A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration (a) is varying with time t as a = k2 rt2, where k is a constant. The power delivered to the particle by the force acting on it is given as
A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration (a) is varying with time t as a = k2 rt2, where k is a constant. The power delivered to the particle by the force acting on it is given as
Option 1 - <p>zero</p>
Option 2 - <p>mk<sup>2</sup> r<sup>2</sup> t<sup>2</sup></p>
Option 3 - <p>mk<sup>2</sup> r<sup>2</sup> t</p>
Option 4 - <p>mk<sup>2</sup> rt</p>
2 Views|Posted 4 months ago
Asked by Shiksha User
1 Answer
R
Answered by
4 months ago
Correct Option - 3
Detailed Solution:
a = k2rt2
? tangential force, Ft = mat = mkr
Note ® Power delivered by centripetal force will be zero.
Similar Questions for you
From A to B the process is isobaric

= W = 2 × R (600 - 350)
= 500 R
Taking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else.
On Shiksha, get access to
66K
Colleges
|
1.2K
Exams
|
6.9L
Reviews
|
1.8M
Answers
Learn more about...

Physics NCERT Exemplar Solutions Class 12th Chapter Eight 2025
View Exam DetailsMost viewed information
SummaryDidn't find the answer you were looking for?
Search from Shiksha's 1 lakh+ Topics
or
Ask Current Students, Alumni & our Experts
Have a question related to your career & education?
or
See what others like you are asking & answering






