A particle is moving in a circle of radius $R$ with constant speed. The time period of the particle is $T$ . In a time $t = \frac{T}{6}$ , average speed of the particle is

Question

A particle is moving in a circle of radius R with constant speed- The time period of the particle T in a time t-dfrac-T-6-

Toppr Admin · Accepted Answer

Correct option is C- -dfrac-2-pi R-T-We know that-theta-2-pi t- t-dfrac-2-pi-T-times t-Rightarrow -theta-dfrac-2-pi-T-times -dfrac-T-6-dfrac-pi-3-average speed -omega R-2-pi fR-dfrac-2-pi-T-times R-dfrac-2-pi R-T-

A particle is moving in a circle of radius $$R$$ with constant speed. The time period of the particle $$T$$ in a time $$t=\dfrac{T}{6}$$.

Correct option is C. $$\dfrac{2\pi R}{T}$$
We know that
$$\theta=2\pi t\ t=\dfrac{2\pi}{T}\times t$$
$$\Rightarrow \theta=\dfrac{2\pi}{T}\times \dfrac{T}{6}=\dfrac{\pi}{3}$$
average speed $$=\omega R=2\pi fR$$
$$=\dfrac{2\pi}{T}\times R=\dfrac{2\pi R}{T}$$

A particle is moving in a circle of radius $$R$$ with constant speed. The time period of the particle $$T$$ in a time $$t=\dfrac{T}{6}$$.

Correct option is C. $$\dfrac{2\pi R}{T}$$We know that$$\theta=2\pi t\ t=\dfrac{2\pi}{T}\times t$$$$\Rightarrow \theta=\dfrac{2\pi}{T}\times \dfrac{T}{6}=\dfrac{\pi}{3}$$average speed $$=\omega R=2\pi fR$$$$=\dfrac{2\pi}{T}\times R=\dfrac{2\pi R}{T}$$

Correct option is C. $$\dfrac{2\pi R}{T}$$
We know that
$$\theta=2\pi t\ t=\dfrac{2\pi}{T}\times t$$
$$\Rightarrow \theta=\dfrac{2\pi}{T}\times \dfrac{T}{6}=\dfrac{\pi}{3}$$
average speed $$=\omega R=2\pi fR$$
$$=\dfrac{2\pi}{T}\times R=\dfrac{2\pi R}{T}$$