Search any question & find its solution
Question:
Answered & Verified by Expert
A disc rotates about its axis of symmetry in a hoizontal plane at a steady rate of $3.5$ revolutions per second. A coin placed at a distance of $1.25 \mathrm{~cm}$ from the axis of rotation remains at rest on the disc. The coefficient of friction between the coin and the disc is $\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right)$
Options:
Solution:
2221 Upvotes
Verified Answer
The correct answer is:
$0.6$
$0.6$
Using, $\mu m g=\frac{m v^2}{r}=m r \omega^2$ $\omega=2 \pi n=2 \pi \times 3.5=7 \pi \mathrm{rad} / \mathrm{sec}$ Radius, $r=1.25 \mathrm{~cm}=1.25 \times 10^{-2} \mathrm{~m}$ Coefficient of friction, $\mu=$ ?
$$
\mu m g=\frac{m(r \omega)^2}{r}(\because v=r \omega)
$$
$$
\begin{gathered}
\Rightarrow \quad \mu=\frac{r \omega^2}{g}=\frac{1.25 \times 10^{-2} \times\left(7 \times \frac{22}{7}\right)^2}{10} \\
=\frac{1.25 \times 10^{-2} \times 22^2}{10}=0.6
\end{gathered}
$$
$$
\mu m g=\frac{m(r \omega)^2}{r}(\because v=r \omega)
$$
$$
\begin{gathered}
\Rightarrow \quad \mu=\frac{r \omega^2}{g}=\frac{1.25 \times 10^{-2} \times\left(7 \times \frac{22}{7}\right)^2}{10} \\
=\frac{1.25 \times 10^{-2} \times 22^2}{10}=0.6
\end{gathered}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.