A field roller in the shape of a cylinder, has a diameter of 2 m and length of 2\(\frac{1}{4}\)m. If the speed at which the roller rolls is 14 revolution per minute, then the maximum area (in m2) that it can roll in 1 hour is? (take \(\pi \) = \(\frac{22}{7}\))
Answer & explanation
Correct answer: option 2
radius = \(\frac{2}{2}\) = 1 meter
height = \(\frac{9}{4}\) meter
Area covered in one revolution = 2\(\pi \)rh (L.S.A.)
Area covered in one minute = 2\(\pi \)rh × 14
= 2 × \(\frac{22}{7}\) × 1 × \(\frac{9}{4}\) × 14 = 198
Area covered in 1 hour = 198 × 60 = 11880