Home
Class 11
PHYSICS
A body weights 98 N on a spring balance ...

A body weights 98 N on a spring balance at the north pole. What will be its weight recorded on the same scale if it is shifted to te equator? Use g=`GM/R^2=9.8m/s^2` and the radius of the earth R=6400 km.

Text Solution

AI Generated Solution

To solve the problem, we need to determine the weight of the body at the equator given its weight at the North Pole and the effects of centrifugal force due to the Earth's rotation. ### Step-by-step Solution: 1. **Determine the Mass of the Body:** The weight of the body at the North Pole is given as 98 N. The weight (W) is related to mass (m) and acceleration due to gravity (g) by the formula: \[ W = m \cdot g ...
Promotional Banner

Topper's Solved these Questions

  • CIRCULAR MOTION

    HC VERMA ENGLISH|Exercise worked out Examples|13 Videos
  • CIRCULAR MOTION

    HC VERMA ENGLISH|Exercise Objective -1|16 Videos
  • CENTRE OF MASS, LINEAR MOMENTUM, COLLISION

    HC VERMA ENGLISH|Exercise Objective -2|11 Videos
  • FLUID MECHANICS

    HC VERMA ENGLISH|Exercise All Questions|90 Videos

Similar Questions

Explore conceptually related problems

The reading of a spring balance corresponds to 100 N while situated at the north pole and a body is kept on it. The weight record on the same scale if it is shifted to the equator, is (take, g = 19 " ms"^(2-) and radius of the earth, R = 6.4 xx 10^(6) m)

A body is weighted by a spring balance to be 1.000 kg at the north pole. How much will it weight at the equator. Account for the earth\'s rotation only.

If weight of an object at pole is 196 N then weight at equator is [ g = 10 m/s^2 , radius of earth = 6400 Km]

Suppose the earth increases its speed of rotation . At what new time period will the weightof a body on the equator becomes zero? Take g=10 m/s^2 and radius of the earth R=6400 km .

If g on the surface of the earth is 9.8 m//s^(2) , find its value at a height of 6400 km. (Radius of the earth = 6400 km)

If 'g' on the surface of the earth is 9.8ms^(-2) , find its value at a depth of 3200 km (radius of the earth = 6400 km)

What is the value of g at a height 8848 m above sea level. Given g on the surface of the earth is 9.8 ms^(-2) . Mean radius of the earth = 6.37 xx 10^(6) m .

A body is taken from equator to pole. What happens to (1) its mass ( 2) its weight.

A particle is dropped down in a deep hole which extends to the centre of the earth. Calculate the velocity at a depth of one km from the surface of this carth ? Assume that g= 10 m//s^2 and radius of the earth = 6400 km.

What is the time period of rotation of the earth around its axis so that the objects at the equator becomes weightless ? ( g = 9.8 m//s^(2) , Radius of earth= 6400 km)