| ⇦ | 
| ⇨ |
The acceleration due to gravity near the surface of a planet of radius R and density d is proportional to
Options
(a) d/R²
(b) dR²
(c) dR
(d) d/R
Correct Answer:
dR
Explanation:
g=GM/R²
(M=Mass of the earth); (R=Distance of body from centre of earth)
g=G Volume x density / R²
Volume of the sphere=4/3 πR³
Therefore, g=G.4/3 πR³.d / R²
g=G 4/3 πRd
g=4πG/3.dR
g is proportional to dR
Related Questions: - Under the action of a force F=cx, the position of a body changes from 0 to x.The work done
- A plane electromagnetic wave of frequency 20 MHz travels through a space along
- Eight drops of a liquid of density ρ and each of radius a are falling through air
- Three identical spheres, each of mass 3 kg are placed touching each other
- The wavelength of the short radio waves, microwaves, ultraviolet waves are λ₁,λ₂ and λ₃
Topics: Gravitation
(63)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Under the action of a force F=cx, the position of a body changes from 0 to x.The work done
- A plane electromagnetic wave of frequency 20 MHz travels through a space along
- Eight drops of a liquid of density ρ and each of radius a are falling through air
- Three identical spheres, each of mass 3 kg are placed touching each other
- The wavelength of the short radio waves, microwaves, ultraviolet waves are λ₁,λ₂ and λ₃
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
Leave a Reply