Question

The diameter of roller 1.5 m long is 84 cm. If it takes 100 revolutions to level a playground, find the cost of levelling this ground at the rate of 50 paise per square metre. 

Answer 1

Given that, 

Diameter of the roller=84cm=0.84m. 

Length of the roller=1.5m 

Radius of the roller =`D/2=0.84/2=0.42.` 

Area covered by the roller on one revolution = covered surface area of roller  

Curved surface of roller= `2pirh=2xx22/7xx0.42xx1.5`

=`0.12xx22xx1.5m^2` 

Area of the playground=`"100xxArea covered by roller in one revolution"` 

=`(100xx0.12xx22xx1.5)m^2`

=`396m^2`

Now , 

Cost of leveling `1 m^2=50P=50/100⇒Re=1/2rs` 

Cost of leveling `396m^2=1/2xx396=Rs.198` 

Hence, cost of leveling `396m^2 is   198`