Answer

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**Hint**

We know that a centripetal force is a force that makes a body follow a curved path. The direction is always orthogonal to the motion of the body and towards the fixed point of the instantaneous centre or curvature of the path. It is a force which is necessary to keep an object moving in a curved path and that is directed inward toward the centre of rotation of a string on the end of which a stone is whirled about exerts centripetal force on the force. Based on this concept we have to answer this question.

**Complete step by step answer**

The time period of earth's rotation or the time that the earth takes to rotate around its own axis is,

$ T=24\text{h}=86400\text{s} $

We know that the angular velocity $ \omega=2 \pi / \mathrm{T} $

Radius of the earth is represented as $ \mathrm{r}=6400 \mathrm{km}=640000 \mathrm{m} $

Now, we can equate the centripetal acceleration as $ \omega^2\text{r}={(\dfrac{2\pi }{\text{T}})}^2 2\text{r}=0.034\text{m}/{{\text{s}}^{2}} $ .

Therefore, the total centripetal acceleration of a point on Earth’s equator is $ 3.4\dfrac{cm}{se{{c}^{2}}} $ .

**Hence, the correct answer is Option (B).**

**Note**

We know that time period is defined as the time that is taken for one complete cycle of vibration to pass a given point. As the frequency of a wave increases, the time period of the wave decreases. Frequency and time period are in a reciprocal relationship that can be expressed mathematically as T = 1 / f.

The rotation period is calculated using the formula rotation period = rotation speed divided by the planet’s circumference, where the circumference is the pi times the diameter.

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