Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
60 views
in Physics by (86.6k points)
closed by
A circular platform is mounted on a frictionless vertical axle. Its radius `R = 2m` and its moment of inertia about the axle is `200 kg m^(2)`. It is initially at rest. A `50 kg` man stands on the edge at the platform and begins to walk along the edge at the speed of `1ms^(-1)` relative to the ground. Time taken by the man to complete one revolution is :
A. `pi sec`
B. `(3 pi)/(2) sec`
C. `2 pi sec`
D. `(pi)/(2) sec`

1 Answer

0 votes
by (79.4k points)
selected by
 
Best answer
Correct Answer - C
( c) From conservation of angular momentum
`I omega = mvr`
`200 xx omega = 50 xx 2 xx 1`
`omega = (1)/(2) rad//s`
`v = r omega = 1 m//s`
:. `t = (2 pi r)/(1 -(-1)) =(2 pi r)/(2) = pi r = 2 pi s`.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...