Question

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

Answer 1

$<b><u>$Given that, 

Diameter of the roller = 84 cm = 0.84 m

Length of the roller = 1.5 m

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 = 2πrh = 2 x 22/7 x 0.42 x 1.5

= 0.12 x 22 x 1.5 m2

 Area of the playground = 100 x Area covered by roller in one revolution

= (100 x 0.12 x 22 x 1.5) m2

= 396 m2

Answer 2

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Here is your answer
Diameter = 84 cm = 84/100 m
= 0.84 m
Radius = 0.84/2
= 0.42 m

Height is 1.5 m

As we know the formula

Curved Surface Area = 2πrh
$= 2 \times \frac{22}{7} \times 0.42 \times 1.5 \\ \\ = 3.96 \: m {}^{2}$
IT takes 100 revolution to label the field

So Total Area of the field = 100 × 3.96
= 396 sq.m

Rate of lavelling the ground is RS 0.50/sq.m

Hence, Cost of levelling the ground = 0.50 × 396

= ₹198


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