| ⇦ | 
| ⇨ |
The weight of a body on surface of earth is 12.6 N. When it is raised to a height half the radius of earth its weight will be
Options
(a) 2.8 N
(b) 5.6 N
(c) 12.6 N
(d) 25.2 N
Correct Answer:
5.6 N
Explanation:
Weight of body on the surface of earth, mg=12.6 N
At height h, the value of g’ is given by
g’=g (R/R+(R/2))²
=g 4/9
Weight at height, h=mg 4/9
h=12.6×4/9 =5.6 N
Related Questions: - A stationary particle explodes into two particles of masses m₁ and m₂ which move
- If F⃗ is the force acting on a particle having position vector r⃗ and ? ⃗ be the torque
- A cycle wheel of radius 4 m completes one revolution in two seconds. Then acceleration
- The power obtained in a reactor using U²³⁵ disintegration is 1000 kW
- A particle moves from position r₁ = 3i + 2j -6k to position r₂ = 14i + 13 j+ 9 k
Topics: Gravitation
(63)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A stationary particle explodes into two particles of masses m₁ and m₂ which move
- If F⃗ is the force acting on a particle having position vector r⃗ and ? ⃗ be the torque
- A cycle wheel of radius 4 m completes one revolution in two seconds. Then acceleration
- The power obtained in a reactor using U²³⁵ disintegration is 1000 kW
- A particle moves from position r₁ = 3i + 2j -6k to position r₂ = 14i + 13 j+ 9 k
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