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When a body is taken from pole to the equator its weight
Remains constant
Increases
Decreases
None
Solution
$g^{\prime}=g-R \omega^2 \cos (2 \lambda)$
$g^{\prime}=g-R \omega 2 \cos 2$ where $\omega \omega$ is the angular velocity of rotation of earth about its polar axis, $R R$ is the radius of the earth and $\lambda \lambda$ is the latitude of a plac
At poles, $\lambda=90$
$\lambda=90$
$\therefore g_p 0 l e=g-R \omega^2 \cos (2 \times 90)$
$\therefore$ gpole $=g-R \omega 2 \cos 2900=g$
at equator $\lambda=0$
gequator $= g – R \omega 2 \cos 200= g – R \omega 2< g$ gequator $= g – R \omega 2 \cos 200= g – R \omega 2$
Thus, the acceleration due to gravity decreases from poles to equator.
Hence, when a body is taken from poles to equator on the earth, its weight decreases.
