When the "wedge and scale" experiment is performed at the equator, we get \(m=M\left(y_{e} / x_{e}\right)\). If the same experiment is performed at the poles, then what is the right equation ? (A) $\mathrm{m}=\mathrm{M}\left(\mathrm{y}_{\mathrm{e}} / \mathrm{x}_{\mathrm{e}}\right) \cdot\left(\mathrm{R}_{\mathrm{e}} / \mathrm{R}_{\mathrm{p}}\right)$ (B) \(m=M\left(y_{e} / x_{e}\right) \cdot\left(R_{p} / R_{e}\right)\) (C) $\mathrm{m}=\mathrm{M}\left(\mathrm{y}_{\mathrm{e}} / \mathrm{x}_{\mathrm{e}}\right) \cdot\left(\mathrm{R}_{\mathrm{p}} / \mathrm{R}_{\mathrm{e}}\right)^{2}$ (D) $\mathrm{m}=\mathrm{M}\left(\mathrm{y}_{\mathrm{e}} / \mathrm{x}_{\mathrm{e}}\right)$ Where \(\mathrm{R}_{\mathrm{e}}\) and \(\mathrm{R}_{\mathrm{p}}\) are the equatorial and polar radius of the earth.

The given equation is \(m = M\left(\frac{y_e}{x_e}\right)\), where \(m\) represents the mass of the object at the equator, \(M\) is the mass of the object, and the fraction \(\frac{y_e}{x_e}\) is the ratio between certain lengths or distances in the experiment that are specific to the equator.

When the experiment is moved to the poles, the Earth's curvature changes, which will affect the distance ratio involved in the experiment. Since the Earth is oblate, the polar radius, \(R_p\), is smaller than the equatorial radius, \(R_e\). As a result, we expect the modified ratio in the equation to involve the ratio between the polar and equatorial radius.

Now that we know the modified ratio should involve the ratio between the polar and equatorial radii, we can examine the answer choices: (A) \(m = M\left(\frac{y_e}{x_e}\right) \cdot\left(\frac{R_e}{R_p}\right)\) (B) \(m = M\left(\frac{y_e}{x_e}\right) \cdot\left(\frac{R_p}{R_e}\right)\) (C) \(m = M\left(\frac{y_e}{x_e}\right) \cdot\left(\frac{R_p}{R_e}\right)^2\) (D) \(m = M\left(\frac{y_e}{x_e}\right)\) Since the fraction \(\frac{y_e}{x_e}\) already takes into account the lengths or distances in the experiment at the equator, our main task is to determine how the ratio of Earth's radii affects the mass at the poles. As Earth's polar radius is smaller than its equatorial radius, it means that the force or effect at the poles should be greater than that at the equator, leading us to choose an answer with a greater term multiplied by the ratio \(\frac{y_e}{x_e}\). Comparing our choices, answer (A) fits this criterion.

Therefore, the right equation when the "wedge and scale" experiment is performed at the poles is given by: (A) \(m = M\left(\frac{y_e}{x_e}\right) \cdot\left(\frac{R_e}{R_p}\right)\)