The linear speed of a particle at the equator of the earth due to its spin motion is ' V '. The linear speed…

1. $\frac{\mathrm{V}}{\sqrt{2}}$
2. $\frac{\mathrm{V}}{2}$
3. $\frac{\sqrt{3}}{2} \mathrm{~V}$
4. V

Solution

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), $\begin{aligned} & \frac{V^{\prime}}{V}=\frac{\frac{\sqrt{3}}{2} R \omega}{R \omega} \\ \therefore \quad & V^{\prime} \\ = & V \frac{\sqrt{3}}{2} \end{aligned}$

Asked in: MHT CET 2024 (16 May Shift 2)

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