Use app×
QUIZARD
QUIZARD
JEE MAIN 2026 Crash Course
NEET 2026 Crash Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
2.7k views
in Physics by (75.2k points)
closed
A particle of mass `m` is placed on centre of curvature of a fixed, uniform semi-circular ring of radius `R` and mass `M` as shown in figure. Calculate:
(a) interaction force between the ring and the particle and
(b) work required to displace the particle from centre of curvature to infinity.
image
A. `(a) F=(2GM)/(piR^(2)), (b) (GM)/R`
B. `(a) F=(2GMm)/(pi^(2)R) , (b) (GMm)/(R^(2))`
C. `(a) F=-(2GMm)/(piR^(2)) (b) -(GMm)/R`
D. `(a) F=(2GMm)/(piR^(2)) (b) (GMm)/R`

1 Answer

0 votes
by (67.7k points)
 
Best answer
Correct Answer - D
image
`dM=(M/(piR))(R d theta)=M/(pi).d theta`
Required force `=int_(0)^(pi//2)2 F sin theta` (towards `C`)
`=int_(0)^(pi//2)(2Gm(dM))/(R^(2))sin theta`
`=int_(0)^(pi)(2GMm)/(piR^(2))sintheta.d theta=(2GMm)/(piR^(2))`
`(b) U=-(GMm)/R`
`:.` Binding energy `=|U|=(GMm)/R`
i.e., this much energy is required to displace the particle from centre of curvature to infinity.

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

...