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Two stones 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 $n t i m e s$ that of the value of heavier stone when they experience same centripetal forces. The value of $n$ is

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$3$

$2$

$1$

$4$

answer is C.

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Detailed Solution

Let $v$ be the tangential speed of the heavier stone. Then, centripetal force experienced by lighter stone is

${(F_{c})}_{l i g h t e r} = \frac{m {(n v)}^{2}}{r}$

and that of the heavier stone is ${(F_{c})}_{h e a v i e r} = \frac{2 m v^{2}}{(r / 2)}$

But ${(F_{c})}_{l i g h t e r} = {(F_{c})}_{h e a v i e r} (g i v e n)$

$\frac{m {(n v)}^{2}}{r} = \frac{2 m v^{2}}{(r / 2)} \Rightarrow n^{2} (\frac{m v^{2}}{r}) = 4 (\frac{m v^{2}}{r}) n^{2} = 4 o r n = 2$

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