For a thin rod of length L, rotating about its center, the radius of gyration is:

This question was previously asked in
RRB JE CE 22 Apr 2025 Shift 2 CBT 2 Official Paper
View all RRB JE Papers >
  1. L
  2. \(\rm \frac{L}{\sqrt 12}\)
  3. \(\rm \frac{L}{\sqrt 3}\)
  4. \(\rm \frac{L}{\sqrt 2}\)

Answer (Detailed Solution Below)

Option 2 : \(\rm \frac{L}{\sqrt 12}\)
Free
General Science for All Railway Exams Mock Test
2.1 Lakh Users
20 Questions 20 Marks 15 Mins

Detailed Solution

Download Solution PDF

Explanation:

The radius of gyration k for a thin rod of length L rotating about its center is given by the formula: \(\rm \frac{L}{\sqrt 12}\)

The radius of gyration represents the distance from the axis of rotation at which the entire mass of the body can be thought of as being concentrated, to give the same moment of inertia.

Additional InformationRadius of Gyration:
  • The radius of gyration is a measure of how far the mass of an object is distributed from its axis of rotation.

  • It helps to simplify complex shapes by representing the distribution of mass with a single distance.

  • It is used in the calculation of the moment of inertia, which describes an object's resistance to rotational motion.

Latest RRB JE Updates

Last updated on Jun 7, 2025

-> RRB JE CBT 2 answer key 2025 for June 4 exam has been released at the official website.

-> Check Your Marks via RRB JE CBT 2 Rank Calculator 2025

-> RRB JE CBT 2 admit card 2025 has been released. 

-> RRB JE CBT 2 city intimation slip 2025 for June 4 exam has been released at the official website.

-> RRB JE CBT 2 Cancelled Shift Exam 2025 will be conducted on June 4, 2025 in offline mode. 

-> RRB JE CBT 2 Exam Analysis 2025 is Out, Candidates analysis their exam according to Shift 1 and 2 Questions and Answers.

-> The RRB JE Notification 2024 was released for 7951 vacancies for various posts of Junior Engineer, Depot Material Superintendent, Chemical & Metallurgical Assistant, Chemical Supervisor (Research) and Metallurgical Supervisor (Research). 

-> The selection process includes CBT 1, CBT 2, and Document Verification & Medical Test.

-> The candidates who will be selected will get an approximate salary range between Rs. 13,500 to Rs. 38,425.

-> Attempt RRB JE Free Current Affairs Mock Test here

-> Enhance your preparation with the RRB JE Previous Year Papers

More Moment of Inertia and Centroid Questions

Get Free Access Now
Hot Links: teen patti stars teen patti lotus mpl teen patti teen patti party