MCQ
The length of seconds hand of a watch is $1\ cm.$ The change in velocity of its tip in $15$ seconds in $cm/ s$ is:
- A$\text{zero}$
- B$\frac{\text{x}}{(30\sqrt{2})}$
- C$\frac{\pi}{30}$
- ✓$\frac{2\pi}{(30\sqrt{2})}$
✓
Answer
Correct option: D.
$\frac{2\pi}{(30\sqrt{2})}$
The angle described at the centre by the length of second hand of a watch in $15$ second
$=90^\circ=\frac{\pi}{2}$ radians Figure.
Linear speed $\text{v}=\text{r}\omega$
$=\frac{\text{r}\theta}{\text{t}}$
$=\frac{1\times\big(\frac{\pi}{2}\big)}{15}$
$=\frac{\pi}{30}\text{cm s}^{-1}$
Magnitude of change in velocity in $15 \sec;$
$|\Delta\vec{\text{v}}|=|\vec{\text{v}}_2-\vec{\text{v}}_1|$
$=\sqrt{\text{v}^2_2-\text{v}_1^2}$
$=\sqrt{\text{v}^2+\text{v}^2}$
$=\sqrt{2}\text{v}$
$=\sqrt{2}\frac{\pi}{30}$
$=\frac{2\pi}{30\sqrt{2}}$
$=90^\circ=\frac{\pi}{2}$ radians Figure.
Linear speed $\text{v}=\text{r}\omega$
$=\frac{\text{r}\theta}{\text{t}}$
$=\frac{1\times\big(\frac{\pi}{2}\big)}{15}$
$=\frac{\pi}{30}\text{cm s}^{-1}$
Magnitude of change in velocity in $15 \sec;$
$|\Delta\vec{\text{v}}|=|\vec{\text{v}}_2-\vec{\text{v}}_1|$
$=\sqrt{\text{v}^2_2-\text{v}_1^2}$
$=\sqrt{\text{v}^2+\text{v}^2}$
$=\sqrt{2}\text{v}$
$=\sqrt{2}\frac{\pi}{30}$
$=\frac{2\pi}{30\sqrt{2}}$
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