When a body is taken from the equator to the poles, its weight:
Correct answer: B. Increases
- A. Remains same
- B. Increases
- C. Decreases
- D. Increase at N-pole & decreases at S-pole
Explanation
The following is the solution: g′=g−Rω2cos(2λ) g′=g−Rω2cos2λ, where ω is the angular velocity of rotation of Earth about its polar axis, R is the radius of the Earth and λ is the latitude of a place, At poles, λ=90, ∴g p ole=g−Rω2cos(2×90) at equator λ=0, g e quator=g−R(ω)2cos(2×0)=g−R(ω)2 Thus, the acceleration due to gravity decreases from the poles to the equator. Hence, when a body is taken from the poles to the equator on the Earth, its weight decreases.
Last updated
Related questions
A 2 kg body falls from infinity to the surface of earth. What will be the kinetic energy of the body on reaching the surface of Earth (Assume escape velocity to be 104 ms-1)?
A bird resting on the floor of an air-tight box which is being carried by a boy, starts flying. The boy will feel that the box is now:
A body in satellite orbiting round the earth is weightless because:
A body is taken from the earth's surface to the moon, the weight of the body will be zero at a point where force of attraction due to:
A body of 2kg is suspended from the ceiling of an elevator moving up with an acceleration 'g' its apparent weight in the elevator will be: