JEE MAIN - Physics (2023 - 31st January Evening Shift - No. 18)
A body is moving with constant speed, in a circle of radius $10 \mathrm{~m}$. The body completes one revolution in $4 \mathrm{~s}$. At the end of 3rd second, the displacement of body (in $\mathrm{m}$ ) from its starting point is :
$15 \pi$
30
$10 \sqrt{2}$
$5 \pi$
Explanation
$$
\begin{aligned}
& \mathrm{\omega}=\frac{2 \pi}{\mathrm{T}}=\frac{2 \pi}{4}=\frac{\pi}{2} \mathrm{rad} / \mathrm{s} \\\\
& \theta=\mathrm{\omega t} \\\\
& \theta=\frac{\pi}{2} \times 3 \\\\
& \theta=\frac{3 \pi}{2} \mathrm{rad}
\end{aligned}
$$
_31st_January_Evening_Shift_en_18_1.png)
$$ \begin{aligned} & r=10 \mathrm{~m} \\\\ & T=4 \mathrm{sec} \\\\ & d=\sqrt{2}(10) \mathrm{m} \end{aligned} $$
$$ \begin{aligned} & r=10 \mathrm{~m} \\\\ & T=4 \mathrm{sec} \\\\ & d=\sqrt{2}(10) \mathrm{m} \end{aligned} $$
Comments (0)
