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 :

[d] \[\text{3}\text{.5 rev/second}\]

\[\text{1rev}\to \text{2}\pi \text{rad}\]

\[3.5rev\to 2\pi \times 3.5rad\]

\[\Rightarrow w=7\pi rad/\sec \]

\[\mu mg=\frac{m{{v}^{2}}}{1.25}\]

\[\mu mg=\frac{m{{(rw)}^{2}}}{r}\]

\[\mu mg=mr{{w}^{2}}\]

\[\mu =\frac{r{{w}^{2}}}{g}=\frac{1.25\times {{10}^{-2}}\times {{\left( 7\times \frac{22}{7} \right)}^{2}}}{10}\]

\[=\frac{1.25\times {{10}^{-2}}\times {{22}^{2}}}{10}\]

\[=0.6\]