The length of the minute hand of a clock is 5 | vedclass

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Question

(A) We know that,in $60\, 	ext{min}$,the minute hand revolves $360^{\circ}$.In $1\, 	ext{min}$,the minute hand revolves $\frac{360^{\circ}}{60} = 6^

Vedclass · Accepted Answer

$45 \frac{5}{6}$

Vedclass · Answer

(A) We know that,in $60\, 	ext{min}$,the minute hand revolves $360^{\circ}$.In $1\, 	ext{min}$,the minute hand revolves $\frac{360^{\circ}}{60} = 6^{\circ}$.Therefore,in the time interval from $6:05\, a.m.$ to $6:40\, a.m.$,the time elapsed is $35\, 	ext{min}$.The angle swept by the minute hand in $35\, 	ext{min}$ is $	heta = 6^{\circ} 	imes 35 = 210^{\circ}$.Given that the length of the minute hand $(r) = 5\, 	ext{cm}$.The area swept by the minute hand is the area of a sector with radius $r = 5\, 	ext{cm}$ and angle $	heta = 210^{\circ}$.Area $= \frac{	heta}{360^{\circ}} 	imes \pi r^2 = \frac{210^{\circ}}{360^{\circ}} 	imes \frac{22}{7} 	imes (5)^2$.Area $= \frac{7}{12} 	imes \frac{22}{7} 	imes 25 = \frac{22 	imes 25}{12} = \frac{11 	imes 25}{6} = \frac{275}{6}$.Converting to a mixed fraction,$\frac{275}{6} = 45 \frac{5}{6}\, 	ext{cm}^2$.Hence,the required area swept by the minute hand is $45 \frac{5}{6}\, 	ext{cm}^2$.

The length of the minute hand of a clock is $5\, cm$. Find the area swept by the minute hand during the time period $6:05\, a.m.$ and $6:40\, a.m.$ (in $cm^2$).

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