MCQ
The difference between angular speed of minute hand and second hand of a clock is
- A$\frac{59 \pi}{900} rad / s$
- ✓$\frac{59 \pi}{1800} rad / s$
- C$\frac{59 \pi}{2400} rad / s$
- D$\frac{59 \pi}{3600} rad / s$
✓
Answer
Correct option: B.
$\frac{59 \pi}{1800} rad / s$
(b) : Angular speed, $\omega=\frac{\text { Angular distance }(\theta)}{\text { Time taken }(t)}$
For minute hand of a clock, $\theta=2 \pi rad , t =3600 s$
$
\therefore \quad \omega_m=\frac{2 \pi}{3600}=\frac{\pi}{1800} rad s ^{-1}
$
For second hand of a clock, $\theta=2 \pi rad , t=60 s$
$
\therefore \quad \omega_s=\frac{2 \pi}{60}=\frac{\pi}{30} rad s ^{-1}
$
Required difference between angular speeds
$
\begin{aligned}
& =\omega s -\omega m \\
& =\frac{\pi}{30}-\frac{\pi}{1800}=\frac{60 \pi-\pi}{1800}=\frac{59 \pi}{1800} rad s ^{-1}
\end{aligned}
$
For minute hand of a clock, $\theta=2 \pi rad , t =3600 s$
$
\therefore \quad \omega_m=\frac{2 \pi}{3600}=\frac{\pi}{1800} rad s ^{-1}
$
For second hand of a clock, $\theta=2 \pi rad , t=60 s$
$
\therefore \quad \omega_s=\frac{2 \pi}{60}=\frac{\pi}{30} rad s ^{-1}
$
Required difference between angular speeds
$
\begin{aligned}
& =\omega s -\omega m \\
& =\frac{\pi}{30}-\frac{\pi}{1800}=\frac{60 \pi-\pi}{1800}=\frac{59 \pi}{1800} rad s ^{-1}
\end{aligned}
$
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