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We are given that the minute hand of a clock is 1.5 cm long.

We need to find the distance covered by its tip in 40 minutes. Its tip is obviously the last point on the hand which moves in a circular motion. Hence, it will cover the perimeter. So, we basically need to find that part of circumference, which can cover in 40 minutes.

We know that $Circumference = 2\pi r$, where r is the radius of the circle. So, when we cover $2\pi = {360^ \circ }$, we get $Circumference = 2\pi r$.

Hence, r is multiplied to the angle covered.

Hence, now we have: \[l = r.\theta \], where $l$ is the arc length, r is the radius that is the length of the minute hand and $\theta $ is the angle covered.

Now, we know about the formula a bit more.

We also know that 1 hr = 60 minutes.

Hence, a minute hand will cover the whole \[{360^ \circ }\] in 60 minutes.

So, 60 minutes are equivalent to \[{360^ \circ }\].

$ \Rightarrow $ 1 minute is equivalent to $\dfrac{{{{360}^ \circ }}}{{60}} = {6^ \circ }$.

$ \Rightarrow $ 40 minutes are equivalent to $40 \times {6^ \circ } = {240^ \circ }$.

Now, we need the angle in radians.

${180^ \circ } = \pi $

$ \Rightarrow {1^ \circ } = \dfrac{\pi }{{180}}$

$ \Rightarrow {240^ \circ } = \dfrac{\pi }{{180}} \times 240 = \dfrac{{4\pi }}{3}$ ……….(1)

Now, coming to the formula \[l = r.\theta \].

$ \Rightarrow l = (1.5) \times \dfrac{{4\pi }}{3}$ (Using (1))

$ \Rightarrow l = (1.5) \times \dfrac{{4 \times 3.14}}{3} = 6.28cm$ (Since, $\pi = 3.14$)

Hence, $l = 6.28cm$ is the distance covered by its tip in 40 minutes.

The students might forget to put the unit of arc length at the end, but they must remember that length is measured in units only. If I write 5 that would not signify anything. It would just be a number, not length or volume or area.