Question

a roller is 120 cm long and has diameter 84 cm. if it takes 500 complete revolutions to level a playground, determine the cost of levelling it at the rate of 30 paise per m

Answer 1

Diameter = 84cm

Radius = 42cm

Height = 120cm

1 revolution = LSA of roller

= 2 x (22/7) x 42  6 x 120

= 31680 cm2

So, 500 revolutions = 31680 x 500

= 15840000cm2

1 cm2 = .0001m2

So,  15840000cm2 = 1584 m2   

Cost = 1584 x .30

= Rs 475.20

Answer 2

AnswEr:

Clearly, the roller is a right circular cylinder of height h = 120 cm and radius of its base r = 42 cm

$\therefore$ $\underline\text{Area\:covered\:by\:the\:roller\:in\:one\:revolutions:}$

$\Rightarrow$ Curved surface area of the roller

$\sf{2}×$$\sf\dfrac{22}{7}\times42\times120\:cm^2$

_________________________

So, Area covered by the roller is 500 revolutions = ( 31680 × 500 )

= $\sf\dfrac{31680\times500}{100\times100}m^2$

= 1584 m².

____________________________

Hence, Cost of levelling the playground

= $\sf{Rs\:1584\times}$$\sf\dfrac{30}{100}=Rs.475.20$