# The acceleration due to gravity on the earth's surface at the poles is g

Question:

The acceleration due to gravity on the earth's surface at the poles is $g$ and angular velocity of the earth about the axis passing through the pole is $\omega$. An object is weighed at the equator and at a height $h$ above the poles by using a spring balance. If the weights are found to be same, then $h$ is : $(h< 1. (1)$\frac{R^{2} \omega^{2}}{2 g}$2. (2)$\frac{R^{2} \omega^{2}}{g}$3. (3)$\frac{R^{2} \omega^{2}}{4 g}$4. (4)$\frac{R^{2} \omega^{2}}{8 g}$Correct Option: , 2 Solution: (2) Value of$g$at equator,$g_{A}=g \cdot-R \omega^{2}$Value of$g$at height$h$above the pole,$g_{B}=g \cdot\left(1-\frac{2 h}{R}\right)$As object is weighed equally at the equator and poles, it means$g$is same at these places.$g_{A}=g_{B}\Rightarrow g-R \omega^{2}=g\left(1-\frac{2 h}{R}\right)\Rightarrow R \omega^{2}=\frac{2 g h}{R} \Rightarrow h=\frac{R^{2} \omega^{2}}{2 g}\$

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