Question

The minute hand of a watch is$1.5cm$ along. How far does its tip move in $40$minutes?

Answer 1

Find the how far does its tip move.

Given:

Minute hand of a watch is$1.5cm$

We know that$\mathbf{}\mathbf{l}\mathbf{=}\mathbf{r\theta }$

Where $l$ -length of the arc, $r-$radius of the circle,$\theta -$angle swipe by minute hand(in radian)

Hence$r=1.5cm$

Now we know that minute hand make $1$ revolution in $1-hour$

Hence we can say

In one hour, the minute hand completes ${360}^{\circ }$

So degree complete in

$1min=\frac{360°}{60}$

So degree complete in

$40\text{min}=\frac{360°}{60}\times 40$

$={240}^{\circ }$

Hence, $\theta ={240}^{\circ }$

Radian measure $=\frac{\mathrm{\pi }}{180°}\times \theta$

$=$ $\frac{\mathrm{\pi }}{180°}\times 240°$

$=$ $\frac{4\mathrm{\pi }}{3}$

Now$\mathbf{}\mathbf{l}\mathbf{=}\mathbf{r\theta }$

$\Rightarrow$ $l=1.5\times \frac{4\mathrm{\pi }}{3}$

$\Rightarrow$ $l=0.5\times 4\pi$

$\Rightarrow$ $l=2\pi$

$=2\times 3.14$ $\left[\mathrm{Taking}\pi =3.14\right]$

$\Rightarrow$ $l=6.28cm$

Hence, in $\mathbf{40}$ minutes, the tip of the minute hand move by$\mathbf{}\mathbf{6}\mathbf{.}\mathbf{28}\mathbf{}\mathbf{cm}\mathbf{.}$