Suppose that the angular velocity of rotation of the Earth is increased. Then, as a consequence:

We begin by recalling the definition of the true (gravitational) weight of a body of mass $$m$$ on the Earth, which is given by $$W_g = m g,$$ where $$g$$ is the magnitude of the acceleration due to gravity directed toward the centre of the Earth.

However, because the Earth is rotating with angular velocity $$\omega,$$ every point on its surface (except the poles) experiences an outward centrifugal acceleration. The formula for the centrifugal acceleration at any latitude is

$$a_c = \omega^2 r,$$

where $$r$$ is the perpendicular distance of the point from the axis of rotation. If the latitude of the point is denoted by $$\phi$$ (zero at the equator and $$90^{\circ}$$ at the poles), then the geometry of the Earth gives

$$r = R \cos\phi,$$

where $$R$$ is the radius of the Earth. Substituting this expression into the centrifugal-acceleration formula yields

$$a_c \;=\; \omega^2 \, (R\cos\phi).$$

The centrifugal acceleration is directed horizontally outward, but we need only the component that opposes gravity, i.e., the component along the radial line from the Earth’s centre. Geometry again tells us that this component is

$$a_{c,\text{radial}} = a_c \cos\phi = \omega^2 R \cos^2\phi.$$

The effective downward acceleration that actually presses the body against the ground (and hence decides the reading on a weighing machine) is therefore

$$g_{\text{eff}} = g \;-\; \omega^2 R \cos^2\phi.$$

Accordingly, the apparent or measured weight becomes

$$W_{\text{apparent}} = m g_{\text{eff}} \;=\; m\Bigl(g - \omega^2 R \cos^2\phi\Bigr).$$

Now imagine that the Earth’s angular velocity of rotation is increased from its present value to some larger value $$\omega'$$ with $$\omega' > \omega.$$ In the expression for $$g_{\text{eff}}$$ we must then replace $$\omega$$ by $$\omega'.$$ Because $$\omega'$$ is larger, the term $$\omega'^2 R\cos^2\phi$$ is also larger, while $$g$$ and $$R$$ remain essentially unchanged. Thus

$$g_{\text{eff,new}} = g - \omega'^2 R \cos^2\phi \;<\; g - \omega^2 R \cos^2\phi = g_{\text{eff,old}}$$

for every latitude where $$\cos\phi \neq 0,$$ that is, for every point except the poles.

At the poles themselves, the latitude is $$\phi = 90^{\circ},$$ so $$\cos\phi = 0.$$ In that special case the centrifugal term vanishes identically, and we have

$$g_{\text{eff,pole}} = g$$

both before and after the increase in $$\omega.$$ Hence there is no change in weight at the poles.

Collecting these observations, we see that increasing the Earth’s spin causes the apparent weight to decrease everywhere except at the two poles, where it remains the same.

Hence, the correct answer is Option 3.