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: Centripetal Force Equality
$F_1 = F_2 \Rightarrow \frac{m_1 v_1^2}{r_1} = \frac{m_2 v_2^2}{r_2}$
Step 2: Substitution of Given Values
Lighter: $m, r, v_1$. Heavier: $3m, r/3, v_2$.
$\frac{m v_1^2}{r} = \frac{3m v_2^2}{r/3} = \frac{9m v_2^2}{r}$
Step 3: Solving for Speed Ratio
$v_1^2 = 9 v_2^2 \Rightarrow v_1 = 3 v_2$
Step 4: Finding n
Given $v_1 = n v_2$, therefore $n = 3$. (Correction: Based on options, check calculation)
Re-check: $\frac{v_1^2}{r} = \frac{9 v_2^2}{r} \to v_1 = 3v_2$.
Note: If $n=1$ is intended, check if angular velocity was equal. For centripetal force equality, $n=3$.
Final Answer:(B) (Updated based on $n=3$ result)