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
149 views
in Physics by (86.6k points)
closed by
From a circular disc of radius R and mass 9 M , a small disc of radius R/3 is removed from the disc. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O is
image
A. `8 mR^(2)`
B. `4 mR^(2)`
C. `(40)/(9) mR^(2)`
D. `(37)/(9) mR^(2)`

1 Answer

0 votes
by (79.4k points)
selected by
 
Best answer
Correct Answer - B
`I_("remaining") = I_("whole") - I_("removed")`
`I =(1)/(2) (9m) (R^(2)) -[(1)/(2)m ((R)/(3))^(2) + m((2R)/(3))^(2)]`
`I = 4mR^(2)`.

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

...