Menu Close

Question-101555




Question Number 101555 by 175 last updated on 03/Jul/20
Answered by EuclidKaBaap last updated on 03/Jul/20
speed of body = n rounds / 60 sec  distance of 1 round = 2Πr  where r is radius of circular track.  so distance covered in 1 sec =         ((n(2Πr))/(60))= ((nΠr)/(30))  distance covered in 50 sec =          ((nΠr)/(30))×50 = ((5nΠr)/3)  thats ur required answer
$${speed}\:{of}\:{body}\:=\:{n}\:{rounds}\:/\:\mathrm{60}\:{sec} \\ $$$${distance}\:{of}\:\mathrm{1}\:{round}\:=\:\mathrm{2}\Pi{r} \\ $$$${where}\:{r}\:{is}\:{radius}\:{of}\:{circular}\:{track}. \\ $$$${so}\:{distance}\:{covered}\:{in}\:\mathrm{1}\:{sec}\:= \\ $$$$\:\:\:\:\:\:\:\frac{{n}\left(\mathrm{2}\Pi{r}\right)}{\mathrm{60}}=\:\frac{{n}\Pi{r}}{\mathrm{30}} \\ $$$${distance}\:{covered}\:{in}\:\mathrm{50}\:{sec}\:= \\ $$$$\:\:\:\:\:\:\:\:\frac{{n}\Pi{r}}{\mathrm{30}}×\mathrm{50}\:=\:\frac{\mathrm{5}{n}\Pi{r}}{\mathrm{3}} \\ $$$${thats}\:{ur}\:{required}\:{answer} \\ $$
Commented by 175 last updated on 03/Jul/20
what is π

Leave a Reply

Your email address will not be published. Required fields are marked *