MENU

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 Physics Question

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. Length of second pendulum is
  2. The equations of motion of a projectile are given by x=36 t metre and 2y=96t-9.8t²
  3. Suppose for some reasons, radius of earth were to shrink by 1% of present value,
  4. A body of mass 0.5 kg travels in a straight line with velocity v=a x³/², where
  5. Surface tension of a solution is 30 x 10⁻² N/m. The radius of the soap bubble is 5 cm

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.

NEETLab Mobile App

Share this page with your friends

Be the first to comment

Leave a Reply

Your email address will not be published.


*