Question

The length of a second hand in a watch is $1\text{cm}$. The magnitude of change in velocity of its tip in $15$ seconds is

• A. $\text{zero}$
• B. $\frac{\pi }{30\sqrt{2}}\text{cm/s}$
• C. $\frac{\pi }{30}\text{cm/s}$
• D. $\frac{\pi \sqrt{2}}{30}\text{cm/s}$

Answer 1

The correct option is D $\frac{\pi \sqrt{2}}{30}\text{cm/s}$

Let us mark the position of the second hand's tip as $1$ and $2$ as shown


So, we have
$\overrightarrow{v_1}=v\hat{i}$
$\overrightarrow{v_2}=-v\hat{j}$
$\overrightarrow{v_2}-\overrightarrow{v_1}=-v(\hat{i}+\hat{j})$
$\Rightarrow |{\overrightarrow{v}}_2-{\overrightarrow{v}}_1|=\sqrt{(-v{)}^2+(-v{)}^2}=\sqrt{2}v$

As we know that $v=\omega r\text{and}\omega =\frac{2\pi }{60}$
$\therefore$ $|{\overrightarrow{v}}_2-{\overrightarrow{v}}_1|=\sqrt{2}v=\sqrt{2}\times \omega r$
$=\sqrt{2}\times \frac{2\pi }{60}\times 1\text{cm}=\frac{\pi \sqrt{2}}{30}\text{cm/s}$