1. Weight shown by spring balance

A spring balance reads the apparent weight.

• At the equator, apparent acceleration due to gravity is reduced by centrifugal effect: $g_{\text{eq}} = g - \omega^2 R$ since at the equator, distance from the rotation axis is $R$.

So, if mass of the object is $m$, weight at equator is $W_{\text{eq}} = m(g - \omega^2 R)$

2. Weight at height $h$ above the poles

At the poles, there is no centrifugal effect because the point lies on the axis of rotation.

Hence, only gravitational acceleration changes with height.

At height $h$ above the pole, $g_h = g\left(\frac{R}{R+h}\right)^2$

Given $h \ll R$, use binomial approximation: $\left(\frac{R}{R+h}\right)^2 = \left(1+\frac{h}{R}\right)^{-2} \approx 1 - \frac{2h}{R}$

Thus, $g_h \approx g\left(1 - \frac{2h}{R}\right)$

So the weight there is $W_h = mg\left(1 - \frac{2h}{R}\right)$

3. Given weights are same

According to the question, $W_{\text{eq}} = W_h$

So, $m(g - \omega^2 R) = mg\left(1 - \frac{2h}{R}\right)$

Cancel $m$: $g - \omega^2 R = g - \frac{2gh}{R}$

$\omega^2 R = \frac{2gh}{R}$

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

4. Match with options

This corresponds to: $\boxed{\frac{R^2\omega^2}{2g}}$

So the correct option is A.