question_answer

A coin is placed on a turn table rotating with \[\omega \], slip at a distance of 4 cm from centre. The distance at which it will slip when co is double, is:

A)  8 cm                                     

B)  2cm                      

C)         6 cm                     

D)         1 cm

Correct Answer: D

Solution :

                \[F=\mu N\]                                                       ?(1) At equilibrium \[F=Mr{{\omega }^{2}}\]                                ?(2) and also               \[N=mg\]                            ?(3) Now, from Eqs. (1), (2) and (3), we obtain \[\mu Mg=Mr{{\omega }^{2}}\]                 or            \[r\propto \frac{1}{{{\omega }^{2}}}\]                 Hence,                  \[\frac{{{r}^{2}}}{{{r}_{1}}}={{\left( \frac{{{\omega }_{1}}}{{{\omega }_{2}}} \right)}^{2}}\] or            \[\frac{{{r}_{2}}}{{{r}_{1}}}=\left( \frac{\omega }{2\omega } \right)=\frac{1}{4}\]                              \[{{r}_{2}}=\frac{{{r}_{1}}}{4}=\frac{4cm}{4}=1\,cm\]