question_answer

A disc rotates about its axis of symmetry in a horizontal plane at a steady rate of 3.5 revolutions per second. A coin placed at a distance of 1.25cm from the axis of rotation remains at rest on the disc. The coefficient of friction between the coin and the disc is\[(g=10m/{{s}^{2}})\]       [JEE Online 15-04-2018 (II)]

A)         0.5                                         

B) 0.7         

C) 0.3                         

D)          0.6

Correct Answer: D

Solution :

\[\text{3}\text{.5 rev/second}\] \[\text{1rev}\to \text{2}\pi \text{rad}\] \[3.5rev\to 2\pi \times 3.5rad\] \[\Rightarrow w=7\pi rad/\sec \] \[\mu mg=\frac{m{{v}^{2}}}{1.25}\] \[\mu mg=\frac{m{{(rw)}^{2}}}{r}\] \[\mu mg=mr{{w}^{2}}\] \[\mu =\frac{r{{w}^{2}}}{g}=\frac{1.25\times {{10}^{-2}}\times {{\left( 7\times \frac{22}{7} \right)}^{2}}}{10}\] \[=\frac{1.25\times {{10}^{-2}}\times {{22}^{2}}}{10}\] \[=0.6\]