the short and long hands of a clock are 8 cm and 12 cm long respectively find the sum of the distances travelled by the tips in 2 days that is in 48 hours (take π=22/7)
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9
For short hand
r= 8cm
since minute hand travels 24 complete circles of clock in 1 day
so in 2 days 48 revolutions
distance moved in 1 revolution = 2 pi r = 2 ×22/7 ×8
Total distance moved = 48 × 2 ×22/7 × 8 = 2413.71 cm
Fir long hand
revolutions covered in 1 day = 1
so in two days = 2
Distance moved in 1 revolution = 2 pi r = 2 ×22/7 × 12
Total distance moved = 2 × 22/7 ×12 × 2 =150.8 cm
r= 8cm
since minute hand travels 24 complete circles of clock in 1 day
so in 2 days 48 revolutions
distance moved in 1 revolution = 2 pi r = 2 ×22/7 ×8
Total distance moved = 48 × 2 ×22/7 × 8 = 2413.71 cm
Fir long hand
revolutions covered in 1 day = 1
so in two days = 2
Distance moved in 1 revolution = 2 pi r = 2 ×22/7 × 12
Total distance moved = 2 × 22/7 ×12 × 2 =150.8 cm
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Answered by
3
as we know in the given question we have short hand as 8cm and long hand as 12cm
•For short hand
radius =8cm
24h=1day
=then for 2 days we have 48hours
=distance moved in one revolution = 2×pi×R
=2×22/7×8 For one day
=total distance for 2days
=48×8×2×22/7
=48×8×44/7
= 2413.71cm
•For long hand
same as short hand
= 2×pi×R
=2×22/7×12×2
=44×12/7
=150.8 cm
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