In $60$ minutes, the minute hand of a watch completes one revolution. Therefore, in $40$ minutes, the minute hand turns through $\frac{2}{3}$ of a revolution. Therefore, ${\theta  = 23 \times {{360}^\circ }}$ or $\frac{4 \pi}{3}$ radian. Hence, the required distance travelled is given by

$l=r \theta=1.5 \times \frac{4 \pi}{3} \,cm =2 \pi \,cm =2 \times 3.14 \,cm =6.28 \,cm$