Two stones of masses $m$ and $2\,m$ are whirled in horizontal circles, the heavier one in a radius $\frac{r}{2}$ and the lighter one in radius $r.$ The tangential speed of lighter stone is $n$ times that of the value of heavier stone when they experience same centripetal forces. The value of $n$ is
AIPMT 2015, Medium
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Let $v$ be tangential speed of heavier stone. Then , centripetal force experienced by lighter stone is
$\left(F_c\right)_{\text {lighter }}=\frac{m(n v)^2}{r} \text { and that of heavier stone is }$
$\left(F_C\right)_{\text {heavier }}=\frac{2 m v^2}{(r / 2)}$
$\text { But }\left(F_c\right)_{\text {lighter }}=\left(F_c\right)_{\text {heavier }}(\text { given })$
$\therefore \frac{m(n v)^2}{r}=\frac{2 m v^2}{(r / 2)}$
$ n^2\left(\frac{m v^2}{r}\right)=4\left(\frac{m v^2}{r}\right)$
$ n^2=4 \text { or } n=2$
$\left(F_c\right)_{\text {lighter }}=\frac{m(n v)^2}{r} \text { and that of heavier stone is }$
$\left(F_C\right)_{\text {heavier }}=\frac{2 m v^2}{(r / 2)}$
$\text { But }\left(F_c\right)_{\text {lighter }}=\left(F_c\right)_{\text {heavier }}(\text { given })$
$\therefore \frac{m(n v)^2}{r}=\frac{2 m v^2}{(r / 2)}$
$ n^2\left(\frac{m v^2}{r}\right)=4\left(\frac{m v^2}{r}\right)$
$ n^2=4 \text { or } n=2$
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