Question

The angular velocity of the earth's rotation about its axis is ω. An object weighed by a spring balance gives the same reading at the equator as at height h above the poles, the value of h will be
A $ \dfrac{{{\omega ^2}{R^2}}}{g} $
B $ \dfrac{{{\omega ^2}{R^2}}}{{2g}} $
C $ \dfrac{{2{\omega ^2}{R^2}}}{g} $
D $ \dfrac{{2{\omega ^2}{R^2}}}{{3g}} $

Answer 1

Hint
as the spring balance gives the same readings at equator and poles so the apparent weight at equator and poles are equal. On comparing the values of both weights, we get the value of height h.

Complete step by step answer
The angular velocity of earth rotation is ω.
As the spring balance gives the same reading at equator and poles so firstly, we write the apparent weight at equator and poles.
Now, at the equator it experiences the pseudo force $ m{\omega ^2}R $ outward due to the rotation of the earth so the apparent weight is
 $ {W_{equator}} = mg - m{\omega ^2}R $
As there is no pseudo force acting at the poles so the apparent weight at the poles is
 $ {W_{poles}} = mg\left( {1 - \dfrac{{2h}}{R}} \right) $
Where, R is the radius of earth
g is acceleration due to gravity
ω is the angular velocity of the earth
now, compare the value of both weights, we get
 $ \Rightarrow {W_{poles}} = {W_{equator}} $
 $ \Rightarrow mg\left( {1 - \dfrac{{2h}}{R}} \right) = mg - m{\omega ^2}R $
On rearranging above equation, we get
 $ \Rightarrow \dfrac{{2hg}}{R} = {\omega ^2}R $
 $ \Rightarrow h = \dfrac{{{\omega ^2}{R^2}}}{{2g}} $
Hence, value of the h is $ \dfrac{{{\omega ^2}{R^2}}}{{2g}} $
Therefore, option (B) is correct.

Note
Here, students must know the apparent weights at poles and equator and it should be noticed that the pseudo force due to the rotation of the earth is only acting at the equator not at the poles. After writing these and rearranging we can easily get the height h.