JEE MAIN - Physics (2023 - 6th April Evening Shift - No. 10)
As shown in the figure, a particle is moving with constant speed $$\pi ~\mathrm{m} / \mathrm{s}$$. Considering its motion from $$\mathrm{A}$$ to $$\mathrm{B}$$, the magnitude of the average velocity is :
_6th_April_Evening_Shift_en_10_1.png)
$$\pi ~\mathrm{m} / \mathrm{s}$$
$$1.5 \sqrt{3} \mathrm{~m} / \mathrm{s}$$
$$\sqrt{3} \mathrm{~m} / \mathrm{s}$$
$$2 \sqrt{3} \mathrm{~m} / \mathrm{s}$$
Explanation
The given situation is shown below.
_6th_April_Evening_Shift_en_10_2.png)
Given, $v=\pi \mathrm{m} / \mathrm{s}$
or
$$ \begin{aligned} R \omega & =\pi \\\\ \omega & =\frac{\pi}{R} \mathrm{rad} / \mathrm{s} \quad[\because v=\omega R] \end{aligned} $$
Angular displacement, $\theta=120^{\circ}$ or $\frac{2 \pi}{3}$
Using $\theta=\omega t$,
$$ t=\frac{\theta}{\omega}=\frac{2 \pi / 3}{\pi / R}=\frac{2 R}{3} $$
Linear displacement,
$$ \begin{aligned} d & =2 R \sin (\theta / 2) \\\\ d & =2 R \sin \left(\frac{120^{\circ}}{2}\right)=2 R \sin 60^{\circ} \\\\ & =2 R \frac{\sqrt{3}}{2} \Rightarrow R \sqrt{3} \end{aligned} $$
Average velocity $=\frac{d}{t}=\frac{R \sqrt{3}}{2 R / 3}=\frac{3 \sqrt{3}}{2}=1.5 \sqrt{3}\mathrm{~m} / \mathrm{s}$
Given, $v=\pi \mathrm{m} / \mathrm{s}$
or
$$ \begin{aligned} R \omega & =\pi \\\\ \omega & =\frac{\pi}{R} \mathrm{rad} / \mathrm{s} \quad[\because v=\omega R] \end{aligned} $$
Angular displacement, $\theta=120^{\circ}$ or $\frac{2 \pi}{3}$
Using $\theta=\omega t$,
$$ t=\frac{\theta}{\omega}=\frac{2 \pi / 3}{\pi / R}=\frac{2 R}{3} $$
Linear displacement,
$$ \begin{aligned} d & =2 R \sin (\theta / 2) \\\\ d & =2 R \sin \left(\frac{120^{\circ}}{2}\right)=2 R \sin 60^{\circ} \\\\ & =2 R \frac{\sqrt{3}}{2} \Rightarrow R \sqrt{3} \end{aligned} $$
Average velocity $=\frac{d}{t}=\frac{R \sqrt{3}}{2 R / 3}=\frac{3 \sqrt{3}}{2}=1.5 \sqrt{3}\mathrm{~m} / \mathrm{s}$
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