The acceleration due to gravity near the surface of a planet of radius R and density d

⇦

⇨

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:

1. 1 Wb/m² equal to
2. The threshold frequency for a certain photosensitive metal is ν₀. When it is illuminated
3. If vectors A=cos wti + sin wtj and B= cos wt/2 i + sin wt/2 j are functions of time,
4. The kirchoff’s first law (∑i=0) and second law (∑iR=∑E), where the symbols
5. In a region, the potential is represented by V(x,y,z)=6x-8xy-8y+6yz, where V

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