| ⇦ | 
| ⇨ |
The angular speed of earth, so that the object on equator may appear weightless is(g = 10 m/s², radius of earth=6400 km)
Options
(a) 1.25 x 10⁻³ rad/s
(b) 1.56 x 10⁻³ rad/s
(c) 1.25 x 10⁻1 rad/s
(d) 1.56 rad/s
Correct Answer:
1.25 x 10⁻³ rad/s
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
Related Questions: - A ball is dropped from the root of a tower of height h
- To draw the maximum current from a combination of cells, how should be the cells
- In India, electricity is supplied for domestic use at 200V. It is supplied at 110 V
- For transistor action (1) Base, emitter and collector regions should have similar size
- A force F=ai+3j+6k is acting at a point r=2i-6j-12k. The value of α for
Topics: Gravitation
(63)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A ball is dropped from the root of a tower of height h
- To draw the maximum current from a combination of cells, how should be the cells
- In India, electricity is supplied for domestic use at 200V. It is supplied at 110 V
- For transistor action (1) Base, emitter and collector regions should have similar size
- A force F=ai+3j+6k is acting at a point r=2i-6j-12k. The value of α for
Topics: Gravitation (63)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Solution:
ω=√(g/R)=√(10/6400*10^3)=1.25*10^(-3) rad/s