Search any question & find its solution
Question:
Answered & Verified by Expert
A constant torque of $200 \mathrm{~N}$ turns a flywheel, which is at rest, about an axis through
its centre and perpendicular to its plane. If its moment of inertia is $50 \mathrm{~kg}-\mathrm{m}^{2}$, then
in 4 second, what will be change in its angular momentum?
Options:
its centre and perpendicular to its plane. If its moment of inertia is $50 \mathrm{~kg}-\mathrm{m}^{2}$, then
in 4 second, what will be change in its angular momentum?
Solution:
1046 Upvotes
Verified Answer
The correct answer is:
$800 \mathrm{~kg}-\mathrm{m}^{2} / \mathrm{s}$
(B)
Change in angular momentum $=\tau . t$
$=200 \times 4=800 \mathrm{~kg} \mathrm{~m}^{2} / \mathrm{s}$
Change in angular momentum $=\tau . t$
$=200 \times 4=800 \mathrm{~kg} \mathrm{~m}^{2} / \mathrm{s}$
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.