Length of a cylindrical roller is 2.5 m and its radius is 1.75 m. When rolled on a road , it covered an area of 5500 m2 . How many revolutions did it make?
Answers
Answered by
333
Hi there !!
The length of the cylindrical pipe is the same as its height.
h = 2.5 m
r = 1.75 m
Total area covered = 5500 m²
Area covered by 1 revolution = C.S.A of the pipe
= 2πrh
= 2 × 22/7 × 1.75 × 2.5
= 27.5 m²
No : of revolutions made = Total area covered ÷ C.S.A
= 5500 ÷ 27.5
= 200 revolutions
The length of the cylindrical pipe is the same as its height.
h = 2.5 m
r = 1.75 m
Total area covered = 5500 m²
Area covered by 1 revolution = C.S.A of the pipe
= 2πrh
= 2 × 22/7 × 1.75 × 2.5
= 27.5 m²
No : of revolutions made = Total area covered ÷ C.S.A
= 5500 ÷ 27.5
= 200 revolutions
Answered by
157
here,
h = 2.5 m
r = 1.75 m
Total area covered = 5500 m²
Area covered by 1 revolution = C.S.A of the pipe
= 2πrh
= 2 × 22/7 × 1.75 × 2.5
= 27.5 m²
No of revolutions = Total area covered / C.S.A
= 5500 / 27.5
= 200 revolutions
h = 2.5 m
r = 1.75 m
Total area covered = 5500 m²
Area covered by 1 revolution = C.S.A of the pipe
= 2πrh
= 2 × 22/7 × 1.75 × 2.5
= 27.5 m²
No of revolutions = Total area covered / C.S.A
= 5500 / 27.5
= 200 revolutions
Similar questions