| ⇦ | 
| ⇨ |
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: - “On flowing current in a conducting wire the magnetic field is produced around it”.
- A generator at a utility company produces 100 A of current at 4000 V. The voltage
- A thin semicircular conducting ring (PQR) of radius r is falling with its plane
- If Q, E and W denote respectively the heat added, change in internal energy
- A long straight wire carries a certain current and produces a magnetic field
Topics: Gravitation
(63)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- “On flowing current in a conducting wire the magnetic field is produced around it”.
- A generator at a utility company produces 100 A of current at 4000 V. The voltage
- A thin semicircular conducting ring (PQR) of radius r is falling with its plane
- If Q, E and W denote respectively the heat added, change in internal energy
- A long straight wire carries a certain current and produces a magnetic field
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