⇦ | ![]() | ⇨ |
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:
- Consider two concentric spherical metal shells of radii r₁ and r₂( r₂>r₁). If the outer
- A beam of light of wavelength 600 nm from a distant source falls
- Two metallic spheres of radii 1 cm and 3 cm are given charges of -1 x 10⁻² C
- A solid cylinder of mass 50kg and radius 0.5 m is, free to rotate about the horizontal axis
- A 220 volts input is supplied to a transformer. The output circuit draws
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