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
6.9k views
in Physics by (50.3k points)

A stone of mass m, tied to the end of a string, is whirled around in a horizontal circle. (Neglect the force due to gravity.) The length of the string is reduced gradually, keeping the angular momentum of the stone about the centre of the circle constant. Then, the tension in the string is given by T = Arn , where A is a constant, r is the instantaneous radius of the circle, and n is 

(a) 1 

(b) -1 

(c) -2 

(d) -3

1 Answer

+1 vote
by (51.2k points)
selected by
 
Best answer

Correct Answer is: (d) -3

L = mvr = constant.

T = mv2 / r = m/r. L/m2r2 = L/m r-3 .

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

...