The length of the minute hand of a clock is $5 c m$ . The area swept by the minute hand during the time period $6 : 05 a m$ and $6 : 40 a m$ is

22 \frac{5}{6} c m^{2}

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9 \frac{5}{6} c m^{2}

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21 \frac{5}{6} c m^{2}

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45 \frac{5}{6} c m^{2}

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Solution

The correct option is D $45 \frac{5}{6} c m^{2}$
$G i v e n - t h e l e n g t h l o f t h e m i n u t e h a n d o f a c l o c k = 5 c m . T o f i n d o u t - t h e a r e a s w e p t b y t h e m i n u t e h a n d d u r i n g t h e p e r i o d 6 : 05 a . m t o 6 : 40 a . m . S o l u t i o n - T h e t i m e s p a n b e t w e e n 6 : 05 a . m t o 6 : 40 a . m = 6 : 05 a . m - 6 : 40 a . m = 35 m i n u t e s . N o w t h e a n g l e d e s c r i b e d b y t h e m i n u t e h a n d i n 1 m i n u t e = 6^{o} . ∴ I n 35 m i n u t e s i t w i l l d e s c r i b e a n a n g l e θ = 35 \times 6^{o} = 210^{o} . s o t h e m i n u t e h a n d w i l l d e s c r i b e a s e c t o r o f r a d i u s r = 5 c m a n d t h e c e n t r a l a n g l e = {θ = 210}^{o} . ∴ a r . s e c t o r = \frac{θ}{360^{o} .} \times π \times r^{2} = \frac{210^{o}}{360^{o} .} \times \frac{22}{7} \times 5^{2} c m^{2} = 45 \frac{5}{6} c m^{2} . A n s - O p t i o n D .$

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Similar questions

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 am and 6: 40 am.

500

6.05

6.40

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