stone's circular path is \( r \).
- The radius of the heavier stone's circular path is \( \frac{r}{2} \).

2. **Write the formula for centripetal force**:
The centripetal force \( F_c \) required to keep an object moving in a circle is given by:
\[ F_c = \frac{mv^2}{r} \]
where \( m \) is the mass, \( v \) is the tangential speed, and \( r \) is the radius of the circular path.

3. **Centripetal force for the lighter stone**:
For the lighter stone (mass \( m \) and radius \( r \)):
\[ F_{c1} = \frac{m v_1^2}{r} \]
where \( v_1 \) is the tangential speed of the lighter stone.

4. **Centripetal force for the heavier stone**:
For the heavier stone (mass \( 2m \) and radius \( \frac{r}{2} \)):
\[ F_{c2} = \frac{2m v_2^2}{\frac{r}{2}} = \frac{4m v_2^2}{r} \]
where \( v_2 \) is the tangential speed of the heavier stone.

5. **Set the centripetal forces equal**:
According to the problem, the centripetal forces are equal:
\[ F_{c1} = F_{c2} \]
Therefore,
\[ \frac{m v_1^2}{r} = \frac{4m v_2^2}{r} \]

6. **Simplify the equation**:
We can cancel \( m \) and \( r \) from both sides (assuming \( m \neq 0 \) and \( r \neq 0 \)):
\[ v_1^2 = 4 v_2^2 \]

7. **Relate the speeds**:
Taking the square root of both sides gives:
\[ v_1 = 2 v_2 \]
This means the speed of the lighter stone is twice that of the heavier stone.

8. **Find the value of \( n \)**:
Given that the speed of the lighter stone \( v_1 \) is \( n \) times that of the heavier stone \( v_2 \):
\[ v_1 = n v_2 \]
From our previous result, we have:
\[ 2 v_2 = n v_2 \]
Dividing both sides by \( v_2 \) (assuming \( v_2 \neq 0 \)):
\[ n = 2 \]

### Final Answer:
The value of \( n \) is \( 2 \).