The linear speed of a particle at the equator of the earth due to its spin motion is ' V '. The linear speed of the particle at latitude \(30^{\circ}\) is \(\left[\begin{array}{l} \sin 30^{\circ}=\cos 60^{\circ}=1 / 2 \ \cos 30^{\circ}=\sin 60^{\circ}=\sqrt{3} / 2 \end{array}\right]\)

The particle on the equator of the earth has linear speed \(V\).
\(\Rightarrow V=R \omega...(i)\)
where R is the radius of the earth.

If the particle is now at \(30^{\circ}\) latitude, the particle will move in a circle of smaller radius.
Let the radius of this smaller circle be r .
\(\Rightarrow \mathrm{r}=\mathrm{R} \cos \theta\)
\(\therefore \quad\) The linear velocity of the particle at \(30^{\circ}\),
\(\begin{aligned}
V^{\prime} & =r \omega \\
& =R \cos \theta \omega=R \cos 30^{\circ} \omega=\frac{\sqrt{3}}{2} R \omega
\end{aligned}...(ii)\)
Dividing equation (ii) by (i),
\(\begin{aligned}
& \frac{V^{\prime}}{V}=\frac{\frac{\sqrt{3}}{2} R \omega}{R \omega} \\
\therefore \quad & V^{\prime} \\
= & V \frac{\sqrt{3}}{2}
\end{aligned}\)