Question

Two stones of masses m and 3 m are whirled in horizontal circles, the heavier one in a radius $(\frac{r}{3})$ and lighter one in a radius r. The tangential speed of lighter stone is 'n' times the value of heavier stone. When the magnitude of centripetal force becomes equal the value of n is

• A. 4
• B. 3
• C. 2
• D. 1

Hint:

$F_c = mv^2/r$. If $r$ decreases and $m$ increases, $v$ must change significantly to keep $F$ constant.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need to find the ratio 'n' of the speeds of two stones such that their centripetal forces are identical.
Step 2: Key Formula or Approach:
Centripetal force $F = \frac{mv^2}{r}$
Step 3: Detailed Explanation:
Let '1' denote the lighter stone and '2' denote the heavier stone.
$m_1 = m, r_1 = r, v_1 = v_1$
$m_2 = 3m, r_2 = r/3, v_2 = v_2$
Given condition: $F_1 = F_2$
\[ \frac{m_1 v_1^2}{r_1} = \frac{m_2 v_2^2}{r_2} \]
\[ \frac{m \cdot v_1^2}{r} = \frac{3m \cdot v_2^2}{r/3} \]
\[ \frac{m v_1^2}{r} = \frac{9 m v_2^2}{r} \]
\[ v_1^2 = 9 v_2^2 \]
\[ v_1 = 3 v_2 \]
The question states $v_1 = n v_2$. Comparing the two, we get $n = 3$.
Step 4: Final Answer:
The value of $n$ is 3.