| ⇦ |
| ⇨ |
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 rod of weight W is supported by two parallel knife edges A and B is in equilibrium
- Heat is not being exchanged in a body. If its internal energy is increased, then
- In an interference experiment, third bright fringe is obtained at a point
- A block of mass m moving at a velocity v collides with another block of mass 2m
- A thermocouple of negligible resistance produces an e.m.f. of 40 µV/⁰C in the linear range
Topics: Gravitation
(63)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A rod of weight W is supported by two parallel knife edges A and B is in equilibrium
- Heat is not being exchanged in a body. If its internal energy is increased, then
- In an interference experiment, third bright fringe is obtained at a point
- A block of mass m moving at a velocity v collides with another block of mass 2m
- A thermocouple of negligible resistance produces an e.m.f. of 40 µV/⁰C in the linear range
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