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
437 views
in Physics by (91.0k points)
closed by
From a circular disc of radius `R`, a triangular portion is cut (sec figure). The distance of the centre of mass of the remainder from the centre of the disc is `-`
image
A. `(4R)/(3(pi-2))`
B. `(2R)/(3(pi-2))`
C. `(5R)/(7(pi-2))`
D. `(R)/(3(pi-1))`

1 Answer

0 votes
by (88.4k points)
selected by
 
Best answer
Correct Answer - D
Ans.(4)
Lot mass per area be `sigma`
image
`Y_(cm)=(M_(1)(0)-M_(2)((R)/(3)))/(M_(1)-M_(2))=(-R)/(3(pi-1)`

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

...