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Q.
The value of $g$ on the surface of earth is smallest at the equator because
NTA AbhyasNTA Abhyas 2020Gravitation
Solution:
Due to rotation of earth the effective acceleration due to gravity $g^{′}=g-R\omega ^{2}c o s^{2}\lambda .$
For a given point on the surface of earth $g$ decreases as $\omega $ increases. The angular speed of earth is maximum at equator hence, the value of $g$ on the surface of the earth is smallest.