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

Options

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

Explanation

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

### Step 1: Identify the given data
- Mass of the lighter stone: \( m \)
- Mass of the heavier stone: \( 2m \)
- Radius for the lighter stone: \( r \)
- Radius for the heavier stone: \( \frac{r}{2} \)
- Let the tangential speed of the heavier stone be \( v_h \).
- Let the tangential speed of the lighter stone be \( v_l = n \cdot v_h \).

### Step 2: Write the expression for centripetal force
The centripetal force \( F_c \) for an object moving in a circle is given by:
\[ F_c = \frac{mv^2}{r} \]

### Step 3: Write the centripetal force equations for both stones
For the lighter stone (mass \( m \) and radius \( r \)):
\[ F_{c,l} = \frac{m v_l^2}{r} \]

For the heavier stone (mass \( 2m \) and radius \( \frac{r}{2} \)):
\[ F_{c,h} = \frac{2m v_h^2}{\frac{r}{2}} = \frac{4m v_h^2}{r} \]

### Step 4: Set the centripetal forces equal
Since both stones experience the same centripetal force:
\[ \frac{m v_l^2}{r} = \frac{4m v_h^2}{r} \]

### Step 5: Simplify the equation
We can cancel \( m \) and \( r \) from both sides:
\[ v_l^2 = 4 v_h^2 \]

### Step 6: Relate \( v_l \) and \( v_h \)
Taking the square root of both sides gives:
\[ v_l = 2 v_h \]

### Step 7: Substitute \( v_l \) in terms of \( n \)
From our earlier definition, we have:
\[ v_l = n v_h \]
Setting the two expressions for \( v_l \) equal gives:
\[ n v_h = 2 v_h \]

### Step 8: Solve for \( n \)
Dividing both sides by \( v_h \) (assuming \( v_h \neq 0 \)):
\[ n = 2 \]

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

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