Question

In one second the earth rotates about its axis by an angle θ where

(a) θ < 10-6  radian

(b) 10-6 radian< θ< 10-5 radian

(c) 10-5 radian < θ < 10-4 radian

(d) θ > 10-4 radian​

Answer 1

Answer:

$10^{-5}$ radian $< \theta < 10^{-4}$ radian.

Explanation:

Step 1: The sidereal period, in which the planet spins once every 23 hours, 56 minutes, and 4.09053 seconds, measures around 40,075 kilometres around the globe. The speed of the earth's surface near the equator is therefore 460 metres per second, or around 1,000 miles per hour.

Let's first find the angular speed of Earth's rotation about its axis:

$\begin{gathered}\omega=\frac{\Delta \theta}{\Delta t}, \\\omega=\frac{2 \pi \mathrm{rad}}{1 d} \cdot \frac{1 d}{24 h \cdot 3600 \frac{\mathrm{s}}{1 \mathrm{~h}}}=7.27 \cdot 10^{-5} \frac{\mathrm{rad}}{\mathrm{s}} .\end{gathered}$

Step 2: Then, in one second the Earth rotates about its axis by an angle $\theta$ equals:

$\begin{gathered}\theta=\omega t, \\\theta=7.27 \cdot 10^{-5} \frac{\mathrm{rad}}{\mathrm{s}} \cdot 1 \mathrm{~s}=7.27 \cdot 10^{-5} \mathrm{rad} .\end{gathered}$

Therefore, the correct answer is (c).

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