Search any question & find its solution
Question:
Answered & Verified by Expert
An elevator starts going up with an acceleration $2 \mathrm{~m} / \mathrm{s}^2$. If radius of the wheel attached to the elevator is $0.1 \mathrm{~m}$, then find out number of revolutions in $10 \mathrm{~s}$.
Options:
Solution:
1085 Upvotes
Verified Answer
The correct answer is:
$160$
$\theta=2 \pi n=\frac{1}{2}(\alpha) t^2$
$n=\frac{1}{2}\left(\frac{a}{r}\right) \frac{t^2}{2 \pi}$
Here, $a=2 \mathrm{~m} / \mathrm{s}^2, r=0.1 \mathrm{~m}, t=10 \mathrm{~s}, n=$ ?
$\therefore n=\frac{1}{2}\left(\frac{2}{0.1}\right) \frac{10^2}{6.28} \approx 160$
$n=\frac{1}{2}\left(\frac{a}{r}\right) \frac{t^2}{2 \pi}$
Here, $a=2 \mathrm{~m} / \mathrm{s}^2, r=0.1 \mathrm{~m}, t=10 \mathrm{~s}, n=$ ?
$\therefore n=\frac{1}{2}\left(\frac{2}{0.1}\right) \frac{10^2}{6.28} \approx 160$
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.