Search any question & find its solution
Question:
Answered & Verified by Expert
A mass of $5 \mathrm{~kg}$ is moving along a circular path of radius 1 $\mathrm{m}$. If the mass moves with $300 \mathrm{rev} / \mathrm{min}$, its kinetic energy would be
Options:
Solution:
1726 Upvotes
Verified Answer
The correct answer is:
$250 \pi^2$
As given that, mass $(m)=5 \mathrm{~kg}$, $n=300$ revolution
$\operatorname{Radius}(R)=1 \mathrm{~m}$ $t=60 \mathrm{sec}$ $\omega=\left(\frac{2 \pi n}{t}\right)=(300 \times 2 \times \pi) \mathrm{rad} / 60 \mathrm{~s}$ $=\frac{600 \times \pi}{60} \mathrm{rad} / \mathrm{s}=10 \pi \mathrm{rad} / \mathrm{s}$
linear speed $(v)=\omega R=(10 \pi \times 1)$ $v=10 \pi \mathrm{m} / \mathrm{s}$
$$
\mathrm{KE}=\frac{1}{2} m v^2
$$
$$
=\frac{1}{2} \times 5 \times(10 \pi)^2=100 \pi^2 \times 5 \times \frac{1}{2}=250 \pi^2 \mathrm{~J}
$$
So, verifies the option (a).
$\operatorname{Radius}(R)=1 \mathrm{~m}$ $t=60 \mathrm{sec}$ $\omega=\left(\frac{2 \pi n}{t}\right)=(300 \times 2 \times \pi) \mathrm{rad} / 60 \mathrm{~s}$ $=\frac{600 \times \pi}{60} \mathrm{rad} / \mathrm{s}=10 \pi \mathrm{rad} / \mathrm{s}$
linear speed $(v)=\omega R=(10 \pi \times 1)$ $v=10 \pi \mathrm{m} / \mathrm{s}$
$$
\mathrm{KE}=\frac{1}{2} m v^2
$$
$$
=\frac{1}{2} \times 5 \times(10 \pi)^2=100 \pi^2 \times 5 \times \frac{1}{2}=250 \pi^2 \mathrm{~J}
$$
So, verifies the option (a).
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.