⇦ | ![]() | ⇨ |
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:
- A planet moving along an elliptical orbit is closest to the sun at a distance of r₁
- The potential differences across the resistance, capacitance and inductance
- A particle executing SHM with amplitude of 0.1 m. At a certain instant,
- The ionization energy of Li⁺⁺ is equal to
- Let A = i Acosθ +j B sinθ be any vector.Another vector B which is normal to A is
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