The acceleration due to gravity on the earth's surface

Question:

The acceleration due to gravity on the earth's surface at the poles is $\mathrm{g}$ 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 $\mathrm{h}$ above the poles by using a spring balance. If the weights are found to be same, then $h$ is : ( $h<

  1. $\frac{\mathrm{R}^{2} \omega^{2}}{8 \mathrm{~g}}$

  2. $\frac{\mathrm{R}^{2} \omega^{2}}{4 \mathrm{~g}}$

  3. $\frac{R^{2} \omega^{2}}{g}$

  4. $\frac{\mathrm{R}^{2} \omega^{2}}{2 \mathrm{~g}}$


Correct Option: , 4

Solution:

$g_{e}=g-R \omega^{2}$

$\mathrm{g}_{2}=\mathrm{g}-\frac{2 \mathrm{gh}}{\mathrm{R}}$

Now $R \omega^{2}=\frac{2 g h}{R}$

$\mathrm{h}=\frac{\mathrm{R}^{2} \omega^{2}}{2 \mathrm{~g}}$

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