Question

The minute hand of a watch is $1.5cm$ long. The distance travelled by the minute hand in $40$ minutes is equal to

• A. $3.28cm$
• B. $4.28cm$
• C. $5.28cm$
• D. $6.28cm$

Answer 1

Answer: (D)

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