Question

A cylindrical roller is of length of 2m and diameter of 1.4 .Find area it will level 25000 revolution

Answer 1

A cylindrical roller has the curved surface area of;
CSA = L x (π x d)
Where L is the length of the roller
and d is the diameter of the base.
Therefore, by substituting the given data;
CSA = 2 x (π x 1.4) = 2.8π m²
This is the area leveled by 1 complete revolution of the roller.
So, to find the area leveled after 25,000 revolutions;
= 25,000 x 2.8π m²
= 70,000π m²
= 2,20,000 m²  (if π = 22/7)
= 219911.4857 m² (for π = 3.14159....)

Hope this helped : )

Answer 2

hiii!!!

here's ur answer....

given the height (length) of the roller is 2m and diameter is 1.4m.

therefore it's radius=1.4/2

=0.7m

CSA of the cylindrical roller=2πrh

=2×22/7×0.7×2

=61.6/7

=8.8m²

area it will cover in 25000 revolutions=8.8×25000

=220000m²

hope this helps..!

☺☺☺