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

Q: 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

b $\begin{array}{l} \,\,\,\,\,\,\,\,\,\,Let\,v\,be\,{\rm{tangential}}\,{\rm{speed}}\,{\rm{of}}\,{\rm{heavier}}\\ {\rm{stone}}{\rm{.}}\,{\rm{Then}}\,{\rm{,}}\,{\rm{centripetal}}\,{\rm{force}}\,{\rm{experienced}}\\ {\rm{by}}\,{\rm{lighter}}\,{\rm{stone}}\,{\rm{is}}\,\\ {\left( {{F_c}} \right)_{lighter}} = \frac{{m{{\left( {nv} \right)}^2}}}{r}\,and\,that\,of\,heavier\,stone\,is\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\left( {{F_C}} \right)_{heavier}} = \frac{{2m{v^2}}}{{\left( {r/2} \right)}} \end{array}$ $\begin{array}{l} But\,{\left( {{F_c}} \right)_{lighter}} = {\left( {{F_c}} \right)_{heavier}}\,(given)\\ \therefore \,\,\,\frac{{m{{\left( {nv} \right)}^2}}}{r} = \frac{{2m{v^2}}}{{\left( {r/2} \right)}}\\ \,\,\,\,\,\,\,\,\,\,{n^2}\left( {\frac{{m{v^2}}}{r}} \right) = 4\left( {\frac{{m{v^2}}}{r}} \right)\\ \,\,\,\,\,\,\,\,\,\,{n^2} = 4\,\,\,or\,\,\,n = 2 \end{array}$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

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

A$1$
B$2$
C$3$
D$4$

AIPMT 2015, Medium

STD 11 Science
NEET
STD 11 - 4.2. friction
Physics

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