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
177 views
in Physics by (76.7k points)
closed by
Initial velocity and acceleration of a particles are as shown in the figure. Acceleration vector of particle remain constant. Then radius of curvature of path of particle `:`
image
A. is `9m` initially
B. is `(9)/(sqrt(3))m` initially
C. will have minimum value of `(9)/(8)m`
D. will have minimum value `(3)/(8)m`

1 Answer

0 votes
by (76.4k points)
selected by
 
Best answer
Correct Answer - A::C
Initially `ROC=(v^(2))/(a sin 30^(@))=(9)/(1)m`
For minimum `ROC=((vsin 30^(@))^(2))/(a)=(9)/(8)m`.
image

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

...