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Q.
The minute hand of a watch is $1.5\, cm$ long. The distance travelled by the minute hand in $40$ minutes is equal to
Trigonometric Functions
Solution:
In $60 \min$, the minute hand of watch completes one revolution. Therefore, in $40 \min$ the minute hand turns through $\frac{2}{3}$ of a revolution. Therefore, $\theta=\frac{2}{3} \times 360^{\circ}$ or $\frac{4 \pi}{3}$ radian.
Hence, the required distance travelled is given by
$I=r \theta=1.5 \times \frac{4 \pi}{3} cm =2 \pi \,cm =2 \times 3.14\, cm =6.28 \,cm$