Two stones of masses m and 3m are whirled in | vedclass

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(C) The centripetal force $F$ is given by the formula $F = \frac{mv^2}{R}$.Let $m_1 = m$ (lighter stone) and $R_1 = r$.Let $m_2 = 3m$ (heavier stone)

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$3$

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(C) The centripetal force $F$ is given by the formula $F = \frac{mv^2}{R}$.Let $m_1 = m$ (lighter stone) and $R_1 = r$.Let $m_2 = 3m$ (heavier stone) and $R_2 = r/3$.Let $v_1$ be the speed of the lighter stone and $v_2$ be the speed of the heavier stone.Given that the centripetal forces are equal, $F_1 = F_2$.Therefore, $\frac{m_1 v_1^2}{R_1} = \frac{m_2 v_2^2}{R_2}$.Substituting the given values: $\frac{m v_1^2}{r} = \frac{3m v_2^2}{r/3}$.Simplifying the equation: $\frac{m v_1^2}{r} = \frac{9m v_2^2}{r}$.$v_1^2 = 9 v_2^2$.Taking the square root on both sides: $v_1 = 3 v_2$.Since $v_1 = n v_2$, we get $n = 3$.

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